Print this page and work through it on paper. Revealed answers are never printed, so you can check one question and still print a clean sheet.

1. 1.Computation · 6 min

   Work with squared norms for as long as possible and take a square root only at the end.

   1. (a)Let $P=(3,-2,1)$ and $Q=(1,2,5)$. Find $\overrightarrow{PQ}$ and $\overrightarrow{QP}$, and the distance from $P$ to $Q$.
   2. (b)Find every value of $t$ for which the point $(2,t,-1)$ is at distance $7$ from $(-1,3,5)$.
   3. (c)Show that no point of the form $(t,3,2)$ is at distance $2$ from $(-1,3,5)$, whatever $t$ is, and say in one sentence what this means geometrically.
   Show the solution to question 1

   (a) Head minus tail. Subtracting the coordinates of the tail $P$ from those of the head $Q$,

   $$\overrightarrow{PQ}=⟨1-3,2-(-2),5-1⟩=⟨-2,4,4⟩,\overrightarrow{QP}=-\overrightarrow{PQ}=⟨2,-4,-4⟩.$$

   The distance is the norm of the displacement. Squares first: $\Vert \overrightarrow{PQ}{\Vert }^2=4+16+16=36$, so the distance is $\sqrt{36}=6$. The reversed vector has the same norm, as it must — distance does not depend on which point you start from.

   (b) Square the distance, not the unknown. The displacement from $(-1,3,5)$ to $(2,t,-1)$ is $⟨3,t-3,-6⟩$, so the squared distance is

   $$9+(t-3{)}^2+36=(t-3{)}^2+45.$$

   Setting this equal to $7^2=49$ gives $(t-3{)}^2=4$, so $t-3=\pm 2$ and $t\in \{1,5\}$.

   Sanity check: at $t=1$ the displacement is $⟨3,-2,-6⟩$ with $9+4+36=49$; at $t=5$ it is $⟨3,2,-6⟩$, same squares. Two answers were expected, because the points $(2,t,-1)$ trace a line and a sphere of radius $7$ about $(-1,3,5)$ reaches that line ($7>\sqrt{45}$), so it must cut it twice.

   (c) A lower bound from one coordinate. The displacement from $(-1,3,5)$ to $(t,3,2)$ is $⟨t+1,0,-3⟩$, so the squared distance is

   $$(t+1{)}^2+0+9\ge 9\text{for every real }t,$$

   because a square is never negative. Hence the distance is at least $3$, and in particular never $2$. Geometrically: the points $(t,3,2)$ form a line parallel to the $x$-axis at height $z=2$, while the fixed point sits at height $z=5$; the closest the line ever comes is $3$ (at $t=-1$, directly below the point), so the sphere of radius $2$ about $(-1,3,5)$ misses the line entirely.

2. 2.Sketch · 7 min

   Let $\mathbf{u}=⟨3,1⟩$ and $\mathbf{v}=⟨-1,2⟩$. Work on squared paper with the same scale on both axes. Every arrow you draw must be labelled with its components.

   1. (a)Draw $\mathbf{u}$ and $\mathbf{v}$ with their tails at the origin and construct $\mathbf{u}+\mathbf{v}$ by the parallelogram rule. Then draw $\mathbf{u}-\mathbf{v}$ as the other diagonal of the same parallelogram, making clear which way its arrowhead points. Compute both vectors to check the drawing, and use their norms to decide which diagonal must be the longer one.
   2. (b)Draw the arrow from $(2,-1)$ to $(5,0)$. Which vector already in your picture is it, and why is it the same vector even though it starts somewhere else?
   3. (c)Compute $\frac{1}{2}(\mathbf{u}+\mathbf{v})$ and the midpoint of the segment joining the heads of $\mathbf{u}$ and $\mathbf{v}$. Mark the point on your sketch and state which fact about parallelograms it illustrates.
   Show the solution to question 2

   (a) The parallelogram and its diagonals. Component-wise, $\mathbf{u}+\mathbf{v}=⟨3-1,1+2⟩=⟨2,3⟩$ and $\mathbf{u}-\mathbf{v}=⟨3+1,1-2⟩=⟨4,-1⟩$.

   The drawing has the four vertices $O=(0,0)$, $U=(3,1)$ (head of $\mathbf{u}$), $V=(-1,2)$ (head of $\mathbf{v}$) and $(2,3)$, the fourth corner, which is where a copy of $\mathbf{v}$ lands when its tail is slid to $U$ (and equally where a copy of $\mathbf{u}$ lands when slid to $V$). The diagonal from $O$ to $(2,3)$ is $\mathbf{u}+\mathbf{v}$: arrowhead at $(2,3)$.

   The other diagonal joins $V=(-1,2)$ and $U=(3,1)$. By head minus tail, $\overrightarrow{VU}=⟨3-(-1),1-2⟩=⟨4,-1⟩=\mathbf{u}-\mathbf{v}$, so $\mathbf{u}-\mathbf{v}$ runs from the head of $\mathbf{v}$ to the head of $\mathbf{u}$: its arrowhead is at $(3,1)$, pointing right and slightly down. (Drawn the other way it would be $\mathbf{v}-\mathbf{u}$.)

   Lengths: $\Vert \mathbf{u}+\mathbf{v}{\Vert }^2=4+9=13$ and $\Vert \mathbf{u}-\mathbf{v}{\Vert }^2=16+1=17$, so the $\mathbf{u}-\mathbf{v}$ diagonal is the longer one. On the paper the segment from $(-1,2)$ to $(3,1)$ should visibly exceed the segment from $(0,0)$ to $(2,3)$; if it does not, an arrowhead has been misplaced. (For reference, $\Vert \mathbf{u}\Vert =\sqrt{10}$ and $\Vert \mathbf{v}\Vert =\sqrt{5}$.)

   (b) A translate is the same vector. The arrow from $(2,-1)$ to $(5,0)$ has components $⟨5-2,0-(-1)⟩=⟨3,1⟩$, which is $\mathbf{u}$. A vector is a displacement — three across, one up — and says nothing about where you start. The arrow from $(2,-1)$ to $(5,0)$ and the arrow from $(0,0)$ to $(3,1)$ are translates of each other, so they are one vector drawn in two places; on the sketch they are parallel, equal in length and point the same way.

   (c) The diagonals bisect each other. $\frac{1}{2}(\mathbf{u}+\mathbf{v})=\frac{1}{2}⟨2,3⟩=⟨1,\frac{3}{2}⟩$, the midpoint of the diagonal from $O$ to $(2,3)$.

   The midpoint of the segment from $V=(-1,2)$ to $U=(3,1)$ is $\left(\frac{-1+3}{2},\frac{2+1}{2}\right)=\left(1,\frac{3}{2}\right)$ — the same point. Mark it at $(1,\frac{3}{2})$, where the two diagonals cross. It illustrates that the diagonals of a parallelogram bisect each other. Algebraically: walking from $V$ halfway along $\mathbf{u}-\mathbf{v}$ gives $\mathbf{v}+\frac{1}{2}(\mathbf{u}-\mathbf{v})=\frac{1}{2}(\mathbf{u}+\mathbf{v})$, using the distributive axioms, so the midpoint of one diagonal is the midpoint of the other.

3. 3.Computation · 7 min

   Let $\mathbf{u}=⟨2,0,3⟩$ and $\mathbf{v}=⟨1,2,1⟩$.

   1. (a)Compute $\mathbf{w}=3\mathbf{u}-2\mathbf{v}$ and the unit vector in the direction of $\mathbf{w}$. State which theorem guarantees your answer has norm $1$.
   2. (b)Decide whether each of $⟨1,6,0⟩$ and $⟨1,8,2⟩$ lies in $\mathrm{span}\{\mathbf{u},\mathbf{v}\}$. Where it does, write it as a combination of $\mathbf{u}$ and $\mathbf{v}$; where it does not, say exactly which equation fails.
   3. (c)For which values of $k$ is the list $\mathbf{u},\mathbf{v},⟨3,k,4⟩$ linearly dependent? Prove both directions: that your value of $k$ gives dependence, and that every other value gives independence.
   Show the solution to question 3

   (a) Scale, then subtract. $3\mathbf{u}=⟨6,0,9⟩$ and $2\mathbf{v}=⟨2,4,2⟩$, so

   $$\mathbf{w}=⟨6-2,0-4,9-2⟩=⟨4,-4,7⟩.$$

   Then $\Vert \mathbf{w}{\Vert }^2=16+16+49=81$, so $\Vert \mathbf{w}\Vert =9$ and

   $$\hat{\mathbf{w}}=\frac{1}{9}\mathbf{w}=⟨\frac{4}{9},-\frac{4}{9},\frac{7}{9}⟩.$$

   By homogeneity of the norm, $\Vert \hat{\mathbf{w}}\Vert =\left|\frac{1}{9}\right|\Vert \mathbf{w}\Vert =\frac{1}{9}\cdot 9=1$; directly, $\frac{16+16+49}{81}=1$. Normalising was legitimate because $\mathbf{w}\ne 0$.

   (b) Solve component by component. A general combination is

   $$a\mathbf{u}+b\mathbf{v}=⟨2a+b,2b,3a+b⟩.$$

   For $⟨1,6,0⟩$: the middle component forces $b=3$; then $2a+3=1$ forces $a=-1$. Neither was a choice, so the third component is a consistency check: $3(-1)+3=0$, which matches. Hence $⟨1,6,0⟩=-\mathbf{u}+3\mathbf{v}$, and it is in the span. (Recompute: $⟨-2,0,-3⟩+⟨3,6,3⟩=⟨1,6,0⟩$.)

   For $⟨1,8,2⟩$: the middle component forces $b=4$, then $2a+4=1$ forces $a=-\frac{3}{2}$. The third component would need $3a+b=2$, but $3\left(-\frac{3}{2}\right)+4=-\frac{1}{2}\ne 2$. That equation fails, so no $a,b$ exist and $⟨1,8,2⟩$ is not in the span — it lies off the plane through the origin spanned by $\mathbf{u}$ and $\mathbf{v}$.

   (c) Reduce to a span question. First, $\mathbf{u}$ and $\mathbf{v}$ are independent: $c\mathbf{u}=\mathbf{v}$ would need $2c=1$ and $0\cdot c=2$, impossible, so neither is a multiple of the other. Write ${\mathbf{w}}_k=⟨3,k,4⟩$.

   Claim: the list is dependent if and only if ${\mathbf{w}}_k\in \mathrm{span}\{\mathbf{u},\mathbf{v}\}$. If ${\mathbf{w}}_k=a\mathbf{u}+b\mathbf{v}$, then $a\mathbf{u}+b\mathbf{v}-{\mathbf{w}}_k=0$ is a nontrivial relation (its last coefficient is $-1$), so the list is dependent — this is the proposition that dependence means redundancy. Conversely suppose $c_1\mathbf{u}+c_2\mathbf{v}+c_3{\mathbf{w}}_k=0$ with not all $c_i$ zero. If $c_3=0$ the relation reads $c_1\mathbf{u}+c_2\mathbf{v}=0$, and independence of $\mathbf{u},\mathbf{v}$ forces $c_1=c_2=0$ too, contradicting nontriviality. So $c_3\ne 0$ and ${\mathbf{w}}_k=-\frac{c_1}{c_3}\mathbf{u}-\frac{c_2}{c_3}\mathbf{v}$ is in the span.

   Now solve $⟨2a+b,2b,3a+b⟩=⟨3,k,4⟩$. The first and third components give $2a+b=3$ and $3a+b=4$; subtracting, $a=1$, and then $b=1$. The middle component then demands $k=2b=2$.

   So for $k=2$ the list is dependent, and explicitly $\mathbf{u}+\mathbf{v}=⟨3,2,4⟩$, i.e. $\mathbf{u}+\mathbf{v}-{\mathbf{w}}_2=0$. For every $k\ne 2$ the first and third components still force $a=b=1$, and the middle component then reads $k=2$, which is false; so ${\mathbf{w}}_k$ is not in the span and, by the claim, the list is independent. The only value is $k=2$.

4. 4.Computation · 7 min

   Give exact values throughout: fractions, roots and angles in degrees or as multiples of $\pi$.

   1. (a)Find the angle between $⟨1,1,1,1⟩$ and $⟨1,1,0,0⟩$ in ${\mathbb{R}}^4$. Before computing, say what the sign of the dot product must be for the angle to come out acute, and afterwards confirm that your cosine obeys Cauchy–Schwarz.
   2. (b)Find every $k$ for which $⟨k,1,2⟩$ and $⟨k,-3,k⟩$ are orthogonal, and verify each answer by a direct dot product.
   3. (c)Suppose $\Vert \mathbf{u}\Vert =2$, $\Vert \mathbf{v}\Vert =5$ and the angle between $\mathbf{u}$ and $\mathbf{v}$ is $120^{\circ }$. Without introducing components, find $\mathbf{u}\cdot \mathbf{v}$, $\Vert \mathbf{u}+\mathbf{v}\Vert$ and $\Vert 2\mathbf{u}-\mathbf{v}\Vert$, and check the second of these against the triangle inequality.
   Show the solution to question 4

   (a) Definition of the angle. The dot product is $1+1+0+0=2$. The norms are $\sqrt{1+1+1+1}=2$ and $\sqrt{1+1}=\sqrt{2}$. Since $\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert >0$, the sign of the cosine is the sign of the dot product, so a positive dot product means an acute angle — and $2>0$, so the angle will be acute. Then

   $$\cos \theta =\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2},\theta =\frac{\pi }{4}=45^{\circ }.$$

   Cauchy–Schwarz demands $|\mathbf{u}\cdot \mathbf{v}|\le \Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert$, i.e. $2\le 2\sqrt{2}$, which holds; equivalently the cosine $\frac{\sqrt{2}}{2}$ lies in $[-1,1]$. There is no picture of ${\mathbb{R}}^4$, but the two vectors span a plane inside it and within that plane the angle is an ordinary $45^{\circ }$.

   (b) One scalar equation, quadratic in $k$.

   $$⟨k,1,2⟩\cdot ⟨k,-3,k⟩=k^2-3+2k=k^2+2k-3=(k+3)(k-1).$$

   This vanishes exactly for $k=1$ and $k=-3$. Two solutions arise because $k$ appears in both vectors, so the condition is quadratic rather than linear.

   Verification: for $k=1$, $⟨1,1,2⟩\cdot ⟨1,-3,1⟩=1-3+2=0$; for $k=-3$, $⟨-3,1,2⟩\cdot ⟨-3,-3,-3⟩=9-3-6=0$.

   (c) Identities only. By the geometric form of the dot product,

   $$\mathbf{u}\cdot \mathbf{v}=\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \cos 120^{\circ }=2\cdot 5\cdot \left(-\frac{1}{2}\right)=-5.$$

   Expanding a squared sum with bilinearity,

   $$\Vert \mathbf{u}+\mathbf{v}{\Vert }^2=\Vert \mathbf{u}{\Vert }^2+2\mathbf{u}\cdot \mathbf{v}+\Vert \mathbf{v}{\Vert }^2=4-10+25=19,\Vert \mathbf{u}+\mathbf{v}\Vert =\sqrt{19}.$$

   For the last one, expand $(2\mathbf{u}-\mathbf{v})\cdot (2\mathbf{u}-\mathbf{v})$ using symmetry, distributivity and homogeneity:

   $$\Vert 2\mathbf{u}-\mathbf{v}{\Vert }^2=4\mathbf{u}\cdot \mathbf{u}-4\mathbf{u}\cdot \mathbf{v}+\mathbf{v}\cdot \mathbf{v}=16+20+25=61,\Vert 2\mathbf{u}-\mathbf{v}\Vert =\sqrt{61}.$$

   Triangle inequality: $\Vert \mathbf{u}+\mathbf{v}\Vert \le \Vert \mathbf{u}\Vert +\Vert \mathbf{v}\Vert =7$, and $\sqrt{19}<\sqrt{49}=7$, so it holds. It also exceeds $|5-2|=3=\sqrt{9}$, as the reverse triangle inequality requires. Notice the negative dot product made the sum shorter than $\sqrt{29}$, the value a right angle would give: the vectors broadly disagree, so they partly cancel.

5. 5.Proof · 9 min

   Prove each statement for all $\mathbf{u},\mathbf{v}\in {\mathbb{R}}^n$. Quote by name every property of the dot product or of the norm that you use; do not introduce components.

   1. (a)The reverse triangle inequality: $|\Vert \mathbf{u}\Vert -\Vert \mathbf{v}\Vert |\le \Vert \mathbf{u}-\mathbf{v}\Vert$.
   2. (b)$\Vert \mathbf{u}\Vert =\Vert \mathbf{v}\Vert$ if and only if $(\mathbf{u}+\mathbf{v})\cdot (\mathbf{u}-\mathbf{v})=0$. Then say what this tells you about the diagonals of a rhombus.
   3. (c)If $\mathbf{u}\cdot \mathbf{w}=\mathbf{v}\cdot \mathbf{w}$ for every $\mathbf{w}\in {\mathbb{R}}^n$, then $\mathbf{u}=\mathbf{v}$. Explain, with an example, why the hypothesis cannot be weakened to a single $\mathbf{w}$.
   Show the solution to question 5

   (a) Two applications of the triangle inequality. Write $\mathbf{u}=(\mathbf{u}-\mathbf{v})+\mathbf{v}$. By the triangle inequality,

   $$\Vert \mathbf{u}\Vert \le \Vert \mathbf{u}-\mathbf{v}\Vert +\Vert \mathbf{v}\Vert ,\text{so}\Vert \mathbf{u}\Vert -\Vert \mathbf{v}\Vert \le \Vert \mathbf{u}-\mathbf{v}\Vert .$$

   Exchanging the roles of $\mathbf{u}$ and $\mathbf{v}$ gives $\Vert \mathbf{v}\Vert -\Vert \mathbf{u}\Vert \le \Vert \mathbf{v}-\mathbf{u}\Vert$, and $\Vert \mathbf{v}-\mathbf{u}\Vert =\Vert (-1)(\mathbf{u}-\mathbf{v})\Vert =|-1|\Vert \mathbf{u}-\mathbf{v}\Vert =\Vert \mathbf{u}-\mathbf{v}\Vert$ by absolute homogeneity. So both $\Vert \mathbf{u}\Vert -\Vert \mathbf{v}\Vert$ and its negative are at most $\Vert \mathbf{u}-\mathbf{v}\Vert$, which is exactly the statement $|\Vert \mathbf{u}\Vert -\Vert \mathbf{v}\Vert |\le \Vert \mathbf{u}-\mathbf{v}\Vert$. $■$

   (b) Expand the product. By distributivity and symmetry of the dot product,

   $$(\mathbf{u}+\mathbf{v})\cdot (\mathbf{u}-\mathbf{v})=\mathbf{u}\cdot \mathbf{u}-\mathbf{u}\cdot \mathbf{v}+\mathbf{v}\cdot \mathbf{u}-\mathbf{v}\cdot \mathbf{v}=\Vert \mathbf{u}{\Vert }^2-\Vert \mathbf{v}{\Vert }^2,$$

   using positive definiteness in the form $\mathbf{x}\cdot \mathbf{x}=\Vert \mathbf{x}{\Vert }^2$ and the cancellation of the two cross terms. Hence the product is $0$ exactly when $\Vert \mathbf{u}{\Vert }^2=\Vert \mathbf{v}{\Vert }^2$, and since norms are non-negative that is exactly when $\Vert \mathbf{u}\Vert =\Vert \mathbf{v}\Vert$. $■$

   In the parallelogram with sides $\mathbf{u},\mathbf{v}$ the diagonals are $\mathbf{u}+\mathbf{v}$ and $\mathbf{u}-\mathbf{v}$. The statement says: the diagonals are orthogonal if and only if the sides have equal length — a parallelogram has perpendicular diagonals precisely when it is a rhombus. Both directions are proved at once, in every dimension.

   (c) Choose the witness $\mathbf{w}=\mathbf{u}-\mathbf{v}$. The hypothesis gives $\mathbf{u}\cdot \mathbf{w}-\mathbf{v}\cdot \mathbf{w}=0$, i.e. $(\mathbf{u}-\mathbf{v})\cdot \mathbf{w}=0$ by distributivity, for every $\mathbf{w}$. Since it holds for every $\mathbf{w}$, it holds for $\mathbf{w}=\mathbf{u}-\mathbf{v}$:

   $$(\mathbf{u}-\mathbf{v})\cdot (\mathbf{u}-\mathbf{v})=\Vert \mathbf{u}-\mathbf{v}{\Vert }^2=0.$$

   By positive definiteness the only vector of norm $0$ is $0$, so $\mathbf{u}-\mathbf{v}=0$ and $\mathbf{u}=\mathbf{v}$. $■$

   A single $\mathbf{w}$ does not suffice, because $\mathbf{u}\cdot \mathbf{w}=\mathbf{v}\cdot \mathbf{w}$ only says that $\mathbf{u}-\mathbf{v}$ is orthogonal to $\mathbf{w}$, and there are plenty of nonzero vectors orthogonal to a given one. Example: $\mathbf{u}=⟨1,0⟩$, $\mathbf{v}=⟨0,1⟩$, $\mathbf{w}=⟨1,1⟩$ give $\mathbf{u}\cdot \mathbf{w}=1=\mathbf{v}\cdot \mathbf{w}$ with $\mathbf{u}\ne \mathbf{v}$. You cannot cancel a common factor from a dot product equation; what rescues the argument in the theorem is the freedom to pick $\mathbf{w}$ after seeing $\mathbf{u}-\mathbf{v}$.

6. 6.Critique · 10 min

   Two arguments follow. Each contains a genuine error. For each: identify the exact sentence that fails, explain what has been misused, and then give a correct treatment.

   1. (a)“Let $\mathbf{u}=⟨1,2,2⟩$ and $\mathbf{v}=⟨-2,-4,-4⟩$. Then $\mathbf{u}\cdot \mathbf{v}=-2-8-8=-18$, while $\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert =3\cdot 6=18$. So $|\mathbf{u}\cdot \mathbf{v}|=\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert$: this is the equality case of Cauchy–Schwarz. The triangle inequality is proved from Cauchy–Schwarz, so it is an equality as well: $\Vert \mathbf{u}+\mathbf{v}\Vert =\Vert \mathbf{u}\Vert +\Vert \mathbf{v}\Vert =9$.”
   2. (b)“Cauchy–Schwarz is trivial. By the geometric form of the dot product, $\mathbf{u}\cdot \mathbf{v}=\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \cos \theta$, and $|\cos \theta |\le 1$ for every angle $\theta$. Therefore $|\mathbf{u}\cdot \mathbf{v}|\le \Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert$ in every ${\mathbb{R}}^n$.” Say for which $n$ this reasoning is acceptable and for which it is not, and supply a proof that works for every $n$.
   Show the solution to question 6

   (a) The failing sentence is “so it is an equality as well”. Everything before it is correct: $\mathbf{v}=-2\mathbf{u}$, so the two vectors are multiples of each other and Cauchy–Schwarz really is an equality, $|-18|=18$.

   What is misused is the structure of the triangle inequality's proof. That proof bounds the cross term by two inequalities in a row,

   $$\mathbf{u}\cdot \mathbf{v}\le |\mathbf{u}\cdot \mathbf{v}|\le \Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert ,$$

   and the conclusion is an equality only if both are. Cauchy–Schwarz equality makes the second one an equality; the first one is an equality only when $\mathbf{u}\cdot \mathbf{v}\ge 0$. Here $\mathbf{u}\cdot \mathbf{v}=-18<18$, so the first inequality is strict and the triangle inequality is strict too. The correct equality condition is $\mathbf{u}\cdot \mathbf{v}=+\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert$: the vectors must be parallel and point the same way.

   Correct treatment: $\mathbf{u}+\mathbf{v}=⟨-1,-2,-2⟩$, so $\Vert \mathbf{u}+\mathbf{v}\Vert =\sqrt{1+4+4}=3$. Via the expansion, $\Vert \mathbf{u}+\mathbf{v}{\Vert }^2=9+2(-18)+36=9$, the same. Not only is $3<9$; in fact $3=|3-6|=|\Vert \mathbf{u}\Vert -\Vert \mathbf{v}\Vert |$, which is the equality case of the reverse triangle inequality — exactly what opposite directions produce. The picture is a flat triangle in which $\mathbf{v}$ doubles back over $\mathbf{u}$.

   (b) The failing sentence is “by the geometric form of the dot product, $\mathbf{u}\cdot \mathbf{v}=\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \cos \theta$”, when $n\ge 4$. In ${\mathbb{R}}^n$ there is no protractor. The angle between two nonzero vectors is defined as the unique $\theta \in [0,\pi ]$ with $\cos \theta =\frac{\mathbf{u}\cdot \mathbf{v}}{\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert }$, and that definition only makes sense because the quotient is known to lie in $[-1,1]$ — which is precisely Cauchy–Schwarz. So the argument uses the theorem to build the angle, and then uses the angle to prove the theorem: it is circular. The identity $\mathbf{u}\cdot \mathbf{v}=\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \cos \theta$ is a rearrangement of the definition, not an independent fact.

   In ${\mathbb{R}}^2$ and ${\mathbb{R}}^3$ the reasoning is acceptable, because there the geometric angle $\phi$ exists on its own and the law of cosines proves $\mathbf{u}\cdot \mathbf{v}=\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \cos \phi$ without assuming anything about the dot product's size. For $n\ge 4$ it proves nothing.

   A proof for every $n$, using only the four algebraic properties of the dot product. If $\mathbf{v}=0$ both sides are $0$; assume $\mathbf{v}\ne 0$. For every real $t$, positive definiteness gives $\Vert \mathbf{u}+t\mathbf{v}{\Vert }^2\ge 0$, and by bilinearity

   $$q(t)=\Vert \mathbf{u}+t\mathbf{v}{\Vert }^2=\Vert \mathbf{v}{\Vert }^2{t}^2+2(\mathbf{u}\cdot \mathbf{v})t+\Vert \mathbf{u}{\Vert }^2\ge 0\text{for all }t.$$

   The leading coefficient $\Vert \mathbf{v}{\Vert }^2$ is strictly positive, so $q$ is an upward parabola that never dips below the axis; therefore it has at most one real root and its discriminant is at most $0$:

   $$4(\mathbf{u}\cdot \mathbf{v}{)}^2-4\Vert \mathbf{u}{\Vert }^2\Vert \mathbf{v}{\Vert }^2\le 0⟹(\mathbf{u}\cdot \mathbf{v}{)}^2\le \Vert \mathbf{u}{\Vert }^2\Vert \mathbf{v}{\Vert }^2.$$

   Taking non-negative square roots gives $|\mathbf{u}\cdot \mathbf{v}|\le \Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert$. $■$ No angle was mentioned, so nothing was assumed — and this is what makes the definition of angle in ${\mathbb{R}}^n$ legitimate in the first place.

7. 7.Computation · 8 min

   Let $\mathbf{u}=⟨4,1,7⟩$ and $\mathbf{v}=⟨2,-1,2⟩$.

   1. (a)Find ${\mathrm{proj}}_{\mathbf{v}}\mathbf{u}$ and ${\mathrm{comp}}_{\mathbf{v}}\mathbf{u}$, and write $\mathbf{u}={\mathbf{u}}_{\parallel }+{\mathbf{u}}_{\perp }$ with ${\mathbf{u}}_{\parallel }$ parallel to $\mathbf{v}$ and ${\mathbf{u}}_{\perp }$ orthogonal to it. Verify the orthogonality directly, and verify that the squared norms of the two pieces add up to $\Vert \mathbf{u}{\Vert }^2$.
   2. (b)Find the distance from the point $(4,1,7)$ to the line through the origin in the direction $\mathbf{v}$, naming the theorem that justifies your method.
   3. (c)Without recomputing anything, write down ${\mathrm{proj}}_{-3\mathbf{v}}\mathbf{u}$ and ${\mathrm{comp}}_{-3\mathbf{v}}\mathbf{u}$, and explain from the formulas why one of them changed and the other did not.
   Show the solution to question 7

   (a) Two dot products do all the work. $\mathbf{v}\ne 0$, so projection onto it is defined.

   $$\mathbf{u}\cdot \mathbf{v}=8-1+14=21,\mathbf{v}\cdot \mathbf{v}=4+1+4=9,\Vert \mathbf{v}\Vert =3.$$

   Hence

   $${\mathbf{u}}_{\parallel }={\mathrm{proj}}_{\mathbf{v}}\mathbf{u}=\frac{21}{9}\mathbf{v}=\frac{7}{3}⟨2,-1,2⟩=⟨\frac{14}{3},-\frac{7}{3},\frac{14}{3}⟩,{\mathrm{comp}}_{\mathbf{v}}\mathbf{u}=\frac{21}{3}=7.$$

   The perpendicular piece is what is left:

   $${\mathbf{u}}_{\perp }=\mathbf{u}-{\mathbf{u}}_{\parallel }=⟨4-\frac{14}{3},1+\frac{7}{3},7-\frac{14}{3}⟩=⟨-\frac{2}{3},\frac{10}{3},\frac{7}{3}⟩.$$

   Orthogonality: ${\mathbf{u}}_{\perp }\cdot \mathbf{v}=\frac{-4-10+14}{3}=0$. Lengths: $\Vert \mathbf{u}{\Vert }^2=16+1+49=66$; $\Vert {\mathbf{u}}_{\parallel }{\Vert }^2={\left(\frac{7}{3}\right)}^2\cdot 9=49$, which also equals $({\mathrm{comp}}_{\mathbf{v}}\mathbf{u}{)}^2$ as it must; $\Vert {\mathbf{u}}_{\perp }{\Vert }^2=\frac{4+100+49}{9}=\frac{153}{9}=17$. And $49+17=66$, as the Pythagorean theorem requires for orthogonal pieces. The positive scalar projection says $\mathbf{u}$ leans towards $\mathbf{v}$, at an acute angle.

   (b) Best approximation on a line. The line is $\{t\mathbf{v}\}$ and the point has position vector $\mathbf{u}$. By the best-approximation theorem, the nearest point of the line to $\mathbf{u}$ is ${\mathrm{proj}}_{\mathbf{v}}\mathbf{u}$, and the distance is $\Vert \mathbf{u}-{\mathrm{proj}}_{\mathbf{v}}\mathbf{u}\Vert =\Vert {\mathbf{u}}_{\perp }\Vert =\sqrt{17}$.

   Sanity check: $\sqrt{17}<\sqrt{66}=\Vert \mathbf{u}\Vert$, the distance to the particular point $0$ of the line, as it must be.

   (c) The projection sees only the line; the scalar projection sees the direction. In the formula ${\mathrm{proj}}_{\mathbf{w}}\mathbf{u}=\frac{\mathbf{u}\cdot \mathbf{w}}{\mathbf{w}\cdot \mathbf{w}}\mathbf{w}$, put $\mathbf{w}=-3\mathbf{v}$: by homogeneity the numerator becomes $-3(\mathbf{u}\cdot \mathbf{v})$, the denominator becomes $9(\mathbf{v}\cdot \mathbf{v})$, and the trailing vector becomes $-3\mathbf{v}$. The factors $-3$ and $-3$ cancel the $9$, so

   $${\mathrm{proj}}_{-3\mathbf{v}}\mathbf{u}={\mathrm{proj}}_{\mathbf{v}}\mathbf{u}=⟨\frac{14}{3},-\frac{7}{3},\frac{14}{3}⟩,$$

   unchanged: $-3\mathbf{v}$ spans the same line, and the projection depends only on the line.

   For the scalar projection, ${\mathrm{comp}}_{\mathbf{w}}\mathbf{u}=\frac{\mathbf{u}\cdot \mathbf{w}}{\Vert \mathbf{w}\Vert }$ with $\mathbf{w}=-3\mathbf{v}$ gives $\frac{-3(\mathbf{u}\cdot \mathbf{v})}{|-3|\Vert \mathbf{v}\Vert }=-{\mathrm{comp}}_{\mathbf{v}}\mathbf{u}=-7$. The length $3$ cancels but the sign survives: the scalar projection is the signed length of the shadow measured along the direction you project onto, and $-3\mathbf{v}$ points the opposite way, so the same shadow now reads as negative.

8. 8.Proof · 9 min

   Fix a nonzero $\mathbf{v}\in {\mathbb{R}}^n$. Prove each statement for all $\mathbf{u},\mathbf{w}\in {\mathbb{R}}^n$ and scalars $c$, naming the dot-product property or theorem used at each step.

   1. (a)${\mathrm{proj}}_{\mathbf{v}}(\mathbf{u}+\mathbf{w})={\mathrm{proj}}_{\mathbf{v}}\mathbf{u}+{\mathrm{proj}}_{\mathbf{v}}\mathbf{w}$ and ${\mathrm{proj}}_{\mathbf{v}}(c\mathbf{u})=c{\mathrm{proj}}_{\mathbf{v}}\mathbf{u}$.
   2. (b)${\mathrm{proj}}_{\mathbf{v}}\left({\mathrm{proj}}_{\mathbf{v}}\mathbf{u}\right)={\mathrm{proj}}_{\mathbf{v}}\mathbf{u}$. Explain in one sentence why this had to be true geometrically.
   3. (c)$\Vert {\mathrm{proj}}_{\mathbf{v}}\mathbf{u}\Vert \le \Vert \mathbf{u}\Vert$, with equality if and only if $\mathbf{u}$ is a scalar multiple of $\mathbf{v}$.
   Show the solution to question 8

   Throughout, $\mathbf{v}\cdot \mathbf{v}=\Vert \mathbf{v}{\Vert }^2>0$ by positive definiteness and $\mathbf{v}\ne 0$, so the projection formula is defined.

   (a) Linearity comes from bilinearity. By definition and then distributivity of the dot product,

   $${\mathrm{proj}}_{\mathbf{v}}(\mathbf{u}+\mathbf{w})=\frac{(\mathbf{u}+\mathbf{w})\cdot \mathbf{v}}{\mathbf{v}\cdot \mathbf{v}}\mathbf{v}=\frac{\mathbf{u}\cdot \mathbf{v}+\mathbf{w}\cdot \mathbf{v}}{\mathbf{v}\cdot \mathbf{v}}\mathbf{v}=\frac{\mathbf{u}\cdot \mathbf{v}}{\mathbf{v}\cdot \mathbf{v}}\mathbf{v}+\frac{\mathbf{w}\cdot \mathbf{v}}{\mathbf{v}\cdot \mathbf{v}}\mathbf{v},$$

   the last step being the vector-space axiom $(a+b)\mathbf{v}=a\mathbf{v}+b\mathbf{v}$. The right side is ${\mathrm{proj}}_{\mathbf{v}}\mathbf{u}+{\mathrm{proj}}_{\mathbf{v}}\mathbf{w}$. For scalars, homogeneity of the dot product gives $(c\mathbf{u})\cdot \mathbf{v}=c(\mathbf{u}\cdot \mathbf{v})$, so

   $${\mathrm{proj}}_{\mathbf{v}}(c\mathbf{u})=\frac{c(\mathbf{u}\cdot \mathbf{v})}{\mathbf{v}\cdot \mathbf{v}}\mathbf{v}=c\left(\frac{\mathbf{u}\cdot \mathbf{v}}{\mathbf{v}\cdot \mathbf{v}}\mathbf{v}\right)=c{\mathrm{proj}}_{\mathbf{v}}\mathbf{u},$$

   using the compatibility axiom $a(b\mathbf{v})=(ab)\mathbf{v}$. $■$

   (b) Projecting twice is projecting once. Put $\mathbf{p}={\mathrm{proj}}_{\mathbf{v}}\mathbf{u}=k\mathbf{v}$ with $k=\frac{\mathbf{u}\cdot \mathbf{v}}{\mathbf{v}\cdot \mathbf{v}}$. Then by homogeneity

   $${\mathrm{proj}}_{\mathbf{v}}\mathbf{p}=\frac{(k\mathbf{v})\cdot \mathbf{v}}{\mathbf{v}\cdot \mathbf{v}}\mathbf{v}=\frac{k(\mathbf{v}\cdot \mathbf{v})}{\mathbf{v}\cdot \mathbf{v}}\mathbf{v}=k\mathbf{v}=\mathbf{p}.■$$

   Geometrically, $\mathbf{p}$ already lies on the line through $\mathbf{v}$, and the shadow of something lying on the floor is itself. (Equivalently, $\mathbf{p}=\mathbf{p}+0$ is a decomposition into a multiple of $\mathbf{v}$ plus a vector orthogonal to $\mathbf{v}$, and by the uniqueness clause of the orthogonal decomposition theorem the parallel part of $\mathbf{p}$ must be $\mathbf{p}$.)

   (c) Pythagoras on the decomposition. By the orthogonal decomposition theorem, $\mathbf{u}={\mathbf{u}}_{\parallel }+{\mathbf{u}}_{\perp }$ with ${\mathbf{u}}_{\parallel }={\mathrm{proj}}_{\mathbf{v}}\mathbf{u}=k\mathbf{v}$ and ${\mathbf{u}}_{\perp }\cdot \mathbf{v}=0$. The two pieces are orthogonal to each other, since ${\mathbf{u}}_{\perp }\cdot (k\mathbf{v})=k({\mathbf{u}}_{\perp }\cdot \mathbf{v})=0$ by homogeneity. So the Pythagorean theorem in ${\mathbb{R}}^n$ gives

   $$\Vert \mathbf{u}{\Vert }^2=\Vert {\mathbf{u}}_{\parallel }{\Vert }^2+\Vert {\mathbf{u}}_{\perp }{\Vert }^2\ge \Vert {\mathbf{u}}_{\parallel }{\Vert }^2,$$

   because $\Vert {\mathbf{u}}_{\perp }{\Vert }^2\ge 0$. Taking non-negative square roots, $\Vert {\mathrm{proj}}_{\mathbf{v}}\mathbf{u}\Vert \le \Vert \mathbf{u}\Vert$.

   Equality holds exactly when $\Vert {\mathbf{u}}_{\perp }{\Vert }^2=0$, which by positive definiteness means ${\mathbf{u}}_{\perp }=0$, i.e. $\mathbf{u}={\mathbf{u}}_{\parallel }=k\mathbf{v}$ is a scalar multiple of $\mathbf{v}$. Conversely, if $\mathbf{u}=c\mathbf{v}$ then ${\mathrm{proj}}_{\mathbf{v}}\mathbf{u}=\frac{c(\mathbf{v}\cdot \mathbf{v})}{\mathbf{v}\cdot \mathbf{v}}\mathbf{v}=c\mathbf{v}=\mathbf{u}$, so the two norms agree. $■$

   (A second route: $\Vert {\mathrm{proj}}_{\mathbf{v}}\mathbf{u}\Vert =\frac{|\mathbf{u}\cdot \mathbf{v}|}{\Vert \mathbf{v}{\Vert }^2}\Vert \mathbf{v}\Vert =\frac{|\mathbf{u}\cdot \mathbf{v}|}{\Vert \mathbf{v}\Vert }\le \frac{\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert }{\Vert \mathbf{v}\Vert }=\Vert \mathbf{u}\Vert$ by Cauchy–Schwarz, with equality in the Cauchy–Schwarz equality case. Both routes say the same thing: a shadow is never longer than the object casting it, and is exactly as long only when the object lies flat along the line.)

9. 9.Computation · 9 min

   Let $A=(1,1,0)$, $B=(3,0,2)$ and $C=(0,2,1)$.

   1. (a)Compute $\mathbf{n}=\overrightarrow{AB}\times \overrightarrow{AC}$, check it by dotting with both factors, and find the area of triangle $ABC$.
   2. (b)Let $D=(2,3,5)$. Find the volume of the parallelepiped with adjacent edges $\overrightarrow{AB}$, $\overrightarrow{AC}$ and $\overrightarrow{AD}$.
   3. (c)For which value of $k$ does $D=(2,3,k)$ lie in the plane through $A$, $B$ and $C$? Name the test you use, then confirm the answer by finding an equation of that plane and substituting.
   Show the solution to question 9

   (a) Edge vectors at $A$, then the cross product. Head minus tail:

   $$\overrightarrow{AB}=⟨2,-1,2⟩,\overrightarrow{AC}=⟨-1,1,1⟩.$$

   Component by component, watching the middle sign ($u_3{v}_1-u_1{v}_3$):

   $$\mathbf{n}=⟨(-1)(1)-(2)(1),(2)(-1)-(2)(1),(2)(1)-(-1)(-1)⟩=⟨-3,-4,1⟩.$$

   Checks: $\mathbf{n}\cdot \overrightarrow{AB}=-6+4+2=0$ and $\mathbf{n}\cdot \overrightarrow{AC}=3-4+1=0$, so $\mathbf{n}$ is orthogonal to both factors as the orthogonality theorem requires. Then $\Vert \mathbf{n}\Vert =\sqrt{9+16+1}=\sqrt{26}$, which is the area of the parallelogram on the two edges, and the triangle is half of it:

   $$\text{Area}(ABC)=\frac{1}{2}\sqrt{26}.$$

   (b) The scalar triple product. $\overrightarrow{AD}=⟨1,2,5⟩$, and the volume is the absolute value of

   $$\overrightarrow{AD}\cdot (\overrightarrow{AB}\times \overrightarrow{AC})=⟨1,2,5⟩\cdot ⟨-3,-4,1⟩=-3-8+5=-6.$$

   So the volume is $|-6|=6$. As a determinant with rows $\overrightarrow{AD},\overrightarrow{AB},\overrightarrow{AC}$, expanding along the first row: $1((-1)(1)-(2)(1))-2((2)(1)-(2)(-1))+5((2)(1)-(-1)(-1))=-3-8+5=-6$, the same. The negative sign only records that this ordering of the edges is left-handed; the volume is $6$.

   (c) The coplanarity test. With $D=(2,3,k)$, $\overrightarrow{AD}=⟨1,2,k⟩$ and the triple product is

   $$⟨1,2,k⟩\cdot ⟨-3,-4,1⟩=-3-8+k=k-11.$$

   By the coplanarity test, the three edge vectors lie in a common plane exactly when this vanishes, i.e. $k=11$. Since $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are not parallel, their span is a plane, so coplanarity of the three vectors means $\overrightarrow{AD}$ lies in that plane, which is the same as $D$ lying in the plane through $A,B,C$. (Consistently, part (b) is the case $k=5$: volume $|5-11|=6$.)

   Confirmation by the plane's equation. The normal is $\mathbf{n}=⟨-3,-4,1⟩$ and the plane passes through $A$, so the point–normal form is $-3(x-1)-4(y-1)+(z-0)=0$, i.e.

   $$3x+4y-z=7.$$

   Check: $B$ gives $9+0-2=7$ and $C$ gives $0+8-1=7$, so $B$ and $C$ lie on it. Substituting $D=(2,3,k)$: $6+12-k=7$ gives $k=11$, agreeing with the triple-product test.

10. 10.Proof · 11 min

    Throughout, $\mathbf{u},\mathbf{v},\mathbf{w}\in {\mathbb{R}}^3$. Argue from the algebraic properties of the cross product (anticommutativity, $\mathbf{x}\times \mathbf{x}=0$, distributivity, compatibility with scalars), the theorem that $\mathbf{x}\times \mathbf{y}$ is orthogonal to both factors, and the magnitude theorem. Do not expand anything into components.

    1. (a)Prove that $(\mathbf{u}-\mathbf{v})\times (\mathbf{u}+\mathbf{v})=2(\mathbf{u}\times \mathbf{v})$. Then interpret the identity as a comparison of two areas.
    2. (b)Prove that if $\mathbf{w}=a\mathbf{u}+b\mathbf{v}$ for some scalars $a,b$, then $\mathbf{u}\cdot (\mathbf{v}\times \mathbf{w})=0$.
    3. (c)Prove that if $\mathbf{u}\ne 0$, $\mathbf{u}\times \mathbf{v}=\mathbf{u}\times \mathbf{w}$ and $\mathbf{u}\cdot \mathbf{v}=\mathbf{u}\cdot \mathbf{w}$, then $\mathbf{v}=\mathbf{w}$. Then give two examples showing that neither hypothesis on its own is enough.
    Show the solution to question 10

    A preliminary: distributivity in the left slot. The stated distributive law is $\mathbf{x}\times (\mathbf{y}+\mathbf{z})=\mathbf{x}\times \mathbf{y}+\mathbf{x}\times \mathbf{z}$. Anticommutativity transfers it to the other slot:

    $$(\mathbf{y}+\mathbf{z})\times \mathbf{x}=-(\mathbf{x}\times (\mathbf{y}+\mathbf{z}))=-(\mathbf{x}\times \mathbf{y})-(\mathbf{x}\times \mathbf{z})=\mathbf{y}\times \mathbf{x}+\mathbf{z}\times \mathbf{x}.$$

    Also $(-\mathbf{y})\times \mathbf{x}=((-1)\mathbf{y})\times \mathbf{x}=-(\mathbf{y}\times \mathbf{x})$ by compatibility with scalars, and likewise in the right slot.

    (a) Expand and cancel. Using left distributivity, then right distributivity in each term,

    $$(\mathbf{u}-\mathbf{v})\times (\mathbf{u}+\mathbf{v})=\mathbf{u}\times \mathbf{u}+\mathbf{u}\times \mathbf{v}-\mathbf{v}\times \mathbf{u}-\mathbf{v}\times \mathbf{v}.$$

    The first and last terms are $0$, and $-\mathbf{v}\times \mathbf{u}=\mathbf{u}\times \mathbf{v}$ by anticommutativity, so the right side is $\mathbf{u}\times \mathbf{v}+\mathbf{u}\times \mathbf{v}=2(\mathbf{u}\times \mathbf{v})$. $■$

    Interpretation: the parallelogram with sides $\mathbf{u},\mathbf{v}$ has diagonals $\mathbf{u}+\mathbf{v}$ and $\mathbf{u}-\mathbf{v}$. Taking norms, the parallelogram whose sides are those two diagonals has area $\Vert (\mathbf{u}-\mathbf{v})\times (\mathbf{u}+\mathbf{v})\Vert =2\Vert \mathbf{u}\times \mathbf{v}\Vert$ — exactly twice the area of the original. (It is also non-degenerate exactly when the original is, since one cross product vanishes iff the other does.)

    (b) A dependent triple has zero triple product. By distributivity and compatibility with scalars,

    $$\mathbf{v}\times \mathbf{w}=\mathbf{v}\times (a\mathbf{u}+b\mathbf{v})=a(\mathbf{v}\times \mathbf{u})+b(\mathbf{v}\times \mathbf{v})=a(\mathbf{v}\times \mathbf{u}).$$

    Hence, by homogeneity of the dot product, $\mathbf{u}\cdot (\mathbf{v}\times \mathbf{w})=a\mathbf{u}\cdot (\mathbf{v}\times \mathbf{u})$, and $\mathbf{v}\times \mathbf{u}$ is orthogonal to both of its factors, in particular to $\mathbf{u}$. So the triple product is $a\cdot 0=0$. $■$ This is one direction of the coplanarity test, proved without a determinant: if the third edge lies in the plane of the first two, the box is flat.

    (c) Cancellation needs both products. Put $\mathbf{x}=\mathbf{v}-\mathbf{w}$. By distributivity (and $-\mathbf{w}=(-1)\mathbf{w}$),

    $$\mathbf{u}\times \mathbf{x}=\mathbf{u}\times \mathbf{v}-\mathbf{u}\times \mathbf{w}=0,\mathbf{u}\cdot \mathbf{x}=\mathbf{u}\cdot \mathbf{v}-\mathbf{u}\cdot \mathbf{w}=0.$$

    Suppose, for contradiction, that $\mathbf{x}\ne 0$. Then $\mathbf{u}$ and $\mathbf{x}$ are both nonzero with $\mathbf{u}\times \mathbf{x}=0$, so by the magnitude theorem they are parallel: $\mathbf{x}=t\mathbf{u}$ for some scalar $t$. Then

    $$0=\mathbf{u}\cdot \mathbf{x}=\mathbf{u}\cdot (t\mathbf{u})=t\Vert \mathbf{u}{\Vert }^2,$$

    and $\Vert \mathbf{u}{\Vert }^2>0$ because $\mathbf{u}\ne 0$ (positive definiteness), so $t=0$ and $\mathbf{x}=0$ — a contradiction. Hence $\mathbf{x}=0$, that is, $\mathbf{v}=\mathbf{w}$. $■$

    The idea: the dot product with $\mathbf{u}$ sees only the component of a vector along $\mathbf{u}$, and the cross product with $\mathbf{u}$ sees only the component perpendicular to $\mathbf{u}$. Either alone is blind to the other component; together they see everything.

    Examples. Take $\mathbf{u}=\mathbf{i}$. With $\mathbf{v}=\mathbf{j}$ and $\mathbf{w}=\mathbf{i}+\mathbf{j}$: $\mathbf{i}\times \mathbf{j}=\mathbf{k}$ and $\mathbf{i}\times (\mathbf{i}+\mathbf{j})=0+\mathbf{k}=\mathbf{k}$, so the cross products agree, but $\mathbf{v}\ne \mathbf{w}$ (their dot products with $\mathbf{i}$ are $0$ and $1$). With $\mathbf{v}=\mathbf{j}$ and $\mathbf{w}=\mathbf{k}$: $\mathbf{i}\cdot \mathbf{j}=0=\mathbf{i}\cdot \mathbf{k}$, so the dot products agree, but $\mathbf{v}\ne \mathbf{w}$ (their cross products with $\mathbf{i}$ are $\mathbf{k}$ and $-\mathbf{j}$).

11. 11.Computation · 11 min

    Consider the lines

    $$L_1:\mathbf{r}(t)=⟨1,0,2⟩+t⟨1,2,2⟩,L_2:\mathbf{r}(s)=⟨3,1,-1⟩+s⟨2,1,-2⟩.$$

    1. (a)Show that $L_1$ and $L_2$ are neither parallel nor intersecting, and find the angle between their directions.
    2. (b)Find an equation of the plane $\Pi$ that contains $L_1$ and is parallel to $L_2$. Verify both properties from your equation.
    3. (c)Show that every point of $L_2$ is at the same distance from $\Pi$, explain why that distance is the shortest distance between any point of $L_1$ and any point of $L_2$, and compute it. (You may take for granted that the shortest distance is actually attained.)
    Show the solution to question 11

    (a) Skew, and in fact perpendicular. The directions are ${\mathbf{d}}_1=⟨1,2,2⟩$ and ${\mathbf{d}}_2=⟨2,1,-2⟩$. They are not parallel: $c{\mathbf{d}}_1={\mathbf{d}}_2$ would need $c=2$ from the first components and $c=\frac{1}{2}$ from the second.

    A common point would need a single pair $(t,s)$ with

    $$1+t=3+2s,2t=1+s,2+2t=-1-2s.$$

    The second gives $s=2t-1$. Substituting into the first, $1+t=3+4t-2$, so $t=0$ and $s=-1$. The third then reads $2+0=-1+2$, i.e. $2=1$, which is false. The system is inconsistent, so the lines do not meet. Not parallel and not intersecting: they are skew.

    Angle: ${\mathbf{d}}_1\cdot {\mathbf{d}}_2=2+2-4=0$, so the directions are orthogonal and the angle between the lines is $90^{\circ }$ (no absolute value question arises, since the cosine is $0$).

    (b) A normal orthogonal to both directions. $\Pi$ contains $L_1$, so it contains the direction ${\mathbf{d}}_1$; it is parallel to $L_2$, so its normal must also be orthogonal to ${\mathbf{d}}_2$. The cross product manufactures exactly such a vector:

    $${\mathbf{d}}_1\times {\mathbf{d}}_2=⟨(2)(-2)-(2)(1),(2)(2)-(1)(-2),(1)(1)-(2)(2)⟩=⟨-6,6,-3⟩.$$

    Check: $⟨-6,6,-3⟩\cdot {\mathbf{d}}_1=-6+12-6=0$ and $⟨-6,6,-3⟩\cdot {\mathbf{d}}_2=-12+6+6=0$. Any nonzero multiple is also a normal, so take $\mathbf{n}=⟨2,-2,1⟩$ (divide by $-3$), which has the convenient norm $\Vert \mathbf{n}\Vert =\sqrt{4+4+1}=3$. Through the point $(1,0,2)$ of $L_1$, the point–normal form is $2(x-1)-2(y-0)+(z-2)=0$, i.e.

    $$\Pi :2x-2y+z=4.$$

    Verification. Along $L_1$: $2(1+t)-2(2t)+(2+2t)=2+2t-4t+2+2t=4$ for every $t$, so all of $L_1$ lies in $\Pi$. Parallel to $L_2$: $\mathbf{n}\cdot {\mathbf{d}}_2=4-2-2=0$, so the direction of $L_2$ is orthogonal to the normal; and the base point $(3,1,-1)$ of $L_2$ gives $6-2-1=3\ne 4$, so $L_2$ is parallel to $\Pi$ without lying in it.

    (c) Distance between skew lines as a point–plane distance. Along $L_2$ the plane's left-hand side reads

    $$2(3+2s)-2(1+s)+(-1-2s)=6+4s-2-2s-1-2s=3,$$

    independent of $s$. So by the point–plane distance formula every point of $L_2$ is at distance

    $$D=\frac{|3-4|}{\Vert \mathbf{n}\Vert }=\frac{1}{3}$$

    from $\Pi$. (Equivalently, $L_2$ lies in the parallel plane $2x-2y+z=3$.)

    Why this is the distance between the lines: take any $P$ on $L_1$ and any $Q$ on $L_2$. Then $P$ lies in $\Pi$, and by the point–plane theorem the nearest point of $\Pi$ to $Q$ is the foot of the perpendicular, at distance $\frac{1}{3}$; so $\Vert \overrightarrow{PQ}\Vert \ge \frac{1}{3}$ for every such pair. No two points of the lines are closer than $\frac{1}{3}$, and since the shortest distance is attained, it equals $\frac{1}{3}$.

    Cross-check with the scalar projection: the displacement between the base points is $⟨3,1,-1⟩-⟨1,0,2⟩=⟨2,1,-3⟩$, and its component along the unit normal is $\frac{|⟨2,1,-3⟩\cdot ⟨2,-2,1⟩|}{3}=\frac{|4-2-3|}{3}=\frac{1}{3}$. (For the curious: the closest pair is $t=-\frac{2}{9}$, $s=-\frac{11}{9}$, giving $P=\left(\frac{7}{9},-\frac{4}{9},\frac{14}{9}\right)$ and $Q=\left(\frac{5}{9},-\frac{2}{9},\frac{13}{9}\right)$ with $\overrightarrow{QP}=\frac{1}{9}⟨2,-2,1⟩$, a multiple of $\mathbf{n}$ of length exactly $\frac{1}{3}$.)

12. 12.Critique · 10 min

    Three arguments follow, each wrong. For each: quote the sentence where it goes wrong, explain what has been misunderstood, and give the correct result with justification.

    1. (a)“The plane $2x-4y+4z=8$ has normal $⟨1,-2,2⟩$, since dividing the coefficients by $2$ does not change the direction. So the distance from $P=(3,1,2)$ to the plane is $\frac{|3-2(1)+2(2)-8|}{\Vert ⟨1,-2,2⟩\Vert }=\frac{|-3|}{3}=1$.”
    2. (b)“The line $\mathbf{r}(t)=⟨1,0,3⟩+t⟨2,1,0⟩$ has direction $\mathbf{d}=⟨2,1,0⟩$, and $⟨2,1,0⟩\cdot ⟨1,-2,2⟩=0$, so $\mathbf{d}$ is orthogonal to the normal of the plane $2x-4y+4z=8$. Hence the line lies in that plane.”
    3. (c)“To find the plane through $A=(1,0,0)$, $B=(2,1,1)$ and $C=(3,2,2)$: $\overrightarrow{AB}=⟨1,1,1⟩$ and $\overrightarrow{AC}=⟨2,2,2⟩$, so $\mathbf{n}=\overrightarrow{AB}\times \overrightarrow{AC}=⟨2-2,2-2,2-2⟩=⟨0,0,0⟩$. The plane is therefore $0(x-1)+0y+0z=0$, which every point satisfies, so the plane through $A$, $B$, $C$ is all of ${\mathbb{R}}^3$.”
    Show the solution to question 12

    (a) The sentence that fails is the distance computation, not the normal. $⟨1,-2,2⟩$ is a perfectly good normal to the plane — any nonzero multiple of $⟨2,-4,4⟩$ is. The error is mixing the reduced normal with the unreduced constant $8$. The formula $D=\frac{|a{x}_1+b{y}_1+c{z}_1-d|}{\sqrt{a^2+b^2+c^2}}$ requires the plane written as $ax+by+cz=d$ with the very same $a,b,c$ in the norm; the numerator is $|\mathbf{n}\cdot {\mathbf{r}}_1-d|$ and $d$ is tied to $\mathbf{n}$.

    Correct treatment, either way. Reduce the whole equation: $x-2y+2z=4$, then

    $$D=\frac{|3-2+4-4|}{\sqrt{1+4+4}}=\frac{1}{3}.$$

    Or keep the original: $\mathbf{n}=⟨2,-4,4⟩$ with $\Vert \mathbf{n}\Vert =\sqrt{4+16+16}=6$ and $D=\frac{|6-4+8-8|}{6}=\frac{2}{6}=\frac{1}{3}$. Both consistent choices agree; the mixed one gave $1$, three times too large.

    Sanity check: at $P$ the reduced left-hand side reads $5>4$, so step $\frac{1}{3}$ against the unit normal $\frac{1}{3}⟨1,-2,2⟩$: the point $\left(3-\frac{1}{9},1+\frac{2}{9},2-\frac{2}{9}\right)=\left(\frac{26}{9},\frac{11}{9},\frac{16}{9}\right)$ gives $\frac{26-22+32}{9}=\frac{36}{9}=4$. It lands on the plane, so the distance is indeed $\frac{1}{3}$.

    (b) The sentence that fails is “hence the line lies in that plane”. The computation $\mathbf{d}\cdot \mathbf{n}=2-2+0=0$ is right, and it shows the line is parallel to the plane: moving along $\mathbf{d}$ never changes the reading $2x-4y+4z$. But a line parallel to a plane either lies in it or misses it entirely, and which one is decided by a single point. Test the base point $(1,0,3)$: $2(1)-4(0)+4(3)=14\ne 8$. So the line is not in the plane.

    Correct result: along the line the reading is $2(1+2t)-4(t)+4(3)=2+4t-4t+12=14$ for every $t$, constant as parallelism predicts, and every point of the line is at distance

    $$\frac{|14-8|}{\Vert ⟨2,-4,4⟩\Vert }=\frac{6}{6}=1$$

    from the plane. The line runs parallel to the plane at height $1$ above it and never meets it. “Lies in” needs two things — direction orthogonal to the normal and one point on the plane — and only the first was checked.

    (c) The sentence that fails is “the plane is therefore $0(x-1)+0y+0z=0$”. The cross product came out $0$ because $\overrightarrow{AC}=2\overrightarrow{AB}$: the three points are collinear, and the recipe for a plane through three points requires them not to be. A normal vector is by definition nonzero, so $0$ is not a normal, and $0=0$ is not the equation of a plane (the theorem on the general equation $ax+by+cz=d$ requires $⟨a,b,c⟩\ne 0$). The cross product was not malfunctioning — it was reporting that the hypothesis fails.

    Correct result: $A$, $B$, $C$ all lie on the line $\mathbf{r}(t)=⟨1,0,0⟩+t⟨1,1,1⟩$ ($B$ at $t=1$, $C$ at $t=2$), and three collinear points do not determine a plane: every plane containing that line passes through all three, and there are infinitely many. Any normal orthogonal to $⟨1,1,1⟩$ produces one. For instance $⟨1,-1,0⟩$ gives the plane $x-y=1$, which contains $A$ ($1-0$), $B$ ($2-1$) and $C$ ($3-2$); and $⟨0,1,-1⟩$ gives $y-z=0$, which also contains all three ($0-0$, $1-1$, $2-2$). Two different planes through the same three points — so “the plane through $A,B,C$” does not exist, and it is certainly not ${\mathbb{R}}^3$.