Vectors & Euclidean Space

(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.

(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.

(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 $k=2$.

(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.

(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 $\mathbf{u}-\mathbf{v}$.

(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.

(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.

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.)

(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.

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 $-\mathbf{j}$).

(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 $\frac{1}{3}$.)

(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 ${\mathbb{R}}^3$.