Contents / Calculus / Vectors, Space Curves, and Motion

Chapter 10

Vectors, Space Curves, and Motion

Vectors in three dimensions, lines and planes, quadric surfaces, vector functions, arc length, curvature, and motion in space.

Introduction

Everything so far has lived on a line or in a plane. A thrown ball, a satellite, a drone following a survey route: these move through space, and describing them takes three coordinates at once. This chapter builds the language for that — points and vectors in $\mathbb{R}^3$ , the two products that turn geometry into arithmetic, the lines, planes and curved surfaces those products describe — and then lets the coordinates depend on time, so that calculus can act on them.

The payoff is that a single-variable derivative, applied to each coordinate, becomes a velocity vector; a single-variable integral becomes arc length; and the second derivative splits cleanly into the part that changes speed and the part that steers. The chapter ends with the equations of a projectile and a glimpse of why planets move in ellipses.

Two ideas do most of the work. The dot product $\mathbf{a}\cdot\mathbf{b}$ measures how much two vectors point the same way, and the cross product $\mathbf{a}\times\mathbf{b}$ manufactures a vector perpendicular to both. Learn what each one computes and almost every formula in the chapter becomes something you could rederive rather than memorise.

10.1Three-dimensional coordinates

To locate a point in space, fix an origin $O$ and three mutually perpendicular axes through it. The orientation is always the right-handed one: curl the fingers of your right hand from the positive $x$ -axis toward the positive $y$ -axis, and your thumb points along the positive $z$ -axis. Every pair of axes spans a coordinate plane — the $xy$ -plane, the $yz$ -plane and the $xz$ -plane — and the three planes cut space into eight octants. The first octant is the one where all three coordinates are positive.

A point $P$ is described by the ordered triple $(a,b,c)$ : drop a perpendicular from $P$ to the $xy$ -plane, landing at $(a,b,0)$ , and $c$ is the signed height of $P$ above that landing point. The three numbers are the coordinates of $P$ , and $\mathbb{R}^3$ is the set of all such triples.

An equation in $x$ , $y$ and $z$ describes a surface, not a curve. The equation $z = 3$ in $\mathbb{R}^3$ is the set of all points three units above the $xy$ -plane — a horizontal plane. Likewise $y = 5$ is a vertical plane parallel to the $xz$ -plane. Only one equation has been imposed, so two degrees of freedom remain, and two degrees of freedom is a surface.

Intuition. Stand in the corner of a room. The two walls and the floor are the coordinate planes, the corner is the origin, and every object in the room has three numbers: how far along one wall, how far along the other, how high off the floor. The point $(2,3,1)$ is two metres along the first wall, three along the second, one metre up. The equation $z = 1$ is not a point or a line — it is the whole invisible sheet one metre above the floor.

Theorem 10.1 (Distance formula). The distance between $P_1(x_1,y_1,z_1)$ and $P_2(x_2,y_2,z_2)$ is

|P_1P_2| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}.

Proof. Build the box with $P_1$ and $P_2$ at opposite corners and edges parallel to the axes, of lengths $|x_2-x_1|$ , $|y_2-y_1|$ and $|z_2-z_1|$ . Let $A$ be the corner at $(x_2,y_2,z_1)$ , at the same height as $P_1$ . The triangle $P_1AP_2$ has a right angle at $A$ , so by Pythagoras

|P_1P_2|^2 = |P_1A|^2 + |AP_2|^2 = |P_1A|^2 + (z_2-z_1)^2.

The segment $P_1A$ lies in the horizontal plane $z = z_1$ , where the plane distance formula gives $|P_1A|^2 = (x_2-x_1)^2 + (y_2-y_1)^2$ . Substitute and take the square root.∎

The proof is worth remembering because it is the whole reason the formula has that shape: Pythagoras once in the floor of the box, then once more up the vertical edge. The formula in $\mathbb{R}^2$ is the special case $z_1 = z_2$ .

Definition 10.2 (Sphere). The sphere with centre $C(h,k,l)$ and radius $r$ is the set of points at distance $r$ from $C$ . By the distance formula, its equation is

(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2.

Expanding the squares produces an equation of the form $x^{2}+y^{2}+z^{2} + Dx + Ey + Fz + G = 0$ . Conversely, any equation of that form is a sphere (or a single point, or nothing at all), and completing the square in each variable recovers the centre and radius.

Example 10.3 (A distance and a sphere). Find the distance from $P(2,-1,7)$ to $Q(1,-3,5)$ , and write the equation of the sphere with centre $P$ that passes through $Q$ .

Solution.

The differences of coordinates are $1-2 = -1$ , $-3-(-1) = -2$ and $5-7 = -2$ .
By the distance formula, $|PQ| = \sqrt{(-1)^2 + (-2)^2 + (-2)^2} = \sqrt{1+4+4} = 3$ .
The sphere centred at $P$ through $Q$ has radius $|PQ| = 3$ , so its equation is $(x-2)^2 + (y+1)^2 + (z-7)^2 = 9$ .
Sanity check: substituting $Q$ gives $1 + 4 + 4 = 9$ , as it must.

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Example 10.4 (Completing the square). Show that $x^{2} + y^{2} + z^{2} + 4x - 6y + 2z + 6 = 0$ is a sphere, and find its centre and radius.

Solution.

Group the variables: $(x^{2}+4x) + (y^{2}-6y) + (z^{2}+2z) = -6$ .
Complete each square, adding the same constants on the right: $(x^{2}+4x+4) + (y^{2}-6y+9) + (z^{2}+2z+1) = -6 + 4 + 9 + 1$ .
So $(x+2)^2 + (y-3)^2 + (z+1)^2 = 8$ .
Comparing with the definition, the centre is $(-2,3,-1)$ and the radius is $\sqrt{8} = 2\sqrt{2}$ . The right-hand side came out positive, which is what makes this a genuine sphere rather than a point or the empty set.

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Example 10.5 (Describing a region). What region of $\mathbb{R}^3$ is described by the inequalities $1 \le x^{2}+y^{2}+z^{2} \le 4$ together with $z \le 0$ ?

Solution.

Since $x^{2}+y^{2}+z^{2}$ is the squared distance from the origin, $1 \le x^{2}+y^{2}+z^{2} \le 4$ says the distance lies between $1$ and $2$ : the solid shell between two concentric spheres.
The condition $z \le 0$ keeps only the points on or below the $xy$ -plane.
The region is the lower half of that shell — a hemispherical shell of inner radius $1$ and outer radius $2$ .

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Pitfall. The same equation means different things in different dimensions. In $\mathbb{R}^2$ , $x^{2} + y^{2} = 1$ is a circle. In $\mathbb{R}^3$ it is a circular cylinder: nothing constrains $z$ , so every horizontal slice is the unit circle and the slices stack into an infinite tube. Always ask which space an equation lives in before you picture it.

Exercises for §10.1 →

10.2Vectors

A vector is a quantity with both magnitude and direction: a displacement, a velocity, a force. Geometrically it is an arrow, and two arrows represent the same vector when they have the same length and point the same way, wherever they start. The arrow from $A$ to $B$ is written $\overrightarrow{AB}$ ; the vector itself is written in bold, $\mathbf{v}$ , or by hand as $\vec{v}$ .

Definition 10.6 (Components, magnitude, zero vector). Place a vector $\mathbf{a}$ with its tail at the origin. Its head lands at some point $(a_1,a_2,a_3)$ , and the numbers $a_1$ , $a_2$ , $a_3$ are the components of $\mathbf{a}$ :

\mathbf{a} = \langle a_1, a_2, a_3 \rangle.

The magnitude (or length) of $\mathbf{a}$ is the length of the arrow,

|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2},

and the zero vector $\mathbf{0} = \langle 0,0,0 \rangle$ is the only vector of length $0$ . It has no direction.

The angle brackets distinguish the vector $\langle 1,2,3\rangle$ from the point $(1,2,3)$ . They are related — the point is where the arrow lands when its tail is at the origin — but a point has a location and a vector does not. The vector from $A(x_1,y_1,z_1)$ to $B(x_2,y_2,z_2)$ has components

\overrightarrow{AB} = \langle x_2-x_1,\; y_2-y_1,\; z_2-z_1 \rangle,

head minus tail, and its magnitude is exactly the distance $|AB|$ from the distance formula.

Definition 10.7 (Vector addition and scalar multiplication). For $\mathbf{a} = \langle a_1,a_2,a_3\rangle$ , $\mathbf{b} = \langle b_1,b_2,b_3\rangle$ and a real number $c$ ,

\mathbf{a}+\mathbf{b} = \langle a_1+b_1,\; a_2+b_2,\; a_3+b_3\rangle, \qquad c\mathbf{a} = \langle ca_1,\; ca_2,\; ca_3\rangle.

The difference is $\mathbf{a}-\mathbf{b} = \mathbf{a} + (-1)\mathbf{b}$ .

Geometrically, $\mathbf{a}+\mathbf{b}$ is the arrow you get by placing the tail of $\mathbf{b}$ at the head of $\mathbf{a}$ and drawing from the tail of $\mathbf{a}$ to the head of $\mathbf{b}$ (the triangle law), or equivalently the diagonal of the parallelogram spanned by the two arrows from a common tail (the parallelogram law). Multiplying by $c$ stretches the arrow by a factor $|c|$ and reverses it if $c < 0$ ; two nonzero vectors are parallel exactly when one is a scalar multiple of the other. The difference $\mathbf{a} - \mathbf{b}$ is the arrow from the head of $\mathbf{b}$ to the head of $\mathbf{a}$ , the vector you must add to $\mathbf{b}$ to reach $\mathbf{a}$ .

Intuition. Walk three blocks east and four blocks north. Your displacement is the vector $\langle 3,4\rangle$ : it does not matter which corner you started from, and its length is the five blocks a crow would fly. Walk it twice and you have $2\langle 3,4\rangle = \langle 6,8\rangle$ ; walk it backwards and you have $\langle -3,-4\rangle$ . Adding vectors is doing one walk after another, and the components just keep separate tallies of east and north.

Theorem 10.8 (Properties of vector algebra). For all vectors $\mathbf{a},\mathbf{b},\mathbf{c}$ in $\mathbb{R}^3$ and all scalars $c,d$ :

\begin{aligned} &\mathbf{a}+\mathbf{b} = \mathbf{b}+\mathbf{a}, &\quad &\mathbf{a}+(\mathbf{b}+\mathbf{c}) = (\mathbf{a}+\mathbf{b})+\mathbf{c},\\ &\mathbf{a}+\mathbf{0} = \mathbf{a}, &\quad &\mathbf{a}+(-\mathbf{a}) = \mathbf{0},\\ &c(\mathbf{a}+\mathbf{b}) = c\mathbf{a}+c\mathbf{b}, &\quad &(c+d)\mathbf{a} = c\mathbf{a}+d\mathbf{a},\\ &(cd)\mathbf{a} = c(d\mathbf{a}), &\quad &1\mathbf{a} = \mathbf{a}. \end{aligned}

Proof. Each identity is checked one component at a time and reduces to a property of real numbers. For the first,

\mathbf{a}+\mathbf{b} = \langle a_1+b_1, a_2+b_2, a_3+b_3\rangle = \langle b_1+a_1, b_2+a_2, b_3+a_3\rangle = \mathbf{b}+\mathbf{a},

because addition of real numbers is commutative. For the distributive law, the first component of $c(\mathbf{a}+\mathbf{b})$ is $c(a_1+b_1) = ca_1 + cb_1$ , which is the first component of $c\mathbf{a}+c\mathbf{b}$ , and the other two components are identical calculations. The remaining six lines are the same pattern.∎

Three vectors deserve names. The standard basis vectors

\mathbf{i} = \langle 1,0,0\rangle, \qquad \mathbf{j} = \langle 0,1,0\rangle, \qquad \mathbf{k} = \langle 0,0,1\rangle

have length $1$ and point along the positive axes, and every vector is a combination of them:

\langle a_1,a_2,a_3\rangle = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}.

The two notations are interchangeable; $\mathbf{i}\mathbf{j}\mathbf{k}$ form is convenient when many components are zero.

Proposition 10.9 (Scaling and unit vectors). For any scalar $c$ and vector $\mathbf{a}$ , $|c\mathbf{a}| = |c|\,|\mathbf{a}|$ . Consequently, if $\mathbf{a} \ne \mathbf{0}$ then

\mathbf{u} = \frac{\mathbf{a}}{|\mathbf{a}|}

is a unit vector — a vector of length $1$ — pointing in the same direction as $\mathbf{a}$ .

Proof. $|c\mathbf{a}|^2 = (ca_1)^2 + (ca_2)^2 + (ca_3)^2 = c^{2}(a_1^2+a_2^2+a_3^2) = c^{2}|\mathbf{a}|^2$ , and taking square roots gives $|c\mathbf{a}| = |c||\mathbf{a}|$ . With $c = 1/|\mathbf{a}| > 0$ the direction is preserved and $|\mathbf{u}| = |\mathbf{a}|/|\mathbf{a}| = 1$ .∎

Dividing a vector by its own length is called normalising it, and it is the standard way to separate a vector into "which way" (the unit vector) and "how much" (the magnitude): $\mathbf{a} = |\mathbf{a}|\,\mathbf{u}$ .

Example 10.10 (A vector between two points). Find the vector from $A(2,-3,4)$ to $B(-2,1,1)$ , its magnitude, and the unit vector in its direction.

Solution.

Head minus tail: $\overrightarrow{AB} = \langle -2-2,\; 1-(-3),\; 1-4\rangle = \langle -4, 4, -3\rangle$ .
Magnitude: $|\overrightarrow{AB}| = \sqrt{16+16+9} = \sqrt{41}$ .
Unit vector: $\mathbf{u} = \dfrac{1}{\sqrt{41}}\langle -4,4,-3\rangle = \left\langle -\dfrac{4}{\sqrt{41}}, \dfrac{4}{\sqrt{41}}, -\dfrac{3}{\sqrt{41}}\right\rangle$ .
Check: $|\mathbf{u}|^2 = \frac{16+16+9}{41} = 1$ .

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Example 10.11 (Vector arithmetic). Let $\mathbf{a} = \langle 4,0,3\rangle$ and $\mathbf{b} = \langle -2,1,5\rangle$ . Compute $|\mathbf{a}|$ , $\mathbf{a}+\mathbf{b}$ , $2\mathbf{a}-3\mathbf{b}$ , and $|\mathbf{a}-\mathbf{b}|$ .

Solution.

$|\mathbf{a}| = \sqrt{16+0+9} = 5$ .
$\mathbf{a}+\mathbf{b} = \langle 4-2,\; 0+1,\; 3+5\rangle = \langle 2,1,8\rangle$ .
$2\mathbf{a}-3\mathbf{b} = \langle 8,0,6\rangle - \langle -6,3,15\rangle = \langle 14,-3,-9\rangle$ .
$\mathbf{a}-\mathbf{b} = \langle 6,-1,-2\rangle$ , so $|\mathbf{a}-\mathbf{b}| = \sqrt{36+1+4} = \sqrt{41}$ .

Note that $|\mathbf{a}-\mathbf{b}| \ne |\mathbf{a}| - |\mathbf{b}|$ ; the length of a difference is a distance between heads, not a difference of lengths.□

Example 10.12 (Tension in two wires). A $100$ -newton weight hangs from a point where two wires meet. The left wire makes an angle of $60^\circ$ with the horizontal ceiling and the right wire makes $30^\circ$ . Find the tension in each wire.

Solution.

Put the junction at the origin with $x$ horizontal and $y$ vertical. The left wire pulls up and to the left with tension $T_1$ , the right wire up and to the right with tension $T_2$ , and the weight pulls straight down: $\mathbf{w} = \langle 0,-100\rangle$ .
Resolve each tension into components. The left wire's direction makes $60^\circ$ with the negative $x$ -axis, so $\mathbf{T}_1 = \langle -T_1\cos 60^\circ,\; T_1\sin 60^\circ\rangle = \left\langle -\tfrac{1}{2}T_1,\; \tfrac{\sqrt3}{2}T_1\right\rangle$ . Similarly $\mathbf{T}_2 = \left\langle \tfrac{\sqrt3}{2}T_2,\; \tfrac12 T_2\right\rangle$ .
Equilibrium means $\mathbf{T}_1+\mathbf{T}_2+\mathbf{w} = \mathbf{0}$ . The $x$ -components give $-\tfrac12 T_1 + \tfrac{\sqrt3}{2}T_2 = 0$ , so $T_1 = \sqrt3\,T_2$ . The $y$ -components give $\tfrac{\sqrt3}{2}T_1 + \tfrac12 T_2 = 100$ .
Substitute: $\tfrac{\sqrt3}{2}\cdot\sqrt3\,T_2 + \tfrac12 T_2 = \tfrac32 T_2 + \tfrac12 T_2 = 2T_2 = 100$ , so $T_2 = 50$ N and $T_1 = 50\sqrt3 \approx 86.6$ N.
Sanity check: the two wires are perpendicular to each other, so they split the load like the legs of a right triangle with hypotenuse $100$ : $50^2 + (50\sqrt3)^2 = 2500 + 7500 = 100^2$ . The steeper wire carries more.

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Pitfall. Magnitude is not additive. $|\mathbf{a}+\mathbf{b}| \le |\mathbf{a}| + |\mathbf{b}|$ with equality only when the vectors point the same way — this is the triangle inequality, and it is the reason a detour is never shorter than the direct route. Writing $|\mathbf{a}+\mathbf{b}| = |\mathbf{a}|+|\mathbf{b}|$ for vectors that are not parallel is the most common slip in this section.

Exercises for §10.2 →

10.3The dot product

Adding vectors and scaling them is not enough to do geometry; we also need lengths and angles. The dot product delivers both from the components.

Definition 10.13 (Dot product). For $\mathbf{a} = \langle a_1,a_2,a_3\rangle$ and $\mathbf{b} = \langle b_1,b_2,b_3\rangle$ , the dot product is the scalar

\mathbf{a}\cdot\mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3.

The output is a number, not a vector — the dot product is sometimes called the scalar product for exactly that reason. Its algebra is as tame as ordinary multiplication.

Theorem 10.14 (Properties of the dot product). For vectors $\mathbf{a},\mathbf{b},\mathbf{c}$ and a scalar $c$ :

\mathbf{a}\cdot\mathbf{a} = |\mathbf{a}|^2, \qquad \mathbf{a}\cdot\mathbf{b} = \mathbf{b}\cdot\mathbf{a}, \qquad \mathbf{a}\cdot(\mathbf{b}+\mathbf{c}) = \mathbf{a}\cdot\mathbf{b} + \mathbf{a}\cdot\mathbf{c},

(c\mathbf{a})\cdot\mathbf{b} = c(\mathbf{a}\cdot\mathbf{b}) = \mathbf{a}\cdot(c\mathbf{b}), \qquad \mathbf{0}\cdot\mathbf{a} = 0.

Proof. The first is $a_1^2+a_2^2+a_3^2 = |\mathbf{a}|^2$ by the definition of magnitude. The second holds because $a_ib_i = b_ia_i$ . For the third,

\mathbf{a}\cdot(\mathbf{b}+\mathbf{c}) = a_1(b_1+c_1) + a_2(b_2+c_2) + a_3(b_3+c_3) = (a_1b_1+a_2b_2+a_3b_3) + (a_1c_1+a_2c_2+a_3c_3),

which is $\mathbf{a}\cdot\mathbf{b}+\mathbf{a}\cdot\mathbf{c}$ . The last two are equally direct.∎

The definition looks arbitrary until you see what it measures.

Theorem 10.15 (Geometric form of the dot product). If $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$ (with $0 \le \theta \le \pi$ , tails together), then

\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}|\,|\mathbf{b}|\cos\theta.

Proof. Draw $\mathbf{a}$ and $\mathbf{b}$ from a common tail. The third side of the triangle they form is $\mathbf{a}-\mathbf{b}$ , and the angle between the first two sides is $\theta$ . The law of cosines states

|\mathbf{a}-\mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2|\mathbf{a}||\mathbf{b}|\cos\theta.

Now expand the left-hand side with the algebraic properties just proved:

|\mathbf{a}-\mathbf{b}|^2 = (\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{b}) = \mathbf{a}\cdot\mathbf{a} - 2\,\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{b} = |\mathbf{a}|^2 - 2\,\mathbf{a}\cdot\mathbf{b} + |\mathbf{b}|^2.

Equating the two expressions and cancelling $|\mathbf{a}|^2+|\mathbf{b}|^2$ from both sides leaves $-2\,\mathbf{a}\cdot\mathbf{b} = -2|\mathbf{a}||\mathbf{b}|\cos\theta$ . If either vector is $\mathbf{0}$ , both sides are $0$ for any $\theta$ .∎

Corollary 10.16 (Angle and orthogonality). For nonzero $\mathbf{a}$ and $\mathbf{b}$ ,

\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|},

and $\mathbf{a}$ and $\mathbf{b}$ are perpendicular (orthogonal) if and only if $\mathbf{a}\cdot\mathbf{b} = 0$ .

The sign of the dot product alone tells you whether the angle is acute, right or obtuse, because $|\mathbf{a}||\mathbf{b}| > 0$ and the sign is carried entirely by $\cos\theta$ . The zero vector is declared orthogonal to everything, which keeps the "if and only if" clean.

Two special cases are worth noting. The angles $\alpha$ , $\beta$ , $\gamma$ that a nonzero vector $\mathbf{a}$ makes with the positive $x$ -, $y$ - and $z$ -axes are its direction angles, and taking $\mathbf{b} = \mathbf{i}$ , $\mathbf{j}$ , $\mathbf{k}$ in the corollary gives the direction cosines

\cos\alpha = \frac{a_1}{|\mathbf{a}|}, \qquad \cos\beta = \frac{a_2}{|\mathbf{a}|}, \qquad \cos\gamma = \frac{a_3}{|\mathbf{a}|},

so that $\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1$ and $\langle \cos\alpha,\cos\beta,\cos\gamma\rangle$ is exactly the unit vector $\mathbf{a}/|\mathbf{a}|$ .

Intuition. On a sunny day with the sun directly overhead, a tilted pole casts a shadow on the ground. The dot product of the pole's vector with a unit vector along the ground is the length of that shadow — positive if the pole leans forward, zero if it stands straight up, negative if it leans back. That is all $|\mathbf{a}|\cos\theta$ is: the shadow of $\mathbf{a}$ on the line of $\mathbf{b}$ . Multiplying by $|\mathbf{b}|$ just rescales.

The shadow picture is worth making precise, because it is the tool behind distances to planes, work, and the decomposition of acceleration later on.

Definition 10.17 (Scalar and vector projections). Let $\mathbf{b} \ne \mathbf{0}$ . The scalar projection of $\mathbf{a}$ onto $\mathbf{b}$ is the signed length of the shadow of $\mathbf{a}$ on the line through $\mathbf{b}$ ,

\operatorname{comp}_{\mathbf{b}}\mathbf{a} = |\mathbf{a}|\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|},

and the vector projection is that shadow as a vector along $\mathbf{b}$ ,

\operatorname{proj}_{\mathbf{b}}\mathbf{a} = \left(\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\right)\frac{\mathbf{b}}{|\mathbf{b}|} = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\,\mathbf{b}.

The two formulas come straight from the geometric form: $|\mathbf{a}|\cos\theta = |\mathbf{a}||\mathbf{b}|\cos\theta/|\mathbf{b}| = \mathbf{a}\cdot\mathbf{b}/|\mathbf{b}|$ , and the vector projection is that signed length times the unit vector $\mathbf{b}/|\mathbf{b}|$ . The remainder $\mathbf{a} - \operatorname{proj}_{\mathbf{b}}\mathbf{a}$ is the component of $\mathbf{a}$ perpendicular to $\mathbf{b}$ ; the two pieces add back to $\mathbf{a}$ and are orthogonal to each other.

Projection is also what physics means by work. A constant force $\mathbf{F}$ moving an object along the displacement $\mathbf{D}$ does work equal to the component of the force in the direction of motion times the distance moved:

W = (|\mathbf{F}|\cos\theta)\,|\mathbf{D}| = \mathbf{F}\cdot\mathbf{D}.

A force perpendicular to the motion does no work at all, which the dot product records as zero.

Example 10.18 (The angle between two vectors). Find the angle between $\mathbf{a} = \langle 2,2,-1\rangle$ and $\mathbf{b} = \langle 5,-3,2\rangle$ .

Solution.

$\mathbf{a}\cdot\mathbf{b} = 2(5) + 2(-3) + (-1)(2) = 10 - 6 - 2 = 2$ .
$|\mathbf{a}| = \sqrt{4+4+1} = 3$ and $|\mathbf{b}| = \sqrt{25+9+4} = \sqrt{38}$ .
$\cos\theta = \dfrac{2}{3\sqrt{38}}$ , so $\theta = \arccos\!\left(\dfrac{2}{3\sqrt{38}}\right) \approx 1.46$ radians, about $84^\circ$ .
The dot product is positive but small compared with $|\mathbf{a}||\mathbf{b}| \approx 18.5$ , so the angle should be acute and close to a right angle. It is.

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Example 10.19 (Testing for perpendicularity). Show that $2\mathbf{i}+2\mathbf{j}-\mathbf{k}$ is perpendicular to $5\mathbf{i}-4\mathbf{j}+2\mathbf{k}$ , and find all values of $t$ for which $\langle t, 2, -1\rangle$ is perpendicular to $\langle 3, 1, 5\rangle$ .

Solution.

$(2)(5) + (2)(-4) + (-1)(2) = 10 - 8 - 2 = 0$ , so the vectors are perpendicular by the corollary.
For the second pair, $\langle t,2,-1\rangle\cdot\langle 3,1,5\rangle = 3t + 2 - 5 = 3t - 3$ .
This vanishes exactly when $t = 1$ . For $t > 1$ the angle is acute, for $t < 1$ it is obtuse.

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Example 10.20 (Scalar and vector projections). Find the scalar and vector projections of $\mathbf{a} = \langle 1,1,2\rangle$ onto $\mathbf{b} = \langle -2,3,1\rangle$ , and the component of $\mathbf{a}$ orthogonal to $\mathbf{b}$ .

Solution.

$\mathbf{a}\cdot\mathbf{b} = -2 + 3 + 2 = 3$ and $|\mathbf{b}| = \sqrt{4+9+1} = \sqrt{14}$ .
Scalar projection: $\operatorname{comp}_{\mathbf{b}}\mathbf{a} = \dfrac{3}{\sqrt{14}}$ .
Vector projection: $\operatorname{proj}_{\mathbf{b}}\mathbf{a} = \dfrac{3}{14}\langle -2,3,1\rangle = \left\langle -\dfrac{3}{7}, \dfrac{9}{14}, \dfrac{3}{14}\right\rangle$ .
Orthogonal component: $\mathbf{a} - \operatorname{proj}_{\mathbf{b}}\mathbf{a} = \left\langle 1+\dfrac{3}{7},\; 1-\dfrac{9}{14},\; 2-\dfrac{3}{14}\right\rangle = \left\langle \dfrac{10}{7}, \dfrac{5}{14}, \dfrac{25}{14}\right\rangle$ .
Check that it really is orthogonal to $\mathbf{b}$ : $-\dfrac{20}{7} + \dfrac{15}{14} + \dfrac{25}{14} = \dfrac{-40+15+25}{14} = 0$ .

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Example 10.21 (Work done by a force). A crate is dragged $100$ m along a level floor by a rope that pulls with a force of $70$ N at $30^\circ$ above the horizontal. How much work is done? Then compute the work done by the force $\mathbf{F} = \langle 3,4,5\rangle$ newtons in moving a particle from $(2,1,0)$ to $(4,6,2)$ , distances in metres.

Solution.

For the crate, $W = |\mathbf{F}||\mathbf{D}|\cos\theta = 70\cdot 100\cdot\cos 30^\circ = 7000\cdot\dfrac{\sqrt3}{2} = 3500\sqrt3 \approx 6062$ J.
Only the horizontal part of the pull moves the crate; the vertical part merely lightens it. The cosine is what discards the vertical part.
For the particle, the displacement is $\mathbf{D} = \langle 4-2,\; 6-1,\; 2-0\rangle = \langle 2,5,2\rangle$ .
$W = \mathbf{F}\cdot\mathbf{D} = 6 + 20 + 10 = 36$ J.

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Pitfall. The dot product of two vectors is a scalar, so an expression like $\mathbf{a}\cdot\mathbf{b}\cdot\mathbf{c}$ is meaningless — after the first dot there is no vector left to dot with. And there is no cancellation law: $\mathbf{a}\cdot\mathbf{b} = \mathbf{a}\cdot\mathbf{c}$ does not force $\mathbf{b} = \mathbf{c}$ , only that $\mathbf{b}-\mathbf{c}$ is perpendicular to $\mathbf{a}$ .

Exercises for §10.3 →

10.4The cross product

The dot product answers "how aligned are these two vectors?". The cross product answers a different question that geometry asks constantly: given two vectors, produce a third perpendicular to both. Unlike the dot product it exists only in three dimensions, and its output is a vector.

Definition 10.22 (Cross product). For $\mathbf{a} = \langle a_1,a_2,a_3\rangle$ and $\mathbf{b} = \langle b_1,b_2,b_3\rangle$ ,

\mathbf{a}\times\mathbf{b} = \langle a_2b_3 - a_3b_2,\; a_3b_1 - a_1b_3,\; a_1b_2 - a_2b_1\rangle.

Nobody remembers that formula as written. The reliable way to compute it is the symbolic determinant

\mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = \begin{vmatrix} a_2 & a_3 \\ b_2 & b_3\end{vmatrix}\mathbf{i} - \begin{vmatrix} a_1 & a_3 \\ b_1 & b_3\end{vmatrix}\mathbf{j} + \begin{vmatrix} a_1 & a_2 \\ b_1 & b_2\end{vmatrix}\mathbf{k},

expanded along the first row, where each $2\times 2$ determinant is $\begin{vmatrix} p & q \\ r & s\end{vmatrix} = ps - qr$ . The minus sign on the $\mathbf{j}$ term is the one everybody forgets; cover a column of the $2\times 3$ block of numbers, compute the little determinant that remains, and alternate the signs $+,-,+$ .

Theorem 10.23 (The cross product is perpendicular to both factors). $\mathbf{a}\times\mathbf{b}$ is orthogonal to $\mathbf{a}$ and to $\mathbf{b}$ .

Proof. Dot the definition with $\mathbf{a}$ :

(\mathbf{a}\times\mathbf{b})\cdot\mathbf{a} = a_1(a_2b_3 - a_3b_2) + a_2(a_3b_1 - a_1b_3) + a_3(a_1b_2 - a_2b_1).

Expanding, $a_1a_2b_3 - a_1a_3b_2 + a_2a_3b_1 - a_1a_2b_3 + a_1a_3b_2 - a_2a_3b_1$ , and every term cancels against another: the sum is $0$ . The computation for $\mathbf{b}$ is identical with the roles swapped.∎

That settles the direction up to a sign, and the sign is fixed by the right-hand rule: curl the fingers of your right hand from $\mathbf{a}$ toward $\mathbf{b}$ through the smaller angle, and your thumb points along $\mathbf{a}\times\mathbf{b}$ . The definition was arranged so that this matches the orientation of the axes; you can check it on $\mathbf{i}\times\mathbf{j} = \mathbf{k}$ directly from the determinant. What remains is the length.

Theorem 10.24 (Length of the cross product). If $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$ , with $0 \le \theta \le \pi$ , then

|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}|\,|\mathbf{b}|\sin\theta.

Proof. Square the components in the definition and add. Multiplying out and regrouping gives the identity

|\mathbf{a}\times\mathbf{b}|^2 = (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2) - (a_1b_1+a_2b_2+a_3b_3)^2 = |\mathbf{a}|^2|\mathbf{b}|^2 - (\mathbf{a}\cdot\mathbf{b})^2.

The identity is verified by brute expansion: each side is a sum of the six terms $a_i^{2}b_j^{2}$ with $i\ne j$ minus the three cross terms $2a_ia_jb_ib_j$ with $i < j$ , and the three terms $a_i^{2}b_i^{2}$ cancel between the two products on the right. Now substitute $\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$ :

|\mathbf{a}\times\mathbf{b}|^2 = |\mathbf{a}|^2|\mathbf{b}|^2(1-\cos^2\theta) = |\mathbf{a}|^2|\mathbf{b}|^2\sin^2\theta.

Since $\sin\theta \ge 0$ on $[0,\pi]$ , taking square roots gives the result.∎

Corollary 10.25 (Parallel vectors and area). Two nonzero vectors are parallel if and only if $\mathbf{a}\times\mathbf{b} = \mathbf{0}$ . For any $\mathbf{a}$ and $\mathbf{b}$ , $|\mathbf{a}\times\mathbf{b}|$ is the area of the parallelogram with sides $\mathbf{a}$ and $\mathbf{b}$ .

Proof. $\sin\theta = 0$ exactly when $\theta = 0$ or $\pi$ , which is parallelism. For the area, the parallelogram has base $|\mathbf{a}|$ and height $|\mathbf{b}|\sin\theta$ , so its area is $|\mathbf{a}||\mathbf{b}|\sin\theta$ .∎

Intuition. Push open a heavy door. How effective your push is depends on two things: how hard you push and how far from the hinge you push. Push straight toward the hinge and nothing happens — that is $\sin\theta = 0$ . Push perpendicular to the door at the far edge and it swings easily — that is $\sin\theta = 1$ with $|\mathbf{r}|$ as large as possible. The cross product $\mathbf{r}\times\mathbf{F}$ packages all of this: its length is the turning effect and its direction is the axis the door turns about.

The products of the basis vectors follow the cyclic pattern $\mathbf{i}\to\mathbf{j}\to\mathbf{k}\to\mathbf{i}$ :

\mathbf{i}\times\mathbf{j} = \mathbf{k}, \quad \mathbf{j}\times\mathbf{k} = \mathbf{i}, \quad \mathbf{k}\times\mathbf{i} = \mathbf{j}, \qquad \mathbf{j}\times\mathbf{i} = -\mathbf{k}, \quad \mathbf{k}\times\mathbf{j} = -\mathbf{i}, \quad \mathbf{i}\times\mathbf{k} = -\mathbf{j},

and $\mathbf{i}\times\mathbf{i} = \mathbf{j}\times\mathbf{j} = \mathbf{k}\times\mathbf{k} = \mathbf{0}$ . Going around the cycle gives a plus sign, going against it a minus sign.

Theorem 10.26 (Algebra of the cross product). For vectors $\mathbf{a},\mathbf{b},\mathbf{c}$ and a scalar $c$ :

\mathbf{a}\times\mathbf{b} = -\,\mathbf{b}\times\mathbf{a}, \qquad (c\mathbf{a})\times\mathbf{b} = c(\mathbf{a}\times\mathbf{b}) = \mathbf{a}\times(c\mathbf{b}),

\mathbf{a}\times(\mathbf{b}+\mathbf{c}) = \mathbf{a}\times\mathbf{b} + \mathbf{a}\times\mathbf{c}, \qquad (\mathbf{a}+\mathbf{b})\times\mathbf{c} = \mathbf{a}\times\mathbf{c} + \mathbf{b}\times\mathbf{c},

\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = (\mathbf{a}\times\mathbf{b})\cdot\mathbf{c}, \qquad \mathbf{a}\times(\mathbf{b}\times\mathbf{c}) = (\mathbf{a}\cdot\mathbf{c})\mathbf{b} - (\mathbf{a}\cdot\mathbf{b})\mathbf{c}.

Proof. Anticommutativity is immediate from the definition: swapping $\mathbf{a}$ and $\mathbf{b}$ swaps the two terms in each component, $a_2b_3 - a_3b_2$ becomes $b_2a_3 - b_3a_2$ , which is the negative. The scalar and distributive laws are checked component by component exactly as for the dot product. The last two identities are verified by writing out both sides in components; the first of them is worth doing once, because it says the scalar triple product below can be computed with the dot and cross in either position.∎

Two properties that ordinary multiplication has are conspicuously absent. The cross product is not commutative — reversing the order reverses the sign — and it is not associative: $\mathbf{i}\times(\mathbf{i}\times\mathbf{j}) = \mathbf{i}\times\mathbf{k} = -\mathbf{j}$ , whereas $(\mathbf{i}\times\mathbf{i})\times\mathbf{j} = \mathbf{0}\times\mathbf{j} = \mathbf{0}$ . Brackets in a cross product are never optional.

Definition 10.27 (Scalar triple product). The scalar triple product of $\mathbf{a}$ , $\mathbf{b}$ , $\mathbf{c}$ is the number

\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix}.

The determinant form follows because $\mathbf{b}\times\mathbf{c}$ has components equal to the three $2\times 2$ minors, and dotting with $\mathbf{a}$ is exactly the first-row expansion.

Theorem 10.28 (Volume of a parallelepiped). The volume of the parallelepiped with edges $\mathbf{a}$ , $\mathbf{b}$ , $\mathbf{c}$ is

V = |\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})|.

In particular, three vectors are coplanar if and only if their scalar triple product is $0$ .

Proof. Take the parallelogram spanned by $\mathbf{b}$ and $\mathbf{c}$ as the base; its area is $A = |\mathbf{b}\times\mathbf{c}|$ . The height $h$ is the distance from the tip of $\mathbf{a}$ to the plane of the base, measured along the base's normal direction $\mathbf{b}\times\mathbf{c}$ . If $\phi$ is the angle between $\mathbf{a}$ and $\mathbf{b}\times\mathbf{c}$ , then $h = |\mathbf{a}||\cos\phi|$ , so

V = Ah = |\mathbf{b}\times\mathbf{c}|\,|\mathbf{a}|\,|\cos\phi| = |\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})|.

The volume is $0$ precisely when $\mathbf{a}$ lies in the plane of $\mathbf{b}$ and $\mathbf{c}$ .∎

The cross product also has a direct physical meaning. When a force $\mathbf{F}$ is applied at the end of a lever arm $\mathbf{r}$ measured from a pivot, the torque about the pivot is

\boldsymbol{\tau} = \mathbf{r}\times\mathbf{F}, \qquad |\boldsymbol{\tau}| = |\mathbf{r}||\mathbf{F}|\sin\theta,

the turning effect of the force. Its direction is the axis of rotation, and its magnitude is largest when the force is perpendicular to the arm.

Example 10.29 (Computing a cross product). Compute $\langle 1,3,4\rangle\times\langle 2,7,-5\rangle$ and verify that the result is perpendicular to both factors.

Solution.

Set up the determinant with $\mathbf{i},\mathbf{j},\mathbf{k}$ in the top row, $1,3,4$ in the second and $2,7,-5$ in the third.
$\mathbf{i}$ component: $3(-5) - 4(7) = -15 - 28 = -43$ .
$\mathbf{j}$ component, with the minus sign: $-\bigl(1(-5) - 4(2)\bigr) = -(-5-8) = 13$ .
$\mathbf{k}$ component: $1(7) - 3(2) = 7 - 6 = 1$ .
So the product is $\langle -43, 13, 1\rangle$ . Check: $\langle -43,13,1\rangle\cdot\langle 1,3,4\rangle = -43 + 39 + 4 = 0$ and $\langle -43,13,1\rangle\cdot\langle 2,7,-5\rangle = -86 + 91 - 5 = 0$ .

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Example 10.30 (A normal vector and the area of a triangle). Find a vector perpendicular to the plane through $P(1,4,6)$ , $Q(-2,5,-1)$ and $R(1,-1,1)$ , and the area of the triangle $PQR$ .

Solution.

Two vectors in the plane are $\overrightarrow{PQ} = \langle -3,1,-7\rangle$ and $\overrightarrow{PR} = \langle 0,-5,-5\rangle$ .
Their cross product is perpendicular to both, hence to the plane:
$\overrightarrow{PQ}\times\overrightarrow{PR} = \langle 1(-5) - (-7)(-5),\; (-7)(0) - (-3)(-5),\; (-3)(-5) - 1(0)\rangle = \langle -40, -15, 15\rangle.$
Any nonzero multiple also works; $\langle -8,-3,3\rangle$ is tidier.
The parallelogram spanned by $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ has area $|\langle -40,-15,15\rangle| = \sqrt{1600+225+225} = \sqrt{2050} = 5\sqrt{82}$ , and the triangle is half of it: $\dfrac{5\sqrt{82}}{2} \approx 22.6$ .

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Example 10.31 (Coplanar vectors). Show that $\mathbf{a} = \langle 1,4,-7\rangle$ , $\mathbf{b} = \langle 2,-1,4\rangle$ and $\mathbf{c} = \langle 0,-9,18\rangle$ are coplanar.

Solution.

Compute the scalar triple product as a determinant, expanding along the first row:
$\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = 1\begin{vmatrix} -1 & 4\\ -9 & 18\end{vmatrix} - 4\begin{vmatrix} 2 & 4\\ 0 & 18\end{vmatrix} + (-7)\begin{vmatrix} 2 & -1\\ 0 & -9\end{vmatrix}.$
The minors are $-18 + 36 = 18$ , $36 - 0 = 36$ and $-18 - 0 = -18$ .
So the triple product is $1(18) - 4(36) - 7(-18) = 18 - 144 + 126 = 0$ .
The parallelepiped has zero volume, so the three vectors lie in a common plane. In fact one is a combination of the other two: solving $\mathbf{c} = s\mathbf{a} + t\mathbf{b}$ from the first two components gives $s = -2$ , $t = 1$ , and the third component confirms it, $-2(-7) + 1(4) = 18$ . So $\mathbf{c} = -2\mathbf{a} + \mathbf{b}$ .

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Example 10.32 (Torque on a bolt). A wrench $0.25$ m long is attached to a bolt, and a force of $40$ N is applied to the end of the handle at an angle of $60^\circ$ to the handle. Find the magnitude of the torque about the bolt.

Solution.

$|\boldsymbol{\tau}| = |\mathbf{r}||\mathbf{F}|\sin\theta = (0.25)(40)\sin 60^\circ$ .
$= 10\cdot\dfrac{\sqrt3}{2} = 5\sqrt3 \approx 8.66$ N $\cdot$ m.
Pushing perpendicular to the handle ( $\theta = 90^\circ$ ) would give the full $10$ N $\cdot$ m; the $60^\circ$ angle wastes the component of the force along the handle.

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Pitfall. Three things the cross product is not. It is not commutative: $\mathbf{b}\times\mathbf{a} = -\mathbf{a}\times\mathbf{b}$ , so order matters when a problem asks for a normal "pointing upward". It is not associative, so never drop the brackets. And it is not a two-dimensional operation: for plane vectors, append a third component $0$ and read off the $\mathbf{k}$ component, which is the signed area $a_1b_2 - a_2b_1$ .

Exercises for §10.4 →

10.5Lines and planes

With a point-plus-direction description of a line and a point-plus-normal description of a plane, every question about how lines and planes sit in space becomes a computation with dot and cross products.

Definition 10.33 (Vector, parametric and symmetric equations of a line). The line through the point $P_0(x_0,y_0,z_0)$ parallel to the nonzero vector $\mathbf{v} = \langle a,b,c\rangle$ consists of the points with position vector

\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}, \qquad t \in \mathbb{R},

where $\mathbf{r}_0 = \langle x_0,y_0,z_0\rangle$ . In components these are the parametric equations

x = x_0 + at, \qquad y = y_0 + bt, \qquad z = z_0 + ct,

and if $a$ , $b$ , $c$ are all nonzero, eliminating $t$ gives the symmetric equations

\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}.

The vector equation is the honest one: start at $\mathbf{r}_0$ and travel $t$ copies of $\mathbf{v}$ . The numbers $a,b,c$ are the direction numbers of the line. Neither the point nor the direction vector is unique — any point on the line and any nonzero multiple of $\mathbf{v}$ give the same line — so two correct answers to "find the equation of the line" may look quite different. If one direction number is $0$ , say $c = 0$ , the symmetric form becomes $\frac{x-x_0}{a} = \frac{y-y_0}{b}$ together with $z = z_0$ : the line lies in a horizontal plane.

Restricting the parameter gives a segment. The segment from $\mathbf{r}_0$ to $\mathbf{r}_1$ is

\mathbf{r}(t) = (1-t)\mathbf{r}_0 + t\mathbf{r}_1, \qquad 0 \le t \le 1,

since the direction is $\mathbf{r}_1 - \mathbf{r}_0$ and $\mathbf{r}_0 + t(\mathbf{r}_1-\mathbf{r}_0)$ regroups to the formula above.

Definition 10.34 (Skew lines). Two lines in $\mathbb{R}^3$ are skew if they are not parallel and do not intersect. Skew lines do not lie in a common plane.

In the plane, two distinct lines either meet or are parallel. In space there is a third possibility — think of a road passing over a railway bridge — and it is the generic one.

Definition 10.35 (Equations of a plane). The plane through $P_0(x_0,y_0,z_0)$ with normal vector $\mathbf{n} = \langle a,b,c\rangle \ne \mathbf{0}$ consists of all points $P$ for which $\overrightarrow{P_0P}$ is perpendicular to $\mathbf{n}$ :

\mathbf{n}\cdot(\mathbf{r}-\mathbf{r}_0) = 0, \qquad\text{that is,}\qquad a(x-x_0) + b(y-y_0) + c(z-z_0) = 0.

Collecting constants gives the linear equation $ax + by + cz + d = 0$ with $d = -(ax_0+by_0+cz_0)$ .

A line is determined by a point and a direction along it; a plane is determined by a point and a direction perpendicular to it. That asymmetry is why the plane needs a dot product and the line does not.

Theorem 10.36 (Linear equations are planes). If $a,b,c$ are not all zero, the set of solutions of $ax+by+cz+d = 0$ is a plane with normal vector $\langle a,b,c\rangle$ .

Proof. Pick any solution $(x_0,y_0,z_0)$ , which exists because some coefficient is nonzero. Subtracting $ax_0+by_0+cz_0+d = 0$ from $ax+by+cz+d = 0$ gives $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$ , which is the plane through $(x_0,y_0,z_0)$ with normal $\langle a,b,c\rangle$ .∎

So the normal can be read straight off the coefficients. Two planes are parallel when their normals are parallel; otherwise they meet in a line, and the angle between the planes is defined to be the (acute) angle between their normals. The direction of the line of intersection is perpendicular to both normals — that is, it is $\mathbf{n}_1\times\mathbf{n}_2$ .

Intuition. A laser pointer describes a line: one point (the pointer) and one direction (the beam). A sheet of glass describes a plane, but its defining direction is the one it blocks — hold a pencil perpendicular to the glass and every direction you can slide along the glass is at right angles to the pencil. The pencil is the normal vector, and "perpendicular to the pencil" is exactly the equation $\mathbf{n}\cdot(\mathbf{r}-\mathbf{r}_0) = 0$ .

Theorem 10.37 (Distance from a point to a plane). The distance from $P_1(x_1,y_1,z_1)$ to the plane $ax+by+cz+d = 0$ is

D = \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}.

Proof. Let $P_0(x_0,y_0,z_0)$ be any point on the plane and $\mathbf{b} = \overrightarrow{P_0P_1} = \langle x_1-x_0,\; y_1-y_0,\; z_1-z_0\rangle$ . The distance from $P_1$ to the plane is the length of the shadow of $\mathbf{b}$ on the normal direction, the absolute scalar projection

D = |\operatorname{comp}_{\mathbf{n}}\mathbf{b}| = \frac{|\mathbf{n}\cdot\mathbf{b}|}{|\mathbf{n}|} = \frac{|a(x_1-x_0)+b(y_1-y_0)+c(z_1-z_0)|}{\sqrt{a^2+b^2+c^2}}.

Since $P_0$ lies on the plane, $ax_0+by_0+cz_0 = -d$ , and the numerator becomes $|ax_1+by_1+cz_1+d|$ .∎

The same theorem handles two further distances. The distance between parallel planes is the distance from any point of one to the other. The distance between skew lines $L_1$ and $L_2$ is found by noting that they lie in a pair of parallel planes whose common normal is $\mathbf{n} = \mathbf{v}_1\times\mathbf{v}_2$ ; the distance between the lines is the distance from any point of $L_1$ to the plane containing $L_2$ with normal $\mathbf{n}$ , which is $|\mathbf{n}\cdot\overrightarrow{P_1P_2}|/|\mathbf{n}|$ for any points $P_1 \in L_1$ , $P_2 \in L_2$ .

Example 10.38 (A line from a point and a direction). Find vector, parametric and symmetric equations of the line through $(5,1,3)$ parallel to $\mathbf{i}+4\mathbf{j}-2\mathbf{k}$ , and find where it crosses the $xy$ -plane.

Solution.

With $\mathbf{r}_0 = \langle 5,1,3\rangle$ and $\mathbf{v} = \langle 1,4,-2\rangle$ , the vector equation is $\mathbf{r}(t) = \langle 5+t,\; 1+4t,\; 3-2t\rangle$ .
Parametric: $x = 5+t$ , $y = 1+4t$ , $z = 3-2t$ .
Symmetric: $x - 5 = \dfrac{y-1}{4} = \dfrac{z-3}{-2}$ .
The $xy$ -plane is $z = 0$ , so $3 - 2t = 0$ gives $t = \tfrac32$ , and the crossing point is $\left(\tfrac{13}{2},\, 7,\, 0\right)$ . Check in the symmetric form: $\tfrac{13}{2}-5 = \tfrac32$ , $\tfrac{7-1}{4} = \tfrac32$ , $\tfrac{0-3}{-2} = \tfrac32$ .

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Example 10.39 (Skew or intersecting?). Determine whether the lines

L_1:\ x = 1+t,\ y = -2+3t,\ z = 4-t \qquad\text{and}\qquad L_2:\ x = 2s,\ y = 3+s,\ z = -3+4s

are parallel, intersecting or skew. If they are skew, find the distance between them.

Solution.

The direction vectors are $\mathbf{v}_1 = \langle 1,3,-1\rangle$ and $\mathbf{v}_2 = \langle 2,1,4\rangle$ . Neither is a multiple of the other, so the lines are not parallel.
If they meet, some $t$ and $s$ satisfy all three coordinate equations. From $x$ : $1+t = 2s$ . From $y$ : $-2+3t = 3+s$ . Substituting $t = 2s-1$ into the second gives $-2 + 6s - 3 = 3 + s$ , so $5s = 8$ , $s = \tfrac85$ and $t = \tfrac{11}{5}$ .
Test the $z$ -equation: $4 - \tfrac{11}{5} = \tfrac95$ but $-3 + \tfrac{32}{5} = \tfrac{17}{5}$ . The values disagree, so the lines do not meet. They are skew.
For the distance, a common normal is $\mathbf{n} = \mathbf{v}_1\times\mathbf{v}_2 = \langle 3(4)-(-1)(1),\; (-1)(2)-1(4),\; 1(1)-3(2)\rangle = \langle 13,-6,-5\rangle$ .
Take $P_1(1,-2,4)$ on $L_1$ and $P_2(0,3,-3)$ on $L_2$ , so $\overrightarrow{P_1P_2} = \langle -1,5,-7\rangle$ , and
$D = \frac{|\mathbf{n}\cdot\overrightarrow{P_1P_2}|}{|\mathbf{n}|} = \frac{|-13 - 30 + 35|}{\sqrt{169+36+25}} = \frac{8}{\sqrt{230}} \approx 0.53.$

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Example 10.40 (A plane through three points). Find an equation of the plane through $P(1,3,2)$ , $Q(3,-1,6)$ and $R(5,2,0)$ .

Solution.

Two vectors in the plane: $\overrightarrow{PQ} = \langle 2,-4,4\rangle$ and $\overrightarrow{PR} = \langle 4,-1,-2\rangle$ .
A normal is their cross product: $\mathbf{n} = \langle (-4)(-2) - 4(-1),\; 4(4) - 2(-2),\; 2(-1) - (-4)(4)\rangle = \langle 12, 20, 14\rangle$ , or after dividing by $2$ , $\langle 6,10,7\rangle$ .
Through $P$ with this normal: $6(x-1) + 10(y-3) + 7(z-2) = 0$ , which simplifies to $6x + 10y + 7z = 50$ .
Check the other two points: $Q$ gives $18 - 10 + 42 = 50$ and $R$ gives $30 + 20 + 0 = 50$ .

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Example 10.41 (Where a line meets a plane). Find the point at which the line $x = 2+3t$ , $y = -4t$ , $z = 5+t$ meets the plane $4x + 5y - 2z = 18$ .

Solution.

Substitute the parametric equations into the plane: $4(2+3t) + 5(-4t) - 2(5+t) = 18$ .
Expand: $8 + 12t - 20t - 10 - 2t = 18$ , so $-10t - 2 = 18$ and $t = -2$ .
Substitute back: $x = 2 - 6 = -4$ , $y = 8$ , $z = 5 - 2 = 3$ . The point is $(-4,8,3)$ .
Check: $4(-4) + 5(8) - 2(3) = -16 + 40 - 6 = 18$ .

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Example 10.42 (Two planes: their angle and their line of intersection). Find the angle between the planes $x+y+z = 1$ and $x-2y+3z = 1$ , and symmetric equations for their line of intersection.

Solution.

The normals are $\mathbf{n}_1 = \langle 1,1,1\rangle$ and $\mathbf{n}_2 = \langle 1,-2,3\rangle$ , so
$\cos\theta = \frac{\mathbf{n}_1\cdot\mathbf{n}_2}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{1-2+3}{\sqrt3\sqrt{14}} = \frac{2}{\sqrt{42}}, \qquad \theta \approx 72^\circ.$
The line of intersection is perpendicular to both normals, so its direction is $\mathbf{n}_1\times\mathbf{n}_2 = \langle 3+2,\; 1-3,\; -2-1\rangle = \langle 5,-2,-3\rangle$ .
To find a point on the line, set $z = 0$ in both equations: $x + y = 1$ and $x - 2y = 1$ give $y = 0$ , $x = 1$ . So $(1,0,0)$ lies on both planes.
Symmetric equations: $\dfrac{x-1}{5} = \dfrac{y}{-2} = \dfrac{z}{-3}$ .

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Example 10.43 (Distance between parallel planes). Find the distance between the planes $10x + 2y - 2z = 5$ and $5x + y - z = 1$ .

Solution.

The normals $\langle 10,2,-2\rangle$ and $\langle 5,1,-1\rangle$ are parallel, so the planes are parallel and the distance is well defined.
Pick a point on the first plane: with $y = z = 0$ , $x = \tfrac12$ , giving $\left(\tfrac12,0,0\right)$ .
Apply the distance formula to the second plane written as $5x + y - z - 1 = 0$ :
$D = \frac{|5\cdot\tfrac12 + 0 - 0 - 1|}{\sqrt{25+1+1}} = \frac{3/2}{3\sqrt3} = \frac{1}{2\sqrt3} = \frac{\sqrt3}{6}.$
The answer does not depend on which point of the first plane was chosen; try $(0,\tfrac52,0)$ and you get the same $\tfrac{3/2}{3\sqrt3}$ .

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Pitfall. For a plane, the coefficients give a normal vector; for a line, the coefficients of $t$ give a direction vector. Students routinely swap these — reading $\langle 2,-1,4\rangle$ off $2x - y + 4z = 12$ and calling it the plane's direction. A plane has infinitely many directions in it and exactly one normal direction; the coefficients are the normal.

Exercises for §10.5 →

10.6Cylinders and quadric surfaces

Planes are the surfaces of degree one. The next family up — surfaces given by second-degree equations — includes the sphere, the cone, the saddle and a handful of others, and they appear throughout the rest of calculus as the graphs and level surfaces you integrate over. The technique for understanding any of them is the same: slice it with planes parallel to the coordinate planes and look at the cross-sections.

Definition 10.44 (Trace). The trace of a surface in a plane is the curve in which the plane cuts the surface. The traces in the planes $x = k$ , $y = k$ and $z = k$ are found by substituting the constant into the equation of the surface.

Definition 10.45 (Cylinder). A cylinder is a surface made up of all lines (rulings) parallel to a fixed line and passing through a fixed plane curve. In coordinates, an equation in which one of the three variables is missing describes a cylinder whose rulings are parallel to the axis of the missing variable.

The parabolic cylinder $z = x^{2}$ is the parabola $z = x^{2}$ in the $xz$ -plane, extruded along the $y$ -axis; every trace $y = k$ is the same parabola. The circular cylinder $x^{2}+y^{2} = 1$ is the unit circle extruded along the $z$ -axis. The word "cylinder" here is wider than everyday usage: the cross-section can be any plane curve.

Definition 10.46 (Quadric surface). A quadric surface is the graph of a second-degree equation in $x$ , $y$ and $z$ . By translating and rotating the axes, any such equation can be brought to one of two shapes,

Ax^2 + By^2 + Cz^2 + J = 0 \qquad\text{or}\qquad Ax^2 + By^2 + Iz = 0,

and there are six standard non-degenerate surfaces.

The six, with $a,b,c > 0$ , are these. The ellipsoid,

\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1,

whose traces in all three directions are ellipses and which reduces to the sphere when $a = b = c$ . The elliptic paraboloid,

\frac{z}{c} = \frac{x^2}{a^2} + \frac{y^2}{b^2},

a bowl opening along the axis of the linear variable: horizontal traces are ellipses, vertical traces are parabolas. The hyperbolic paraboloid,

\frac{z}{c} = \frac{x^2}{a^2} - \frac{y^2}{b^2},

the saddle: horizontal traces are hyperbolas, vertical traces are parabolas opening in opposite directions in the two coordinate directions. The cone,

\frac{z^2}{c^2} = \frac{x^2}{a^2} + \frac{y^2}{b^2},

with elliptical horizontal traces that shrink to a point at the origin, and vertical traces that are hyperbolas except through the origin, where they are pairs of lines. The hyperboloid of one sheet,

\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1,

a connected surface with a waist, whose horizontal traces are ellipses and vertical traces hyperbolas; the axis is the variable with the minus sign. And the hyperboloid of two sheets,

-\frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1,

two separate bowls opening away from each other along the axis of the variable with the plus sign, with no points at all where $|z| < c$ .

Read the signs. Three squared terms all positive with a $1$ on the right is an ellipsoid. One negative sign is a hyperboloid of one sheet; two negative signs, two sheets; a $0$ instead of the $1$ is a cone. A linear variable in place of a squared one gives a paraboloid — elliptic if the two squared terms agree in sign, hyperbolic if they disagree.

Intuition. Picture slicing a loaf of bread. Each slice is a trace, and the shape of the loaf is reconstructed from the sequence of slices. Slice a bowl (an elliptic paraboloid) horizontally and the slices are ovals that grow as you go up. Slice a saddle horizontally and you get hyperbolas — two arcs — that swap their orientation as you pass the level of the seat. Slice an hourglass (a hyperboloid of one sheet) and the ovals shrink to a waist and grow again. Three or four slices in each direction usually pin the surface down.

Method 10.47 (Identifying a quadric).

Complete the square in any variable that appears both squared and linearly, to locate the centre or vertex.
Divide through so that the right-hand side is $1$ (or $0$ , or a single linear term).
Count the squared terms and their signs, and note which variable, if any, appears only linearly.
Match with the six standard forms; the special axis is the variable that differs from the other two.
Confirm with one or two traces.

Example 10.48 (Sketching an ellipsoid by traces). Use traces to describe the surface $x^{2} + \dfrac{y^{2}}{9} + \dfrac{z^{2}}{4} = 1$ .

Solution.

Setting $z = 0$ gives $x^{2} + \dfrac{y^{2}}{9} = 1$ , an ellipse in the $xy$ -plane with semi-axes $1$ and $3$ .
Setting $z = k$ with $|k| < 2$ gives $x^{2} + \dfrac{y^{2}}{9} = 1 - \dfrac{k^{2}}{4}$ , a smaller ellipse; at $k = \pm 2$ it shrinks to a point, and for $|k| > 2$ there is no trace at all.
The traces $x = 0$ and $y = 0$ are the ellipses $\dfrac{y^{2}}{9} + \dfrac{z^{2}}{4} = 1$ and $x^{2} + \dfrac{z^{2}}{4} = 1$ .
Every trace is an ellipse, so this is an ellipsoid, stretched to $\pm 1$ in $x$ , $\pm 3$ in $y$ and $\pm 2$ in $z$ .

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Example 10.49 (A paraboloid and a saddle). Identify and describe the surfaces $z = 4x^{2} + y^{2}$ and $z = y^{2} - x^{2}$ .

Solution.

For $z = 4x^{2} + y^{2}$ : the trace $z = k > 0$ is the ellipse $4x^{2} + y^{2} = k$ , and $z = 0$ is the single point at the origin. The traces $x = 0$ and $y = 0$ are the parabolas $z = y^{2}$ and $z = 4x^{2}$ , both opening upward.
Ellipses horizontally and parabolas vertically: an elliptic paraboloid with vertex at the origin, opening along the positive $z$ -axis.
For $z = y^{2} - x^{2}$ : the trace $z = k$ is the hyperbola $y^{2} - x^{2} = k$ , opening in the $y$ -direction if $k > 0$ and in the $x$ -direction if $k < 0$ ; at $k = 0$ it is the pair of lines $y = \pm x$ .
The trace $x = 0$ is $z = y^{2}$ , opening up, while $y = 0$ is $z = -x^{2}$ , opening down. A surface that curves up in one direction and down in the other is a saddle: this is a hyperbolic paraboloid, and the origin is its saddle point.

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Example 10.50 (Hyperboloids). Identify $\dfrac{x^{2}}{4} + y^{2} - \dfrac{z^{2}}{4} = 1$ and $4x^{2} - y^{2} + 2z^{2} + 4 = 0$ .

Solution.

The first has three squared terms, one of them negative, and $1$ on the right: a hyperboloid of one sheet. Its axis is the $z$ -axis (the negative variable). The horizontal traces $z = k$ are the ellipses $\dfrac{x^{2}}{4} + y^{2} = 1 + \dfrac{k^{2}}{4}$ , smallest at the waist $z = 0$ .
For the second, move the constant across and divide by $-4$ : $-x^{2} + \dfrac{y^{2}}{4} - \dfrac{z^{2}}{2} = 1$ .
Two negative signs: a hyperboloid of two sheets, with axis the $y$ -axis (the positive variable). Setting $y = k$ gives $x^{2} + \dfrac{z^{2}}{2} = \dfrac{k^{2}}{4} - 1$ , which has solutions only when $|k| \ge 2$ ; the two sheets begin at $y = \pm 2$ and open away from each other.

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Example 10.51 (Completing the square to find a shifted quadric). Identify the surface $x^{2} + 2z^{2} - 6x - y + 10 = 0$ .

Solution.

Complete the square in $x$ : $x^{2} - 6x = (x-3)^2 - 9$ . Then the equation reads $(x-3)^2 + 2z^{2} - y + 1 = 0$ , or $y - 1 = (x-3)^2 + 2z^{2}$ .
Two squared terms with the same sign and one linear variable: an elliptic paraboloid.
The axis is parallel to the $y$ -axis, the vertex is at $(3,1,0)$ , and since the right-hand side is nonnegative the surface opens toward positive $y$ .
Check with a trace: $y = 3$ gives the ellipse $(x-3)^2 + 2z^{2} = 2$ , and $y = 0$ gives nothing, consistent with a bowl whose bottom is at $y = 1$ .

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Pitfall. A hyperboloid of one sheet and a cone look alike near the axis and are told apart by the constant: $x^{2} + y^{2} - z^{2} = 1$ has a waist of radius $1$ , while $x^{2} + y^{2} - z^{2} = 0$ pinches to a point. And the equation $z = x^{2}$ on its own is a cylinder, not a parabola and not a paraboloid — whenever a variable is missing, the surface is extruded along that axis.

Exercises for §10.6 →

10.7Vector functions and space curves

A curve in space is most naturally described by letting each coordinate be a function of one parameter, exactly as parametric curves in the plane were. Packaging the three coordinate functions as a single vector-valued object is what lets the calculus that follows be written once instead of three times.

Definition 10.52 (Vector function). A vector function (or vector-valued function) $\mathbf{r}$ assigns to each real number $t$ in its domain a vector

\mathbf{r}(t) = \langle f(t), g(t), h(t)\rangle = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k}.

The real functions $f$ , $g$ , $h$ are its component functions. Unless stated otherwise, the domain is the set of $t$ for which all three components are defined.

Definition 10.53 (Limit and continuity of a vector function). If the limits of the component functions exist,

\lim_{t\to a}\mathbf{r}(t) = \left\langle \lim_{t\to a}f(t),\; \lim_{t\to a}g(t),\; \lim_{t\to a}h(t)\right\rangle,

and $\mathbf{r}$ is continuous at $a$ if $\lim_{t\to a}\mathbf{r}(t) = \mathbf{r}(a)$ — equivalently, if each component function is continuous at $a$ .

The definition is componentwise, which means every limit law from single-variable calculus applies unchanged. What it means geometrically is that $\mathbf{r}(t)$ approaches $\mathbf{L}$ when the arrow's head approaches the head of $\mathbf{L}$ : the distance $|\mathbf{r}(t)-\mathbf{L}|$ tends to $0$ . The two descriptions agree because that distance is $\sqrt{(f-L_1)^2 + (g-L_2)^2 + (h-L_3)^2}$ , which tends to $0$ exactly when each difference does.

Definition 10.54 (Space curve). Let $f$ , $g$ , $h$ be continuous on an interval $I$ . The set of points $(x,y,z)$ with

x = f(t), \qquad y = g(t), \qquad z = h(t), \qquad t \in I,

is a space curve $C$ , and the equations are its parametric equations. The vector function $\mathbf{r}(t) = \langle f(t),g(t),h(t)\rangle$ is a parametrisation of $C$ : as $t$ runs through $I$ , the head of $\mathbf{r}(t)$ traces out $C$ .

The parameter usually plays the role of time, and the vector $\mathbf{r}(t)$ then records where a moving point is at time $t$ . The same curve has many parametrisations — run along it faster, or backwards, or starting from a different point — and later, when we measure arc length and curvature, we will need to check that the answers do not depend on which parametrisation was chosen.

Intuition. A drone logs its GPS position every second: a longitude, a latitude and an altitude. That log is a vector function of time, and the path drawn by connecting the dots is the space curve. Limits and continuity say the drone does not teleport — as the sampling gets finer, the positions settle down to a definite point at each instant. The log and the path are different things: the same path flown twice as fast is the same curve with a different vector function.

Two curves recur throughout the rest of the chapter and are worth knowing by sight. The circular helix

\mathbf{r}(t) = \langle \cos t, \sin t, t\rangle

has $x^{2} + y^{2} = 1$ , so it lies on the cylinder of radius $1$ about the $z$ -axis, and since $z = t$ it climbs steadily as it winds: a coil spring or the thread of a screw. The twisted cubic

\mathbf{r}(t) = \langle t, t^2, t^3\rangle

projects onto the $xy$ -plane as the parabola $y = x^{2}$ , onto the $xz$ -plane as the cubic $z = x^{3}$ , and onto the $yz$ -plane as $z^{2} = y^{3}$ ; it is the standard example of a curve that does not lie in any plane.

Example 10.55 (Domain and limit of a vector function). Find the domain of $\mathbf{r}(t) = \langle t^{3},\ \ln(3-t),\ \sqrt{t}\rangle$ , and compute $\displaystyle\lim_{t\to 0}\left\langle 1+t^{3},\ te^{-t},\ \frac{\sin t}{t}\right\rangle$ .

Solution.

$t^{3}$ is defined for all $t$ ; $\ln(3-t)$ needs $3 - t > 0$ , that is $t < 3$ ; $\sqrt t$ needs $t \ge 0$ . The domain is the intersection $[0,3)$ .
For the limit, take each component separately: $\lim_{t\to 0}(1+t^{3}) = 1$ , $\lim_{t\to 0}te^{-t} = 0\cdot 1 = 0$ , and $\lim_{t\to 0}\dfrac{\sin t}{t} = 1$ , the standard trigonometric limit.
So the limit is $\langle 1, 0, 1\rangle$ . The middle component being $0$ at $t = 0$ does not matter for the third; each component's limit is its own problem.

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Example 10.56 (The helix and the twisted cubic). Describe the curve $\mathbf{r}(t) = \langle 2\cos t, 2\sin t, 3t\rangle$ , and find the points where the twisted cubic $\langle t, t^{2}, t^{3}\rangle$ meets the plane $x + y + z = 3$ .

Solution.

Since $x^{2} + y^{2} = 4\cos^{2} t + 4\sin^{2} t = 4$ , every point lies on the cylinder of radius $2$ about the $z$ -axis. As $t$ increases by $2\pi$ the point goes once around the cylinder while $z$ rises by $6\pi$ . This is a helix of radius $2$ and pitch $6\pi$ , winding anticlockwise seen from above.
On the twisted cubic, the plane's equation becomes $t + t^{2} + t^{3} = 3$ , or $t^{3} + t^{2} + t - 3 = 0$ .
$t = 1$ is a root, and factoring gives $(t-1)(t^{2} + 2t + 3) = 0$ . The quadratic has discriminant $4 - 12 < 0$ , so $t = 1$ is the only real root.
The curve meets the plane once, at $\mathbf{r}(1) = (1,1,1)$ . Check: $1 + 1 + 1 = 3$ .

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Example 10.57 (A line segment as a vector function). Find a vector function for the line segment from $P(1,3,-2)$ to $Q(2,-1,3)$ .

Solution.

With $\mathbf{r}_0 = \langle 1,3,-2\rangle$ and $\mathbf{r}_1 = \langle 2,-1,3\rangle$ , the segment is $\mathbf{r}(t) = (1-t)\mathbf{r}_0 + t\mathbf{r}_1$ for $0 \le t \le 1$ .
Componentwise: $\mathbf{r}(t) = \langle 1 + t,\; 3 - 4t,\; -2 + 5t\rangle$ , $0 \le t \le 1$ .
Check the endpoints: $\mathbf{r}(0) = \langle 1,3,-2\rangle$ and $\mathbf{r}(1) = \langle 2,-1,3\rangle$ .

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Example 10.58 (A curve as an intersection of surfaces). Find a vector function for the curve in which the cylinder $x^{2} + y^{2} = 1$ meets the plane $y + z = 2$ , and one for the curve in which the cone $z = \sqrt{x^{2}+y^{2}}$ meets the plane $z = 1 + y$ .

Solution.

For the cylinder, the projection of the curve onto the $xy$ -plane is the unit circle, so set $x = \cos t$ , $y = \sin t$ with $0 \le t \le 2\pi$ .
The plane then determines $z = 2 - y = 2 - \sin t$ , giving $\mathbf{r}(t) = \langle \cos t, \sin t, 2 - \sin t\rangle$ . The curve is an ellipse — a tilted plane slicing a round tube.
For the cone and plane, eliminate $z$ : $\sqrt{x^{2}+y^{2}} = 1 + y$ . Squaring, $x^{2} + y^{2} = 1 + 2y + y^{2}$ , so $y = \dfrac{x^{2} - 1}{2}$ — the projection is a parabola.
Let $x = t$ . Then $y = \dfrac{t^{2}-1}{2}$ and $z = 1 + y = \dfrac{t^{2}+1}{2}$ , so $\mathbf{r}(t) = \left\langle t,\ \dfrac{t^{2}-1}{2},\ \dfrac{t^{2}+1}{2}\right\rangle$ . Since $z \ge 0$ always, the squaring introduced nothing spurious. The curve is a parabola in space: a plane parallel to a ruling of the cone cuts a parabola, as in classical conic sections.

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Pitfall. A vector function and its curve are different objects. $\langle \cos t, \sin t, 0\rangle$ and $\langle \cos 2t, \sin 2t, 0\rangle$ trace the same circle, but the second does it twice as fast, and derived quantities like velocity and speed differ. When a problem asks for "the curve", any parametrisation will do; when it asks about motion, the parametrisation is part of the question.

Exercises for §10.7 →

10.8Derivatives and integrals of vector functions

Definition 10.59 (Derivative of a vector function). The derivative of $\mathbf{r}$ at $t$ is

\mathbf{r}'(t) = \frac{d\mathbf{r}}{dt} = \lim_{h\to 0}\frac{\mathbf{r}(t+h) - \mathbf{r}(t)}{h},

whenever this limit exists.

The difference quotient has a picture. The vector $\mathbf{r}(t+h) - \mathbf{r}(t)$ is the secant vector from the point at time $t$ to the point at time $t+h$ ; dividing by $h$ rescales it (and, if $h < 0$ , reverses it so that it still points in the direction of increasing $t$ ). As $h \to 0$ the secant direction settles into the tangent direction. So $\mathbf{r}'(t)$ , when nonzero, is a tangent vector to the curve at $\mathbf{r}(t)$ , pointing the way the curve is being traversed.

Theorem 10.60 (Differentiate componentwise). If $\mathbf{r}(t) = \langle f(t),g(t),h(t)\rangle$ with $f$ , $g$ , $h$ differentiable, then

\mathbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle.

Proof. Limits of vector functions are computed componentwise, so

\mathbf{r}'(t) = \lim_{h\to 0}\left\langle \frac{f(t+h)-f(t)}{h},\ \frac{g(t+h)-g(t)}{h},\ \frac{h(t+h)-h(t)}{h}\right\rangle = \langle f'(t), g'(t), h'(t)\rangle,

each component limit being the ordinary derivative.∎

Definition 10.61 (Unit tangent vector and smooth curve). If $\mathbf{r}'(t) \ne \mathbf{0}$ , the unit tangent vector at $t$ is

\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}.

A curve is smooth on an interval if $\mathbf{r}'$ is continuous and $\mathbf{r}'(t) \ne \mathbf{0}$ throughout it. A smooth curve has no corners or cusps: the tangent direction turns continuously.

The condition $\mathbf{r}'(t) \ne \mathbf{0}$ matters. The curve $\mathbf{r}(t) = \langle 1 + t^{3}, t^{2}\rangle$ has continuous derivative $\langle 3t^{2}, 2t\rangle$ , but at $t = 0$ this is $\mathbf{0}$ and the curve has a cusp at $(1,0)$ : the point slows to a stop, reverses one component, and leaves in a new direction. The tangent line at a point where $\mathbf{r}'(t_0) \ne \mathbf{0}$ is the line through $\mathbf{r}(t_0)$ with direction $\mathbf{r}'(t_0)$ , and the tools of the lines-and-planes section apply to it directly.

Theorem 10.62 (Differentiation rules). Let $\mathbf{u}$ and $\mathbf{v}$ be differentiable vector functions, $c$ a scalar and $f$ a differentiable real function. Then

\begin{aligned} &\frac{d}{dt}\bigl[\mathbf{u}(t)+\mathbf{v}(t)\bigr] = \mathbf{u}'(t)+\mathbf{v}'(t), &\quad &\frac{d}{dt}\bigl[c\,\mathbf{u}(t)\bigr] = c\,\mathbf{u}'(t),\\ &\frac{d}{dt}\bigl[f(t)\,\mathbf{u}(t)\bigr] = f'(t)\,\mathbf{u}(t) + f(t)\,\mathbf{u}'(t), &\quad &\frac{d}{dt}\bigl[\mathbf{u}(f(t))\bigr] = f'(t)\,\mathbf{u}'(f(t)),\\ &\frac{d}{dt}\bigl[\mathbf{u}(t)\cdot\mathbf{v}(t)\bigr] = \mathbf{u}'(t)\cdot\mathbf{v}(t) + \mathbf{u}(t)\cdot\mathbf{v}'(t), &\quad &\frac{d}{dt}\bigl[\mathbf{u}(t)\times\mathbf{v}(t)\bigr] = \mathbf{u}'(t)\times\mathbf{v}(t) + \mathbf{u}(t)\times\mathbf{v}'(t). \end{aligned}

Proof. Every rule follows by writing both sides in components and applying the corresponding single-variable rule. For the dot product rule, with $\mathbf{u} = \langle u_1,u_2,u_3\rangle$ and $\mathbf{v} = \langle v_1,v_2,v_3\rangle$ ,

\frac{d}{dt}(\mathbf{u}\cdot\mathbf{v}) = \frac{d}{dt}\sum_{i=1}^{3}u_iv_i = \sum_{i=1}^{3}\bigl(u_i'v_i + u_iv_i'\bigr) = \mathbf{u}'\cdot\mathbf{v} + \mathbf{u}\cdot\mathbf{v}',

by the ordinary product rule applied to each term. The cross product rule is the same argument on each of the three components $u_2v_3 - u_3v_2$ and so on; the only thing to watch is that the order of the factors is preserved, since the cross product is not commutative.∎

Corollary 10.63 (Constant length forces a perpendicular derivative). If $|\mathbf{r}(t)|$ is constant, then $\mathbf{r}(t)\cdot\mathbf{r}'(t) = 0$ for all $t$ : the derivative is always perpendicular to the position.

Proof. $|\mathbf{r}|^2 = \mathbf{r}\cdot\mathbf{r}$ is constant, so differentiating with the dot product rule gives $0 = \mathbf{r}'\cdot\mathbf{r} + \mathbf{r}\cdot\mathbf{r}' = 2\,\mathbf{r}\cdot\mathbf{r}'$ .∎

This small result is used constantly. A point moving on a sphere has velocity tangent to the sphere; and since the unit tangent $\mathbf{T}$ has constant length $1$ , its derivative $\mathbf{T}'$ is always perpendicular to $\mathbf{T}$ — the fact on which the whole theory of curvature rests.

Definition 10.64 (Integral of a vector function). For continuous $\mathbf{r}(t) = \langle f(t),g(t),h(t)\rangle$ ,

\int_a^b\mathbf{r}(t)\,dt = \left\langle \int_a^b f(t)\,dt,\ \int_a^b g(t)\,dt,\ \int_a^b h(t)\,dt\right\rangle,

and the indefinite integral $\int\mathbf{r}(t)\,dt$ is any antiderivative $\mathbf{R}$ with $\mathbf{R}' = \mathbf{r}$ , determined up to a constant vector $\mathbf{C}$ .

The definition is the Riemann sum definition applied to each component, and the Fundamental Theorem of Calculus carries over verbatim: $\int_a^b\mathbf{r}(t)\,dt = \mathbf{R}(b) - \mathbf{R}(a)$ . The constant of integration is a vector, which means three constants, one per component, and an initial condition $\mathbf{r}(0) = \mathbf{r}_0$ fixes all three at once.

Intuition. Differentiating a vector function is differentiating three ordinary functions and keeping the results in their slots. Nothing about the vector structure gets in the way. The one genuinely new fact is geometric: the three derivatives, read together as a vector, point along the curve. A drone's log of positions, differenced second by second, becomes a log of velocities; integrate the velocities again and, given where it started, you get back the positions.

Example 10.65 (A derivative and a unit tangent). For $\mathbf{r}(t) = \langle 1+t^{3},\ te^{-t},\ \sin 2t\rangle$ , find $\mathbf{r}'(t)$ and the unit tangent vector at $t = 0$ .

Solution.

Differentiate each component: $\mathbf{r}'(t) = \langle 3t^{2},\ (1-t)e^{-t},\ 2\cos 2t\rangle$ , the middle one by the product rule.
At $t = 0$ : $\mathbf{r}'(0) = \langle 0, 1, 2\rangle$ and $|\mathbf{r}'(0)| = \sqrt{0+1+4} = \sqrt5$ .
So $\mathbf{T}(0) = \dfrac{1}{\sqrt5}\langle 0,1,2\rangle = \left\langle 0, \dfrac{1}{\sqrt5}, \dfrac{2}{\sqrt5}\right\rangle$ .

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Example 10.66 (A tangent line to a helix). Find parametric equations for the tangent line to $\mathbf{r}(t) = \langle 2\cos t, \sin t, t\rangle$ at the point $(0, 1, \pi/2)$ .

Solution.

First find the parameter value: $2\cos t = 0$ , $\sin t = 1$ and $t = \pi/2$ agree at $t = \pi/2$ .
$\mathbf{r}'(t) = \langle -2\sin t, \cos t, 1\rangle$ , so the direction of the tangent line is $\mathbf{r}'(\pi/2) = \langle -2, 0, 1\rangle$ .
The tangent line is the line through $(0,1,\pi/2)$ with that direction: $x = -2s$ , $y = 1$ , $z = \dfrac{\pi}{2} + s$ .
The $y$ -component is constant because $\cos(\pi/2) = 0$ : at the top of its sideways swing the curve momentarily moves only in $x$ and $z$ .

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Example 10.67 (Using the product rules). Let $\mathbf{u}(t) = \langle t, \cos t, \sin t\rangle$ and $\mathbf{v}(t) = \langle e^t, 1, t\rangle$ . Compute $\dfrac{d}{dt}\bigl[\mathbf{u}(t)\cdot\mathbf{v}(t)\bigr]$ at $t = 0$ , and show that $\dfrac{d}{dt}\bigl[\mathbf{r}(t)\times\mathbf{r}'(t)\bigr] = \mathbf{r}(t)\times\mathbf{r}''(t)$ for any twice-differentiable $\mathbf{r}$ .

Solution.

$\mathbf{u}'(t) = \langle 1, -\sin t, \cos t\rangle$ and $\mathbf{v}'(t) = \langle e^t, 0, 1\rangle$ .
By the dot product rule, the derivative is $\mathbf{u}'\cdot\mathbf{v} + \mathbf{u}\cdot\mathbf{v}'$ . At $t = 0$ : $\mathbf{u}(0) = \langle 0,1,0\rangle$ , $\mathbf{u}'(0) = \langle 1,0,1\rangle$ , $\mathbf{v}(0) = \langle 1,1,0\rangle$ , $\mathbf{v}'(0) = \langle 1,0,1\rangle$ .
So the value is $\langle 1,0,1\rangle\cdot\langle 1,1,0\rangle + \langle 0,1,0\rangle\cdot\langle 1,0,1\rangle = 1 + 0 = 1$ .
Check by expanding first: $\mathbf{u}\cdot\mathbf{v} = te^t + \cos t + t\sin t$ , whose derivative $e^t + te^t - \sin t + \sin t + t\cos t$ equals $1$ at $t = 0$ .
For the identity, apply the cross product rule: $\dfrac{d}{dt}[\mathbf{r}\times\mathbf{r}'] = \mathbf{r}'\times\mathbf{r}' + \mathbf{r}\times\mathbf{r}''$ . The first term is the cross product of a vector with itself, which is $\mathbf{0}$ since the angle between them is $0$ . What remains is $\mathbf{r}\times\mathbf{r}''$ .

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Example 10.68 (Integrating a vector function). Evaluate $\displaystyle\int_0^{\pi/2}\langle 2\cos t,\ \sin t,\ 2t\rangle\,dt$ , and find $\mathbf{r}(t)$ given $\mathbf{r}'(t) = \langle 1, 2t, 3t^{2}\rangle$ and $\mathbf{r}(0) = \langle 1, 0, -1\rangle$ .

Solution.

Integrate componentwise: $\left[2\sin t\right]_0^{\pi/2} = 2$ , $\left[-\cos t\right]_0^{\pi/2} = 0 - (-1) = 1$ , $\left[t^{2}\right]_0^{\pi/2} = \dfrac{\pi^{2}}{4}$ . The integral is $\left\langle 2, 1, \dfrac{\pi^{2}}{4}\right\rangle$ .
For the second part, antidifferentiate: $\mathbf{r}(t) = \langle t, t^{2}, t^{3}\rangle + \mathbf{C}$ .
The initial condition gives $\mathbf{C} = \mathbf{r}(0) = \langle 1,0,-1\rangle$ , so $\mathbf{r}(t) = \langle t+1,\ t^{2},\ t^{3}-1\rangle$ .
Check: $\mathbf{r}'(t) = \langle 1,2t,3t^{2}\rangle$ and $\mathbf{r}(0) = \langle 1,0,-1\rangle$ .

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Pitfall. The derivative of $|\mathbf{r}(t)|$ is not $|\mathbf{r}'(t)|$ . The first is the rate at which the distance from the origin changes; the second is the speed. A point moving on a circle centred at the origin has the first equal to $0$ and the second positive. When the length of a vector function is needed, differentiate $\mathbf{r}\cdot\mathbf{r}$ and use the product rule.

Exercises for §10.8 →

10.9Arc length and curvature

Two questions about a curve have nothing to do with how fast it is traversed: how long is it, and how sharply does it bend? Both are answered by first finding a formula in terms of a parametrisation and then checking that the answer does not depend on which one was used.

Theorem 10.69 (Arc length of a space curve). If $\mathbf{r}(t) = \langle f(t),g(t),h(t)\rangle$ is smooth on $[a,b]$ and the curve is traversed exactly once as $t$ runs from $a$ to $b$ , its length is

L = \int_a^b|\mathbf{r}'(t)|\,dt = \int_a^b\sqrt{[f'(t)]^2 + [g'(t)]^2 + [h'(t)]^2}\,dt.

Proof. Partition $[a,b]$ into $n$ pieces at $t_0 < t_1 < \cdots < t_n$ and join the points $\mathbf{r}(t_{i-1})$ , $\mathbf{r}(t_i)$ by straight segments. By the distance formula the $i$ -th segment has length $|\mathbf{r}(t_i) - \mathbf{r}(t_{i-1})|$ . The Mean Value Theorem, applied to each component, gives points $t_i^*, t_i^{**}, t_i^{***}$ in $(t_{i-1},t_i)$ with

|\mathbf{r}(t_i) - \mathbf{r}(t_{i-1})| = \sqrt{[f'(t_i^*)]^2 + [g'(t_i^{**})]^2 + [h'(t_i^{***})]^2}\,\Delta t_i.

Because $f'$ , $g'$ , $h'$ are continuous, the three sample points may be replaced by a single one with an error that vanishes as the mesh shrinks, and the sum of the segment lengths becomes a Riemann sum for $\int_a^b|\mathbf{r}'(t)|\,dt$ . Its limit is the length of the curve.∎

The formula is the three-dimensional version of $\int\sqrt{1+[f'(x)]^2}\,dx$ for a graph: parametrise the graph as $\mathbf{r}(x) = \langle x, f(x)\rangle$ and $|\mathbf{r}'(x)| = \sqrt{1 + [f'(x)]^2}$ . Read as physics, it says distance travelled is the integral of speed, which is why the formula does not care which parametrisation is used: a faster traversal has larger $|\mathbf{r}'|$ over a shorter interval of $t$ , and the substitution rule shows the two effects cancel exactly.

Definition 10.70 (Arc length function and reparametrisation). For a smooth curve starting at $t = a$ , the arc length function is

s(t) = \int_a^t|\mathbf{r}'(u)|\,du,

the distance along the curve from $\mathbf{r}(a)$ to $\mathbf{r}(t)$ . By the Fundamental Theorem of Calculus, $\dfrac{ds}{dt} = |\mathbf{r}'(t)|$ . Solving for $t$ as a function of $s$ and substituting gives the parametrisation by arc length, $\mathbf{r}(t(s))$ , in which the parameter is distance travelled and the speed is identically $1$ .

Parametrisation by arc length is the intrinsic description of a curve: it depends only on the curve and the starting point, not on any choice of speed. That makes it the right setting in which to define curvature, even though in practice one almost always computes with the original parameter and the chain rule.

Definition 10.71 (Curvature). The curvature of a smooth curve is

\kappa = \left|\frac{d\mathbf{T}}{ds}\right|,

the rate at which the unit tangent vector turns per unit of arc length.

The unit tangent $\mathbf{T}$ records direction and nothing else, so $|d\mathbf{T}/ds|$ measures how fast the direction changes as one moves along the curve at unit speed. A straight line has constant $\mathbf{T}$ and curvature $0$ ; a tight bend turns $\mathbf{T}$ quickly and has large curvature. Measuring per unit arc length rather than per unit time is what makes the number a property of the curve alone.

Theorem 10.72 (Curvature from any parametrisation). For a smooth curve $\mathbf{r}(t)$ (with $\mathbf{r}''$ existing, for the second formula),

\kappa(t) = \frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|} = \frac{|\mathbf{r}'(t)\times\mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}.

Proof. By the chain rule, $\dfrac{d\mathbf{T}}{dt} = \dfrac{d\mathbf{T}}{ds}\dfrac{ds}{dt}$ , so $\dfrac{d\mathbf{T}}{ds} = \dfrac{\mathbf{T}'(t)}{ds/dt} = \dfrac{\mathbf{T}'(t)}{|\mathbf{r}'(t)|}$ , and taking lengths gives the first formula.

For the second, write $v = |\mathbf{r}'|$ so that $\mathbf{r}' = v\mathbf{T}$ . Differentiating with the scalar product rule, $\mathbf{r}'' = v'\mathbf{T} + v\mathbf{T}'$ , and therefore

\mathbf{r}'\times\mathbf{r}'' = v\mathbf{T}\times(v'\mathbf{T} + v\mathbf{T}') = vv'(\mathbf{T}\times\mathbf{T}) + v^2(\mathbf{T}\times\mathbf{T}') = v^2(\mathbf{T}\times\mathbf{T}'),

since $\mathbf{T}\times\mathbf{T} = \mathbf{0}$ . Now $|\mathbf{T}| = 1$ , so $\mathbf{T}'$ is perpendicular to $\mathbf{T}$ by the constant-length corollary, and $|\mathbf{T}\times\mathbf{T}'| = |\mathbf{T}||\mathbf{T}'|\sin\tfrac{\pi}{2} = |\mathbf{T}'|$ . Hence $|\mathbf{r}'\times\mathbf{r}''| = v^{2}|\mathbf{T}'|$ , and

\kappa = \frac{|\mathbf{T}'|}{v} = \frac{|\mathbf{r}'\times\mathbf{r}''|/v^2}{v} = \frac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|^3}.

∎

Corollary 10.73 (Curvature of a graph). For the plane curve $y = f(x)$ ,

\kappa(x) = \frac{|f''(x)|}{\bigl(1 + [f'(x)]^2\bigr)^{3/2}}.

Proof. Parametrise by $\mathbf{r}(x) = \langle x, f(x), 0\rangle$ , so $\mathbf{r}' = \langle 1, f', 0\rangle$ and $\mathbf{r}'' = \langle 0, f'', 0\rangle$ . Then $\mathbf{r}'\times\mathbf{r}'' = \langle 0, 0, f''\rangle$ , $|\mathbf{r}'\times\mathbf{r}''| = |f''|$ , and $|\mathbf{r}'|^3 = (1 + [f']^2)^{3/2}$ .∎

Intuition. Drive a road at a steady speed and watch the steering wheel. On a straight the wheel is centred: curvature zero. Entering a bend you turn the wheel, and the tighter the bend the more you turn it: that is $\kappa$ . The reason to measure turning per metre rather than per second is that the road's bend is the same whether you drive it fast or slowly — driving faster only makes you turn the wheel sooner, not more. A roundabout of radius $r$ needs the wheel held at a fixed angle the whole way round, and that angle is what $\kappa = 1/r$ encodes.

Example 10.74 (Curvature of a circle). Show that a circle of radius $a$ has curvature $1/a$ at every point.

Solution.

Parametrise by $\mathbf{r}(t) = \langle a\cos t, a\sin t, 0\rangle$ . Then $\mathbf{r}'(t) = \langle -a\sin t, a\cos t, 0\rangle$ and $|\mathbf{r}'(t)| = a$ .
The unit tangent is $\mathbf{T}(t) = \langle -\sin t, \cos t, 0\rangle$ , with $\mathbf{T}'(t) = \langle -\cos t, -\sin t, 0\rangle$ and $|\mathbf{T}'(t)| = 1$ .
By the first formula, $\kappa = \dfrac{|\mathbf{T}'|}{|\mathbf{r}'|} = \dfrac{1}{a}$ .
Small circles bend sharply and have large curvature; as $a \to \infty$ the circle straightens out and $\kappa \to 0$ , which matches the picture.

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Example 10.75 (Arc length, and reparametrising a helix). Find the length of $\mathbf{r}(t) = \langle t^{2}, 2t, \ln t\rangle$ for $1 \le t \le e$ . Then reparametrise the helix $\mathbf{r}(t) = \langle \cos t, \sin t, t\rangle$ by arc length measured from $(1,0,0)$ .

Solution.

$\mathbf{r}'(t) = \left\langle 2t, 2, \dfrac{1}{t}\right\rangle$ , so $|\mathbf{r}'(t)|^2 = 4t^{2} + 4 + \dfrac{1}{t^{2}} = \left(2t + \dfrac1t\right)^2$ , a perfect square.
$L = \displaystyle\int_1^e\left(2t + \frac1t\right)dt = \Bigl[t^{2} + \ln t\Bigr]_1^e = (e^{2} + 1) - (1 + 0) = e^{2}$ .
For the helix, $\mathbf{r}'(t) = \langle -\sin t, \cos t, 1\rangle$ and $|\mathbf{r}'(t)| = \sqrt{\sin^{2} t + \cos^{2} t + 1} = \sqrt2$ , a constant speed.
The arc length from $t = 0$ is $s(t) = \displaystyle\int_0^t\sqrt2\,du = \sqrt2\,t$ , so $t = s/\sqrt2$ and
$\mathbf{r}(t(s)) = \left\langle \cos\frac{s}{\sqrt2},\ \sin\frac{s}{\sqrt2},\ \frac{s}{\sqrt2}\right\rangle.$
Check: the derivative with respect to $s$ is $\dfrac{1}{\sqrt2}\left\langle -\sin\dfrac{s}{\sqrt2}, \cos\dfrac{s}{\sqrt2}, 1\right\rangle$ , of length $\dfrac{1}{\sqrt2}\cdot\sqrt2 = 1$ , as an arc length parametrisation must be.

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Example 10.76 (Curvature of the twisted cubic and of a parabola). Find the curvature of $\mathbf{r}(t) = \langle t, t^{2}, t^{3}\rangle$ at a general point and at the origin. Then find the curvature of $y = x^{2}$ at $(0,0)$ and at $(1,1)$ .

Solution.

$\mathbf{r}'(t) = \langle 1, 2t, 3t^{2}\rangle$ and $\mathbf{r}''(t) = \langle 0, 2, 6t\rangle$ .
$\mathbf{r}'\times\mathbf{r}'' = \langle 2t(6t) - 3t^{2}(2),\ 3t^{2}(0) - 1(6t),\ 1(2) - 2t(0)\rangle = \langle 6t^{2}, -6t, 2\rangle$ , of length $\sqrt{36t^{4} + 36t^{2} + 4} = 2\sqrt{9t^{4} + 9t^{2} + 1}$ .
$|\mathbf{r}'|^3 = (1 + 4t^{2} + 9t^{4})^{3/2}$ , so
$\kappa(t) = \frac{2\sqrt{9t^{4} + 9t^{2} + 1}}{(1 + 4t^{2} + 9t^{4})^{3/2}}, \qquad \kappa(0) = \frac{2}{1} = 2.$
For the parabola, $f(x) = x^{2}$ , $f' = 2x$ , $f'' = 2$ , so $\kappa(x) = \dfrac{2}{(1 + 4x^{2})^{3/2}}$ . At the origin $\kappa = 2$ ; at $(1,1)$ , $\kappa = \dfrac{2}{5^{3/2}} = \dfrac{2}{5\sqrt5} \approx 0.18$ .
Both curvatures at the origin are $2$ , which is no coincidence: near the origin the twisted cubic is $\langle t, t^{2}, t^{3}\rangle \approx \langle t, t^{2}, 0\rangle$ , the parabola itself, since $t^{3}$ is negligible compared with $t^{2}$ .

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Curvature says how much the curve bends, but not in which direction. That needs two more unit vectors.

Definition 10.77 (Normal and binormal vectors). At a point of a smooth curve where $\kappa \ne 0$ , the principal unit normal vector and the binormal vector are

\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}, \qquad \mathbf{B}(t) = \mathbf{T}(t)\times\mathbf{N}(t).

Together, $\mathbf{T}$ , $\mathbf{N}$ , $\mathbf{B}$ form the TNB frame (or Frenet frame): three mutually perpendicular unit vectors that travel along with the curve.

$\mathbf{N}$ is perpendicular to $\mathbf{T}$ by the constant-length corollary, and it points the way the curve is turning. $\mathbf{B}$ is perpendicular to both and has length $|\mathbf{T}||\mathbf{N}|\sin\tfrac{\pi}{2} = 1$ . The plane spanned by $\mathbf{N}$ and $\mathbf{B}$ at a point is the normal plane, the plane of all lines perpendicular to $\mathbf{T}$ there; its normal vector is $\mathbf{T}$ , or equivalently $\mathbf{r}'$ . The plane spanned by $\mathbf{T}$ and $\mathbf{N}$ is the osculating plane, the plane that fits the curve most closely at that point; its normal vector is $\mathbf{B}$ . For a plane curve the osculating plane is the plane of the curve.

Definition 10.78 (Osculating circle). The osculating circle at a point $P$ of a curve with $\kappa \ne 0$ is the circle in the osculating plane that is tangent to the curve at $P$ , lies on the side toward which $\mathbf{N}$ points, and has radius $\rho = 1/\kappa$ . Its centre is $P + \rho\,\mathbf{N}$ .

It is the circle that best approximates the curve at $P$ : same tangent, same curvature, same side. Engineers use it to design roads and railway tracks, where the curvature must change gradually so that the sideways force on a vehicle does not jump.

A space curve can also twist out of its osculating plane, and the rate at which it does so is its torsion. Since $|\mathbf{B}| = 1$ , $\mathbf{B}'$ is perpendicular to $\mathbf{B}$ ; a short computation shows it is also perpendicular to $\mathbf{T}$ , so it is a multiple of $\mathbf{N}$ , and the torsion $\tau$ is defined by $\dfrac{d\mathbf{B}}{ds} = -\tau\,\mathbf{N}$ . In terms of an arbitrary parameter,

\tau = \frac{(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}'''}{|\mathbf{r}'\times\mathbf{r}''|^2}.

A plane curve has $\tau = 0$ . Curvature and torsion together determine a space curve up to a rigid motion — a theorem whose proof belongs to differential geometry — and the three Frenet formulas $\mathbf{T}' = \kappa\mathbf{N}$ , $\mathbf{N}' = -\kappa\mathbf{T} + \tau\mathbf{B}$ , $\mathbf{B}' = -\tau\mathbf{N}$ (derivatives with respect to $s$ ) describe how the frame rotates as one moves along.

Example 10.79 (The TNB frame, normal plane and osculating plane of a helix). For the helix $\mathbf{r}(t) = \langle \cos t, \sin t, t\rangle$ , find $\mathbf{T}$ , $\mathbf{N}$ , $\mathbf{B}$ , the curvature and the torsion. Then find the normal plane and osculating plane at $(0, 1, \pi/2)$ .

Solution.

$\mathbf{r}'(t) = \langle -\sin t, \cos t, 1\rangle$ with $|\mathbf{r}'| = \sqrt2$ , so $\mathbf{T}(t) = \dfrac{1}{\sqrt2}\langle -\sin t, \cos t, 1\rangle$ .
$\mathbf{T}'(t) = \dfrac{1}{\sqrt2}\langle -\cos t, -\sin t, 0\rangle$ with $|\mathbf{T}'| = \dfrac{1}{\sqrt2}$ , so $\mathbf{N}(t) = \langle -\cos t, -\sin t, 0\rangle$ . The normal points horizontally toward the axis of the cylinder, as it should.
$\mathbf{B} = \mathbf{T}\times\mathbf{N} = \dfrac{1}{\sqrt2}\langle \cos t\cdot 0 - 1\cdot(-\sin t),\ 1\cdot(-\cos t) - (-\sin t)\cdot 0,\ (-\sin t)(-\sin t) - \cos t(-\cos t)\rangle = \dfrac{1}{\sqrt2}\langle \sin t, -\cos t, 1\rangle$ .
Curvature: $\kappa = \dfrac{|\mathbf{T}'|}{|\mathbf{r}'|} = \dfrac{1/\sqrt2}{\sqrt2} = \dfrac12$ . The helix bends less sharply than the unit circle it sits above (which has $\kappa = 1$ ), because the climb straightens it.
Torsion: $\mathbf{r}'' = \langle -\cos t, -\sin t, 0\rangle$ , $\mathbf{r}''' = \langle \sin t, -\cos t, 0\rangle$ , and $\mathbf{r}'\times\mathbf{r}'' = \langle \sin t, -\cos t, 1\rangle$ (the same computation as for $\mathbf{B}$ , without the $1/\sqrt2$ ). Then $(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}''' = \sin^{2} t + \cos^{2} t = 1$ and $|\mathbf{r}'\times\mathbf{r}''|^2 = 2$ , so $\tau = \dfrac12$ . Curvature and torsion are both constant — the helix looks the same everywhere.
The point $(0,1,\pi/2)$ is $t = \pi/2$ , where $\mathbf{r}' = \langle -1, 0, 1\rangle$ and $\mathbf{B} = \dfrac{1}{\sqrt2}\langle 1, 0, 1\rangle$ .
Normal plane, with normal $\mathbf{r}'$ : $-1(x - 0) + 0(y - 1) + 1\left(z - \dfrac{\pi}{2}\right) = 0$ , that is $z - x = \dfrac{\pi}{2}$ .
Osculating plane, with normal $\langle 1,0,1\rangle$ : $x + z = \dfrac{\pi}{2}$ .

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Example 10.80 (The osculating circle of a parabola). Find the osculating circle of $y = x^{2}$ at the origin.

Solution.

From the earlier example, $\kappa(0) = 2$ , so the radius is $\rho = \dfrac{1}{\kappa} = \dfrac12$ .
The tangent at the origin is horizontal and the parabola bends upward, so $\mathbf{N} = \langle 0, 1\rangle$ and the centre is $(0,0) + \tfrac12\langle 0,1\rangle = \left(0, \tfrac12\right)$ .
The osculating circle is $x^{2} + \left(y - \tfrac12\right)^2 = \tfrac14$ .
Near the origin the circle's lower arc is $y = \tfrac12 - \sqrt{\tfrac14 - x^{2}} \approx x^{2}$ for small $x$ , confirming that it hugs the parabola to second order.

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Pitfall. The formula $\kappa = |\mathbf{T}'(t)|$ is wrong unless $t$ happens to be arc length. Curvature is $|d\mathbf{T}/ds|$ , and converting from $t$ to $s$ costs a factor $|\mathbf{r}'(t)|$ in the denominator. Forgetting it makes the "curvature" depend on how fast the curve is traversed, which is exactly what the definition was built to avoid.

Exercises for §10.9 →

10.10Motion in space

Let the parameter be time. Then the vector function is the position of a moving particle, and its derivatives are the kinematic quantities.

Definition 10.81 (Velocity, speed, acceleration). If $\mathbf{r}(t)$ is the position of a particle at time $t$ , then its velocity, speed and acceleration are

\mathbf{v}(t) = \mathbf{r}'(t), \qquad v(t) = |\mathbf{v}(t)| = \frac{ds}{dt}, \qquad \mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t).

Velocity is a vector: it points along the tangent and its length is the speed, so $\mathbf{v} = v\,\mathbf{T}$ . Speed is the rate at which arc length accumulates, which is why the arc length integral is the integral of speed. Acceleration is the rate of change of velocity, and since velocity can change in length or in direction, acceleration has something to say even when the speed is constant.

Position is recovered from acceleration by integrating twice, with the initial velocity and initial position supplying the two constant vectors:

\mathbf{v}(t) = \mathbf{v}(t_0) + \int_{t_0}^t\mathbf{a}(u)\,du, \qquad \mathbf{r}(t) = \mathbf{r}(t_0) + \int_{t_0}^t\mathbf{v}(u)\,du.

Newton's second law, $\mathbf{F} = m\mathbf{a}$ , connects this to physics: given the forces on a particle, its acceleration is known, and two integrations produce its trajectory.

Example 10.82 (Velocity and acceleration from position). A particle has position $\mathbf{r}(t) = \langle t^{2}, e^t, te^t\rangle$ . Find its velocity, speed and acceleration at $t = 0$ .

Solution.

$\mathbf{v}(t) = \langle 2t, e^t, (1+t)e^t\rangle$ and $\mathbf{a}(t) = \langle 2, e^t, (2+t)e^t\rangle$ .
At $t = 0$ : $\mathbf{v}(0) = \langle 0,1,1\rangle$ , speed $|\mathbf{v}(0)| = \sqrt2$ , and $\mathbf{a}(0) = \langle 2,1,2\rangle$ .
The acceleration is not parallel to the velocity, so the particle is both changing speed and turning at that instant.

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Example 10.83 (Position from acceleration). A particle starts at $\mathbf{r}(0) = \langle 0, 1, 0\rangle$ with initial velocity $\mathbf{v}(0) = \langle 1, 0, 0\rangle$ , and its acceleration is $\mathbf{a}(t) = \langle 4t, 6t, 1\rangle$ . Find its position at time $t$ .

Solution.

Integrate the acceleration: $\mathbf{v}(t) = \langle 2t^{2}, 3t^{2}, t\rangle + \mathbf{C}$ , and $\mathbf{v}(0) = \langle 1,0,0\rangle$ gives $\mathbf{C} = \langle 1,0,0\rangle$ . So $\mathbf{v}(t) = \langle 2t^{2} + 1, 3t^{2}, t\rangle$ .
Integrate again: $\mathbf{r}(t) = \left\langle \dfrac{2t^{3}}{3} + t,\ t^{3},\ \dfrac{t^{2}}{2}\right\rangle + \mathbf{D}$ , and $\mathbf{r}(0) = \langle 0,1,0\rangle$ gives $\mathbf{D} = \langle 0,1,0\rangle$ .
$\mathbf{r}(t) = \left\langle \dfrac{2t^{3}}{3} + t,\ t^{3} + 1,\ \dfrac{t^{2}}{2}\right\rangle$ .
Check: $\mathbf{r}''(t) = \langle 4t, 6t, 1\rangle$ , and both initial conditions hold.

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The most famous trajectory in the subject is the one under gravity alone.

Theorem 10.84 (Projectile motion). A projectile launched from the origin with initial speed $v_0$ at angle $\alpha$ above the horizontal, subject only to gravity $g$ (air resistance ignored), has position

\mathbf{r}(t) = \left\langle (v_0\cos\alpha)\,t,\ (v_0\sin\alpha)\,t - \tfrac12 gt^2\right\rangle.

Its horizontal range on level ground is

d = \frac{v_0^2\sin 2\alpha}{g},

which is greatest, for a given $v_0$ , when $\alpha = 45^\circ$ .

Proof. Take $x$ horizontal and $y$ vertical. The only force is gravity, $\mathbf{F} = -mg\,\mathbf{j}$ , so by Newton's second law $\mathbf{a} = -g\,\mathbf{j}$ . Integrating once, $\mathbf{v}(t) = -gt\,\mathbf{j} + \mathbf{v}_0$ with $\mathbf{v}_0 = \langle v_0\cos\alpha, v_0\sin\alpha\rangle$ . Integrating again with $\mathbf{r}(0) = \mathbf{0}$ gives $\mathbf{r}(t) = -\tfrac12 gt^{2}\,\mathbf{j} + t\mathbf{v}_0$ , which is the stated formula in components.

The projectile returns to the ground when $y = 0$ : $(v_0\sin\alpha)t - \tfrac12 gt^{2} = 0$ gives $t = 0$ (launch) or $t = \dfrac{2v_0\sin\alpha}{g}$ . Substituting the latter into $x$ ,

d = (v_0\cos\alpha)\cdot\frac{2v_0\sin\alpha}{g} = \frac{v_0^2\cdot 2\sin\alpha\cos\alpha}{g} = \frac{v_0^2\sin 2\alpha}{g}.

Since $\sin 2\alpha \le 1$ with equality at $2\alpha = 90^\circ$ , the range is maximised at $\alpha = 45^\circ$ .∎

Two further facts drop out of the same computation. The time to reach maximum height is where the vertical velocity $v_0\sin\alpha - gt$ vanishes, $t = \dfrac{v_0\sin\alpha}{g}$ , half the flight time; and the maximum height is $y$ at that instant, $\dfrac{v_0^2\sin^{2}\alpha}{2g}$ . Eliminating $t$ between the components shows the path is a parabola.

Example 10.85 (A projectile). A projectile is fired with initial speed $100$ m/s at an angle of $30^\circ$ above the horizontal. Taking $g = 9.8$ m/s $^2$ , find its range, its maximum height and its speed on impact. At what angle should it be fired to land $400$ m away?

Solution.

Range: $d = \dfrac{100^2\sin 60^\circ}{9.8} = \dfrac{10000\cdot\sqrt3/2}{9.8} \approx 883.7$ m.
Maximum height: $\dfrac{100^2\sin^{2} 30^\circ}{2\cdot 9.8} = \dfrac{10000\cdot\frac14}{19.6} \approx 127.6$ m.
Impact speed: by symmetry of the parabola the projectile lands with the same speed it left with, $100$ m/s. Directly: at $t = \dfrac{2\cdot 100\sin 30^\circ}{9.8} = \dfrac{100}{9.8}$ the velocity is $\langle 100\cos 30^\circ,\ 50 - 100\rangle = \langle 50\sqrt3, -50\rangle$ , of length $\sqrt{7500 + 2500} = 100$ .
For a range of $400$ m: $\sin 2\alpha = \dfrac{400\cdot 9.8}{10000} = 0.392$ , so $2\alpha \approx 23.1^\circ$ or $156.9^\circ$ , and $\alpha \approx 11.5^\circ$ or $78.5^\circ$ . Two angles, symmetric about $45^\circ$ , give the same range: a low flat shot and a high lob.

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Intuition. In a car, two things push you around: braking or accelerating pushes you back and forth in your seat, and cornering pushes you sideways. Those are the two components of acceleration. The first, along $\mathbf{T}$ , is the rate of change of the speedometer reading. The second, along $\mathbf{N}$ , depends on how sharp the bend is and how fast you take it — twice the speed means four times the sideways push, because it is $\kappa v^{2}$ . A driver who never touches the brake on a winding road still accelerates the whole time.

Theorem 10.86 (Tangential and normal components of acceleration). For a particle on a smooth path with speed $v = |\mathbf{v}|$ and path curvature $\kappa$ ,

\mathbf{a} = a_T\,\mathbf{T} + a_N\,\mathbf{N}, \qquad a_T = v' = \frac{\mathbf{r}'\cdot\mathbf{r}''}{|\mathbf{r}'|}, \qquad a_N = \kappa v^2 = \frac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|}.

In particular the acceleration always lies in the osculating plane; it has no component along $\mathbf{B}$ .

Proof. Since $\mathbf{v} = v\mathbf{T}$ , the scalar product rule gives $\mathbf{a} = v'\mathbf{T} + v\mathbf{T}'$ . By the curvature formula, $|\mathbf{T}'| = \kappa|\mathbf{r}'| = \kappa v$ , and by the definition of $\mathbf{N}$ , $\mathbf{T}' = |\mathbf{T}'|\mathbf{N} = \kappa v\,\mathbf{N}$ . Substituting,

\mathbf{a} = v'\,\mathbf{T} + \kappa v^2\,\mathbf{N}.

For the computable forms, dot $\mathbf{a}$ with $\mathbf{v} = v\mathbf{T}$ : since $\mathbf{T}\cdot\mathbf{T} = 1$ and $\mathbf{T}\cdot\mathbf{N} = 0$ , $\mathbf{v}\cdot\mathbf{a} = vv'$ , so $a_T = v' = \dfrac{\mathbf{r}'\cdot\mathbf{r}''}{|\mathbf{r}'|}$ . And from the proof of the curvature formula, $|\mathbf{r}'\times\mathbf{r}''| = v^{2}|\mathbf{T}'| = \kappa v^{3}$ , so $a_N = \kappa v^{2} = \dfrac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|}$ .∎

Since $\mathbf{T}$ and $\mathbf{N}$ are perpendicular unit vectors, $|\mathbf{a}|^2 = a_T^{2} + a_N^{2}$ , which is a useful check: compute $a_T$ , and get $a_N$ for free as $\sqrt{|\mathbf{a}|^2 - a_T^{2}}$ . Uniform circular motion, $\mathbf{r}(t) = \langle R\cos\omega t, R\sin\omega t\rangle$ , has $v = R\omega$ constant, so $a_T = 0$ and the whole acceleration is $a_N = \kappa v^{2} = \dfrac{(R\omega)^2}{R} = R\omega^{2}$ , pointing along $\mathbf{N}$ toward the centre: the familiar centripetal acceleration.

Example 10.87 (Splitting the acceleration). For $\mathbf{r}(t) = \langle t^{2}, t^{2}, t^{3}\rangle$ , $t > 0$ , find the tangential and normal components of acceleration, and evaluate them at $t = 1$ .

Solution.

$\mathbf{r}'(t) = \langle 2t, 2t, 3t^{2}\rangle$ and $\mathbf{r}''(t) = \langle 2, 2, 6t\rangle$ . Speed: $|\mathbf{r}'| = \sqrt{8t^{2} + 9t^{4}} = t\sqrt{8 + 9t^{2}}$ .
$\mathbf{r}'\cdot\mathbf{r}'' = 4t + 4t + 18t^{3} = 8t + 18t^{3}$ , so $a_T = \dfrac{8t + 18t^{3}}{t\sqrt{8+9t^{2}}} = \dfrac{8 + 18t^{2}}{\sqrt{8 + 9t^{2}}}$ .
$\mathbf{r}'\times\mathbf{r}'' = \langle 2t(6t) - 3t^{2}(2),\ 3t^{2}(2) - 2t(6t),\ 2t(2) - 2t(2)\rangle = \langle 6t^{2}, -6t^{2}, 0\rangle$ , of length $6\sqrt2\,t^{2}$ . So $a_N = \dfrac{6\sqrt2\,t^{2}}{t\sqrt{8+9t^{2}}} = \dfrac{6\sqrt2\,t}{\sqrt{8 + 9t^{2}}}$ .
At $t = 1$ : $a_T = \dfrac{26}{\sqrt{17}}$ and $a_N = \dfrac{6\sqrt2}{\sqrt{17}}$ .
Check: $a_T^{2} + a_N^{2} = \dfrac{676 + 72}{17} = \dfrac{748}{17} = 44$ , and $|\mathbf{r}''(1)|^2 = 4 + 4 + 36 = 44$ .

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The same machinery, applied to a planet under the sun's gravity, produces Kepler's laws. Write $\mathbf{r}$ for the planet's position relative to the sun; the inverse-square law gives $\mathbf{a} = -\dfrac{GM}{|\mathbf{r}|^3}\mathbf{r}$ , which is parallel to $\mathbf{r}$ . Then by the cross product rule

\frac{d}{dt}(\mathbf{r}\times\mathbf{v}) = \mathbf{v}\times\mathbf{v} + \mathbf{r}\times\mathbf{a} = \mathbf{0} + \mathbf{0},

so $\mathbf{h} = \mathbf{r}\times\mathbf{v}$ is a constant vector. The position is always perpendicular to $\mathbf{h}$ , so the orbit lies in a plane; and since $|\mathbf{r}\times\mathbf{v}|$ is twice the rate at which the radius vector sweeps out area, that rate is constant — Kepler's second law, equal areas in equal times. A longer calculation with the same tools shows the orbit is an ellipse with the sun at a focus (the first law) and relates the period to the semi-major axis (the third). Newton's derivation of these three laws from one force law was the founding achievement of the calculus you have just learned.

Pitfall. The tangential component is the derivative of the speed, $a_T = \dfrac{d}{dt}|\mathbf{v}|$ , not the magnitude of the acceleration. A car rounding a bend at a steady $60$ km/h has $a_T = 0$ and $|\mathbf{a}| = \kappa v^{2} > 0$ . Conversely, $|\mathbf{a}| = 0$ requires both components to vanish: straight-line motion at constant speed.

Summary. In $\mathbb{R}^3$ the two products carry the geometry: $\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$ is a scalar, zero exactly when the nonzero vectors are orthogonal, and gives $\operatorname{proj}_{\mathbf{b}}\mathbf{a} = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\mathbf{b}$ ; $\mathbf{a}\times\mathbf{b}$ is a vector perpendicular to both with $|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$ , the area of the parallelogram they span, vanishing exactly when they are parallel, and $|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})|$ is the volume of the parallelepiped. A line is $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}$ ; a plane is $\mathbf{n}\cdot(\mathbf{r}-\mathbf{r}_0) = 0$ , that is $ax+by+cz+d = 0$ with normal $\langle a,b,c\rangle$ , and a point stands off it by $D = |ax_1+by_1+cz_1+d|/\sqrt{a^{2}+b^{2}+c^{2}}$ . Vector functions are differentiated and integrated componentwise, $\mathbf{T} = \mathbf{r}'/|\mathbf{r}'|$ , and constant $|\mathbf{r}|$ forces $\mathbf{r}\cdot\mathbf{r}' = 0$ . For a smooth curve traversed once,

L = \int_a^b |\mathbf{r}'(t)|\,dt, \qquad s(t) = \int_a^t|\mathbf{r}'(u)|\,du, \qquad \kappa = \left|\frac{d\mathbf{T}}{ds}\right| = \frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|} = \frac{|\mathbf{r}'(t)\times\mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3},

with $\kappa = |f''|/(1+[f']^2)^{3/2}$ for $y = f(x)$ . Where $\kappa \neq 0$ , $\mathbf{N} = \mathbf{T}'/|\mathbf{T}'|$ and $\mathbf{B} = \mathbf{T}\times\mathbf{N}$ complete the frame, and acceleration stays in the osculating plane: $\mathbf{a} = v'\,\mathbf{T} + \kappa v^{2}\,\mathbf{N}$ .

Exercises for §10.10 →

Common Mistakes

Confusing a point with a vector. $(1,2,3)$ is a location; $\langle 1,2,3\rangle$ is a displacement. The vector from $A$ to $B$ is $B - A$ , head minus tail, never tail minus head.
Treating magnitude as additive: $|\mathbf{a}+\mathbf{b}| = |\mathbf{a}| + |\mathbf{b}|$ only when the vectors point the same way.
Forgetting the minus sign on the $\mathbf{j}$ component when expanding the cross product determinant. Always check the result is orthogonal to both factors — it takes ten seconds.
Writing $\mathbf{b}\times\mathbf{a}$ for $\mathbf{a}\times\mathbf{b}$ , or dropping brackets in a triple cross product. The cross product is neither commutative nor associative.
Using $\mathbf{a}\cdot\mathbf{b} = 0$ to conclude one of the vectors is zero. It means they are perpendicular.
Reading the coefficients of $ax + by + cz = d$ as a direction in the plane. They are the normal to the plane. For a line, the coefficients of $t$ are the direction.
Using $\cos\theta = \mathbf{a}\cdot\mathbf{b}/(|\mathbf{a}||\mathbf{b}|)$ to find the angle between two planes and forgetting that the acute angle is usually wanted; if the cosine comes out negative, take its absolute value.
Applying the point-to-plane distance formula with the plane not written as $ax + by + cz + d = 0$ ; a right-hand side of $5$ contributes $d = -5$ , not $+5$ .
Identifying a quadric from the equation without first completing the square and dividing through, so that $z = x^{2}$ is called a parabola (it is a cylinder) or $x^{2} + y^{2} - z^{2} = 0$ a hyperboloid (it is a cone).
Treating a vector function and its curve as the same thing. The curve is a set of points; the function also encodes speed and direction of travel, and derived quantities depend on it.
Computing $\dfrac{d}{dt}|\mathbf{r}(t)|$ as $|\mathbf{r}'(t)|$ . The first is how fast the distance from the origin changes; the second is the speed.
Using $\kappa = |\mathbf{T}'(t)|$ without dividing by $|\mathbf{r}'(t)|$ . Curvature is turning per unit arc length, not per unit time.
Choosing the wrong normal for a plane: the normal plane at a point has normal $\mathbf{T}$ (or $\mathbf{r}'$ ); the osculating plane has normal $\mathbf{B}$ .
Claiming the acceleration is zero because the speed is constant. Constant speed kills $a_T$ only; on any bend $a_N = \kappa v^{2} \ne 0$ .
Forgetting that the projectile range formula $v_0^2\sin 2\alpha/g$ assumes launch and landing at the same height. From a cliff or onto a hill, go back to the parametric equations and solve for the landing time directly.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

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