In $\mathbb{R}^3$ two vectors have a product that is again a vector: $\mathbf{u}\times\mathbf{v}$ is perpendicular to both and as long as the area they span โ the key to normals, planes, volumes and distances.
A solar engineer needs to know which way a sloping roof faces. A game engine must decide, for each of millions of tiny triangles, which way it points so that it can be lit correctly. A structural engineer needs the turning effect of a force applied to a spanner, and a geologist wants the dip of a rock layer from three drill-hole depths. All of these need a vector that is perpendicular to two given vectors โ the "normal" of a surface.
In $\mathbb{R}^3$ (and only there) there is a product that builds such a vector directly: the cross product $\mathbf{u}\times\mathbf{v}$. Its direction is perpendicular to both $\mathbf{u}$ and $\mathbf{v}$, and its length is the area of the parallelogram they span. With it, and with the dot product of Section 4.2, we can write equations of lines and planes in space, find where they meet, and measure distances and volumes.
The Cross Product
We look for a vector $\mathbf{w}$ with $\mathbf{w}\cdot\mathbf{u}=0$ and $\mathbf{w}\cdot\mathbf{v}=0$. Solving these two equations in general (try it!) leads to the following formula, which is easiest to remember as a $3\times3$ determinant (Chapter 3) whose first row holds the standard unit vectors.
๐ Definition 4.8 โ Cross product
The cross product of $\mathbf{u}=(u_1,u_2,u_3)$ and $\mathbf{v}=(v_1,v_2,v_3)$ in $\mathbb{R}^3$ is the vector
$$\mathbf{u}\times\mathbf{v}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\u_1&u_2&u_3\\v_1&v_2&v_3\end{vmatrix}=\begin{vmatrix}u_2&u_3\\v_2&v_3\end{vmatrix}\mathbf{i}-\begin{vmatrix}u_1&u_3\\v_1&v_3\end{vmatrix}\mathbf{j}+\begin{vmatrix}u_1&u_2\\v_1&v_2\end{vmatrix}\mathbf{k}=\begin{bmatrix}u_2v_3-u_3v_2\\u_3v_1-u_1v_3\\u_1v_2-u_2v_1\end{bmatrix}.$$
(The "determinant" is a memory aid: expand along the first row by cofactors as in Chapter 3.) The cross product is defined only for vectors in $\mathbb{R}^3$.
โ๏ธ Example 1 โ Computing a cross product
Find $\mathbf{u}\times\mathbf{v}$ and $\mathbf{v}\times\mathbf{u}$ for $\mathbf{u}=(1,2,-2)$ and $\mathbf{v}=(3,0,1)$, and check that $\mathbf{u}\times\mathbf{v}$ is orthogonal to both vectors.
(f) unit vectors$\mathbf{i}\times\mathbf{j}=\mathbf{k}$, $\mathbf{j}\times\mathbf{k}=\mathbf{i}$, $\mathbf{k}\times\mathbf{i}=\mathbf{j}$
Each follows from Definition 4.8 or from the rules for determinants (Chapter 3): swapping two rows changes the sign, and two equal rows give $0$. The cross product is not associative: $(\mathbf{i}\times\mathbf{i})\times\mathbf{j}=\mathbf{0}$ but $\mathbf{i}\times(\mathbf{i}\times\mathbf{j})=\mathbf{i}\times\mathbf{k}=-\mathbf{j}$.
Which of the two perpendicular directions? There are two directions perpendicular to both $\mathbf{u}$ and $\mathbf{v}$. The cross product follows the right-hand rule: point the fingers of your right hand along $\mathbf{u}$ and curl them towards $\mathbf{v}$ through the smaller angle; your thumb points along $\mathbf{u}\times\mathbf{v}$. That is why $\mathbf{i}\times\mathbf{j}=\mathbf{k}$ points up, and why reversing the order reverses the direction.
How long is it? Expanding both sides in components (a page of algebra) proves Lagrange's identity $\|\mathbf{u}\times\mathbf{v}\|^2=\|\mathbf{u}\|^2\|\mathbf{v}\|^2-(\mathbf{u}\cdot\mathbf{v})^2$. Substituting $\mathbf{u}\cdot\mathbf{v}=\|\mathbf{u}\|\|\mathbf{v}\|\cos\theta$ gives $\|\mathbf{u}\|^2\|\mathbf{v}\|^2(1-\cos^2\theta)=\|\mathbf{u}\|^2\|\mathbf{v}\|^2\sin^2\theta$, and $\sin\theta\ge0$ for $0\le\theta\le\pi$.
๐ Theorem 4.9 โ The geometry of $\mathbf{u}\times\mathbf{v}$
Let $\mathbf{u},\mathbf{v}$ be vectors in $\mathbb{R}^3$, nonzero, with angle $\theta$ between them ($0\le\theta\le\pi$). Then
(a) $\|\mathbf{u}\times\mathbf{v}\|=\|\mathbf{u}\|\,\|\mathbf{v}\|\sin\theta$, which is the area of the parallelogram with sides $\mathbf{u}$ and $\mathbf{v}$ (base $\|\mathbf{u}\|$, height $\|\mathbf{v}\|\sin\theta$). The triangle with these two sides has area $\tfrac12\|\mathbf{u}\times\mathbf{v}\|$.
(b) $\mathbf{u}\times\mathbf{v}=\mathbf{0}$ if and only if $\mathbf{u}$ and $\mathbf{v}$ are parallel (one is a scalar multiple of the other).
โ๏ธ Example 2 โ The area of a triangle in space
Find the area of the triangle with vertices $A(1,0,1)$, $B(2,2,1)$ and $C(0,1,3)$.
๐ก Show solutionโถ
Two sides $\overrightarrow{AB}=(1,2,0)$ and $\overrightarrow{AC}=(-1,1,2)$.
Area $\|(4,-2,3)\|=\sqrt{16+4+9}=\sqrt{29}$, so the area is $\tfrac12\sqrt{29}\approx2.693$ square units.
Check With Theorem 4.9(a) directly: $\|\overrightarrow{AB}\|^2=5$, $\|\overrightarrow{AC}\|^2=6$, $\overrightarrow{AB}\cdot\overrightarrow{AC}=1$, and Lagrange's identity gives $5\cdot6-1^2=29$. โ We will reuse the normal $(4,-2,3)$ in Example 5.
Interactive 1 โ The Cross Product in 3-D
An oblique 3-D view of u, v, the parallelogram they span, and u ร v. Turn the scene with the slider to convince yourself that $\mathbf{u}\times\mathbf{v}$ sticks straight out of the parallelogram. In the first setting Play swings $\mathbf{v}$ round $\mathbf{u}$: the arrow's length follows the shaded area, it vanishes when the vectors are parallel, and it flips over (right-hand rule) when $\mathbf{v}$ passes to the other side of $\mathbf{u}$. In the other settings Play turns the scene.
Volume. Three vectors $\mathbf{u},\mathbf{v},\mathbf{w}$ drawn from one point are the edges of a slanted box called a parallelepiped. Take the parallelogram of $\mathbf{v}$ and $\mathbf{w}$ as its base: its area is $\|\mathbf{v}\times\mathbf{w}\|$, and the height is the length of the projection of $\mathbf{u}$ onto the normal $\mathbf{v}\times\mathbf{w}$, which is $\dfrac{|\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w})|}{\|\mathbf{v}\times\mathbf{w}\|}$. Base times height leaves just the number $|\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w})|$.
๐ Theorem 4.10 โ The scalar triple product
For $\mathbf{u},\mathbf{v},\mathbf{w}$ in $\mathbb{R}^3$, the scalar triple product is
$$\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w})=\det\begin{bmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1&w_2&w_3\end{bmatrix}.$$
Its absolute value is the volume of the parallelepiped with edges $\mathbf{u},\mathbf{v},\mathbf{w}$. In particular, $\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w})=0$ if and only if $\mathbf{u},\mathbf{v},\mathbf{w}$ lie in one plane through the origin (they are coplanar).
โ๏ธ Example 3 โ Volume of a parallelepiped
Find the volume of the parallelepiped with edges $\mathbf{u}=(1,0,2)$, $\mathbf{v}=(0,3,1)$ and $\mathbf{w}=(2,1,0)$.
๐ก Show solutionโถ
Determinant Expand along the first row: $$\det\begin{bmatrix}1&0&2\\0&3&1\\2&1&0\end{bmatrix}=1(3\cdot0-1\cdot1)-0(0\cdot0-1\cdot2)+2(0\cdot1-3\cdot2)=-1-0-12=-13.$$
Volume $V=|-13|=13$ cubic units. The negative sign only says that $\mathbf{u},\mathbf{v},\mathbf{w}$ (in this order) form a left-handed set.
Check $\mathbf{v}\times\mathbf{w}=(3\cdot0-1\cdot1,\ 1\cdot2-0\cdot0,\ 0\cdot1-3\cdot2)=(-1,2,-6)$ and $\mathbf{u}\cdot(-1,2,-6)=-1+0-12=-13$. โ
Lines in $\mathbb{R}^2$ and $\mathbb{R}^3$
A line is fixed by one point on it and its direction. If $\mathbf{p}$ is the position vector of a point $P$ on the line and $\mathbf{d}\ne\mathbf{0}$ points along it, then every point of the line is reached from $P$ by walking some multiple $t\mathbf{d}$ โ forward for $t\gt0$, backward for $t\lt0$.
๐ Definition 4.9 โ Vector and parametric equations of a line
The line through the point $P$ (position vector $\mathbf{p}$) in the direction of the direction vector $\mathbf{d}\ne\mathbf{0}$ is the set of points $\mathbf{x}$ with
$$\mathbf{x}=\mathbf{p}+t\mathbf{d},\qquad t\in\mathbb{R}\qquad\text{(vector equation)}.$$
In $\mathbb{R}^3$, with $\mathbf{p}=(x_0,y_0,z_0)$ and $\mathbf{d}=(a,b,c)$, this is the set of parametric equations
$$x=x_0+ta,\qquad y=y_0+tb,\qquad z=z_0+tc .$$
In $\mathbb{R}^2$ drop the $z$-equation. The same line has many equations: any point on it can serve as $P$, and any nonzero multiple of $\mathbf{d}$ as the direction.
Think of $t$ as time and $\mathbf{x}(t)$ as the position of a drone flying with constant velocity $\mathbf{d}$: at $t=0$ it is at $P$. In $\mathbb{R}^2$ a line can also be written as one equation $ax+by=c$; in $\mathbb{R}^3$ a single linear equation describes a plane, not a line, so lines in space are described parametrically. Two lines in $\mathbb{R}^3$ can be parallel (directions are multiples), intersecting, or skew โ not parallel and never meeting, like a road and a bridge passing over it.
โ๏ธ Example 4 โ The line through two points
Find vector and parametric equations of the line through $P(1,-2,3)$ and $Q(4,0,1)$. Does $R(7,2,-1)$ lie on it? Does $S(7,2,0)$?
๐ก Show solutionโถ
Direction $\mathbf{d}=\overrightarrow{PQ}=(4-1,\,0+2,\,1-3)=(3,2,-2)$.
Test $R$ $x=7$ needs $t=2$; then $y=-2+4=2$ โ and $z=3-4=-1$ โ. All three agree, so $R$ is on the line.
Test $S$ Again $t=2$ from $x$ and $y$, but then $z=-1\ne0$. No single $t$ works, so $S$ is not on the line.
Planes, Distances and Intersections
A plane in $\mathbb{R}^3$ is fixed by one point $P$ on it and a normal vector $\mathbf{n}\ne\mathbf{0}$ perpendicular to it. A point $X$ lies in the plane exactly when the arrow $\overrightarrow{PX}=\mathbf{x}-\mathbf{p}$ lies flat in the plane, i.e. is orthogonal to $\mathbf{n}$.
๐ Definition 4.10 โ Equations of a plane
The plane through $P(x_0,y_0,z_0)$ with normal vector $\mathbf{n}=(a,b,c)\ne\mathbf{0}$ has point-normal equation
$$\mathbf{n}\cdot(\mathbf{x}-\mathbf{p})=0,\qquad\text{i.e.}\qquad a(x-x_0)+b(y-y_0)+c(z-z_0)=0 .$$
Multiplying out gives the general equation $ax+by+cz=d$, where $d=ax_0+by_0+cz_0$. Conversely, every equation $ax+by+cz=d$ with $(a,b,c)\ne(0,0,0)$ is a plane with normal $(a,b,c)$ โ read the normal off the coefficients.
๐งญ Problem-solving strategy โ the plane through three points
Form two vectors in the plane from one of the points: $\overrightarrow{PQ}$ and $\overrightarrow{PR}$.
Take $\mathbf{n}=\overrightarrow{PQ}\times\overrightarrow{PR}$. If $\mathbf{n}=\mathbf{0}$ the points are collinear and there is no single plane โ stop.
Write $\mathbf{n}\cdot(\mathbf{x}-\mathbf{p})=0$ and simplify to $ax+by+cz=d$.
Check by substituting the other two points.
๐ Theorem 4.11 โ Distances from a point
(a) To a plane. The distance from the point $Q(x_1,y_1,z_1)$ to the plane $ax+by+cz=d$ (with $(a,b,c)\ne\mathbf{0}$) is
$$D=\frac{|ax_1+by_1+cz_1-d|}{\sqrt{a^2+b^2+c^2}}.$$
(b) To a line in $\mathbb{R}^3$. The distance from the point $Q$ to the line through $P$ with direction $\mathbf{d}\ne\mathbf{0}$ is $D=\dfrac{\|\overrightarrow{PQ}\times\mathbf{d}\|}{\|\mathbf{d}\|}$.
Why. (a) For any point $P$ in the plane, $D$ is the length of the projection of $\overrightarrow{PQ}$ onto $\mathbf{n}$: $\dfrac{|\mathbf{n}\cdot\overrightarrow{PQ}|}{\|\mathbf{n}\|}$, and $\mathbf{n}\cdot(\mathbf{q}-\mathbf{p})=ax_1+by_1+cz_1-d$. (b) $\|\overrightarrow{PQ}\times\mathbf{d}\|$ is the area of the parallelogram on $\overrightarrow{PQ}$ and $\mathbf{d}$; dividing by the base $\|\mathbf{d}\|$ leaves the height, which is the distance. $\blacksquare$
โ๏ธ Example 5 โ A plane through three points, and a distance
Find the equation of the plane through $A(1,0,1)$, $B(2,2,1)$ and $C(0,1,3)$, and the distance from $S(2,1,4)$ to this plane.
๐ก Show solutionโถ
Normal From Example 2, $\mathbf{n}=\overrightarrow{AB}\times\overrightarrow{AC}=(4,-2,3)$.
Equation Point-normal form through $A$: $4(x-1)-2(y-0)+3(z-1)=0$, i.e. $$4x-2y+3z=7 .$$
Check $B$: $8-4+3=7$ โ. $C$: $0-2+9=7$ โ.
Distance $$D=\frac{|4(2)-2(1)+3(4)-7|}{\sqrt{16+4+9}}=\frac{11}{\sqrt{29}}\approx2.043 .$$ The nearest point of the plane is $\mathbf{s}-\tfrac{11}{29}\mathbf{n}=\left(\tfrac{14}{29},\tfrac{51}{29},\tfrac{83}{29}\right)$; you can see it in Interactive 2.
โ๏ธ Example 6 โ Where lines and planes meet
(a) Find where the line of Example 4, $\mathbf{x}=(1,-2,3)+t(3,2,-2)$, meets the plane $4x-2y+3z=7$ of Example 5. (b) Find the line in which the planes $x+y+z=6$ and $x-y+2z=3$ intersect.
๐ก Show solutionโถ
(a) Substitute Put $x=1+3t$, $y=-2+2t$, $z=3-2t$ into the plane: $4(1+3t)-2(-2+2t)+3(3-2t)=4+12t+4-4t+9-6t=17+2t$.
(a) Solve $17+2t=7$ gives $t=-5$, so the point is $(1-15,\,-2-10,\,3+10)=(-14,-12,13)$. Check: $-56+24+39=7$ โ. (If the $t$-terms had cancelled, the line would be parallel to the plane: no solution, or the whole line in the plane.)
(b) Direction The line lies in both planes, so it is perpendicular to both normals: $$\mathbf{d}=(1,1,1)\times(1,-1,2)=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&1&1\\1&-1&2\end{vmatrix}=(2+1,\ 1-2,\ -1-1)=(3,-1,-2).$$
(b) A point Put $z=0$: $x+y=6$, $x-y=3$, so $x=\tfrac92$, $y=\tfrac32$. The line is $\mathbf{x}=\left(\tfrac92,\tfrac32,0\right)+t(3,-1,-2)$. Check: $\tfrac92+\tfrac32+0=6$ โ, $\tfrac92-\tfrac32+0=3$ โ, and $\mathbf{d}$ is orthogonal to both normals: $3-1-2=0$, $3+1-4=0$ โ.
Interactive 2 โ The Plane through Three Points
The plane through $P$, $Q$, $R$ is drawn as a shaded patch. Its normal $\mathbf{n}=\overrightarrow{PQ}\times\overrightarrow{PR}$ (green) rises from the middle of the triangle. The orange point $S$ drops a perpendicular (dashed) to its nearest point $F$ on the plane; that length is the distance of Theorem 4.11. Play turns the scene โ watch the dashed segment stay parallel to $\mathbf{n}$ from every angle. The last setting shows three points on one line.
$\overrightarrow{PQ}$, $\overrightarrow{PR}$โ
normal $\mathbf{n}$โ
planeโ
distance from $S$โ
โ ๏ธ Common mistakes
1. The middle sign. In the determinant expansion the $\mathbf{j}$-term has a minus sign: the second component of $\mathbf{u}\times\mathbf{v}$ is $u_3v_1-u_1v_3$, not $u_1v_3-u_3v_1$. Always check your answer by dotting it with $\mathbf{u}$ and $\mathbf{v}$.
2. Order matters. $\mathbf{u}\times\mathbf{v}=-\mathbf{v}\times\mathbf{u}$. For areas and normals either order is fine (the length and the line are the same); for torque and orientation it is not.
3. Direction versus normal. In $\mathbf{x}=\mathbf{p}+t\mathbf{d}$ the vector $\mathbf{d}$ runs along the line; in $ax+by+cz=d$ the vector $(a,b,c)$ is perpendicular to the plane. A line is parallel to a plane when its direction is orthogonal to the plane's normal.
4. Forgetting the absolute value in the distance formula, or forgetting to move $d$ to the left-hand side before substituting.
Exercises
I
Cross products, areas and volumes.
1 $(1,2,3)\times(4,5,6)$
2 $(2,-1,0)\times(1,3,-2)$
3 $\mathbf{i}\times\mathbf{k}$ and $(\mathbf{i}+\mathbf{j})\times(\mathbf{i}-\mathbf{j})$, using Theorem 4.8
4 Check that your answer to part 2 is orthogonal to both factors.
5 Find both unit vectors orthogonal to $(1,1,0)$ and $(0,1,1)$.
6 Find the area of the parallelogram with sides $(3,0,1)$ and $(1,2,-1)$.
7 Find the area of the triangle with vertices $P(1,1,0)$, $Q(3,0,2)$, $R(0,2,1)$.
8 Find the volume of the parallelepiped with edges $(1,1,0)$, $(0,1,1)$, $(1,0,1)$.
9 Are $(1,2,3)$, $(2,-1,1)$ and $(3,1,4)$ coplanar?
10 Find the area of the triangle in the plane with vertices $(0,0)$, $(4,1)$, $(1,3)$ by placing it in the plane $z=0$ of $\mathbb{R}^3$.
Check: cross product and triple product calculator
4 $(2,4,7)\cdot(2,-1,0)=4-4+0=0$ โ and $(2,4,7)\cdot(1,3,-2)=2+12-14=0$ โ.
5 $(1,1,0)\times(0,1,1)=(1\cdot1-0\cdot1,\ 0\cdot0-1\cdot1,\ 1\cdot1-1\cdot0)=(1,-1,1)$, of length $\sqrt3$. The unit vectors are $\pm\tfrac{1}{\sqrt3}(1,-1,1)$.
6 $(3,0,1)\times(1,2,-1)=(0\cdot(-1)-1\cdot2,\ 1\cdot1-3(-1),\ 3\cdot2-0\cdot1)=(-2,4,6)$; area $=\sqrt{4+16+36}=\sqrt{56}=2\sqrt{14}\approx7.483$.
7 $\overrightarrow{PQ}=(2,-1,2)$, $\overrightarrow{PR}=(-1,1,1)$, $\overrightarrow{PQ}\times\overrightarrow{PR}=((-1)(1)-2\cdot1,\ 2(-1)-2\cdot1,\ 2\cdot1-(-1)(-1))=(-3,-4,1)$; area $=\tfrac12\sqrt{9+16+1}=\tfrac{\sqrt{26}}{2}\approx2.550$.
8 $\det\begin{bmatrix}1&1&0\\0&1&1\\1&0&1\end{bmatrix}=1(1-0)-1(0-1)+0=2$, so $V=2$.
10 $(4,1,0)\times(1,3,0)=(0,0,4\cdot3-1\cdot1)=(0,0,11)$, so the area is $\tfrac{11}{2}$. (In general the area of the triangle on $(a,b)$ and $(c,d)$ is $\tfrac12|ad-bc|$ โ a $2\times2$ determinant.)
II
Lines.
1 Write vector and parametric equations for the line through $(2,-1,4)$ with direction $(3,0,-2)$.
2 Find parametric equations of the line through $A(1,2,3)$ and $B(4,0,5)$.
3 In $\mathbb{R}^2$, write parametric equations of the line through $(1,-2)$ parallel to $(3,4)$, then eliminate $t$ to get an equation $ax+by=c$.
4 Does $(7,-2,7)$ lie on the line of part 2?
5 Show that the lines $\mathbf{x}=(1,0,2)+t(1,1,-1)$ and $\mathbf{x}=(2,3,0)+s(1,2,1)$ are skew.
6 Show that the lines $\mathbf{x}=(1,1,0)+t(2,-1,1)$ and $\mathbf{x}=(2,-1,3)+s(1,1,-2)$ intersect, and find the point.
7 Find the distance from $Q(1,1,1)$ to the line through the origin with direction $(1,2,2)$.
3 $x=1+3t$, $y=-2+4t$. Then $t=\tfrac{x-1}{3}=\tfrac{y+2}{4}$, so $4(x-1)=3(y+2)$, i.e. $4x-3y=10$. Check: $4(1)-3(-2)=10$ โ; the normal $(4,-3)$ is orthogonal to $(3,4)$ โ.
4 $x=7$ gives $t=2$; then $y=2-4=-2$ โ, $z=3+4=7$ โ. Yes.
5 The directions $(1,1,-1)$ and $(1,2,1)$ are not multiples, so the lines are not parallel. Setting the points equal: $1+t=2+s$, $t=3+2s$, $2-t=s$. The first two give $t=1+s$ and $1+s=3+2s$, so $s=-2$, $t=-1$; the third then says $2-(-1)=3\ne-2$. No common point, so the lines are skew.
6 $1+2t=2+s$, $1-t=-1+s$, $t=3-2s$. Subtracting the second from the first: $3t=3$, $t=1$, then $s=1$; the third: $1=3-2$ โ. The point is $(1+2,\,1-1,\,0+1)=(3,0,1)$ (and $(2+1,\,-1+1,\,3-2)=(3,0,1)$ โ).
1 Find the plane through $(1,-1,2)$ with normal $(3,1,-4)$.
2 Find the plane through $P(1,0,0)$, $Q(0,2,0)$, $R(0,0,3)$.
3 Find the plane through $(2,1,-1)$ parallel to $2x-y+3z=5$.
4 Find the plane that contains the point $(1,2,3)$ and the line $\mathbf{x}=(0,1,0)+t(1,0,1)$.
5 Find the distance from $(3,1,-2)$ to the plane $x+2y-2z=4$.
6 Find the distance between the parallel planes $2x-y+2z=1$ and $2x-y+2z=10$.
7 Find where the line $\mathbf{x}=(1,0,2)+t(2,1,-1)$ meets the plane $x+y+z=6$.
8 Find the line of intersection of the planes $x-y+z=1$ and $2x+y-z=5$.
9 Find the angle between the planes $x+y=1$ and $y+z=2$ (the angle between their normals).
๐ก Show solutionsโถ
1 $3(x-1)+(y+1)-4(z-2)=0$, i.e. $3x+y-4z=-6$. Check: $3-1-8=-6$ โ.
2 $\overrightarrow{PQ}=(-1,2,0)$, $\overrightarrow{PR}=(-1,0,3)$, $\mathbf{n}=(2\cdot3-0\cdot0,\ 0\cdot(-1)-(-1)\cdot3,\ (-1)\cdot0-2\cdot(-1))=(6,3,2)$. Plane: $6(x-1)+3y+2z=0$, i.e. $6x+3y+2z=6$ (equivalently $\tfrac x1+\tfrac y2+\tfrac z3=1$). All three points check โ.
3 Same normal $(2,-1,3)$: $2x-y+3z=2(2)-1+3(-1)=0$.
4 $A(0,1,0)$ is on the line; $\overrightarrow{AP}=(1,1,3)$ and $\mathbf{d}=(1,0,1)$ both lie in the plane, so $\mathbf{n}=\mathbf{d}\times\overrightarrow{AP}=(0\cdot3-1\cdot1,\ 1\cdot1-1\cdot3,\ 1\cdot1-0\cdot1)=(-1,-2,1)$. Plane: $-x-2(y-1)+z=0$, i.e. $x+2y-z=2$. Check $(1,2,3)$: $1+4-3=2$ โ; $(1,1,1)$ ($t=1$): $1+2-1=2$ โ.
5 $D=\dfrac{|3+2+4-4|}{\sqrt{1+4+4}}=\dfrac53$.
6 $(0,-1,0)$ lies on the first plane ($0+1+0=1$). Its distance to the second: $\dfrac{|0+1+0-10|}{\sqrt{4+1+4}}=\dfrac93=3$.
7 $(1+2t)+t+(2-t)=3+2t=6$, so $t=\tfrac32$ and the point is $\left(4,\tfrac32,\tfrac12\right)$. Check: $4+1.5+0.5=6$ โ.
8 $\mathbf{d}=(1,-1,1)\times(2,1,-1)=((-1)(-1)-1\cdot1,\ 1\cdot2-1\cdot(-1),\ 1\cdot1-(-1)\cdot2)=(0,3,3)$, or more simply $(0,1,1)$. Adding the equations: $3x=6$, $x=2$; then $-y+z=-1$, so take $z=0$, $y=1$. Line: $\mathbf{x}=(2,1,0)+t(0,1,1)$. Check: $2-(1+t)+t=1$ โ, $4+(1+t)-t=5$ โ.
9 Normals $(1,1,0)$ and $(0,1,1)$: $\cos\theta=\dfrac{1}{\sqrt2\sqrt2}=\dfrac12$, so $\theta=60^\circ$.
IV
Applications and proofs.
1Engineering. A force $\mathbf{F}=(0,20,15)$ N is applied at the end of a spanner whose position vector from the bolt is $\mathbf{r}=(0.3,0,0)$ m. The torque is $\boldsymbol{\tau}=\mathbf{r}\times\mathbf{F}$. Find $\boldsymbol{\tau}$ and its magnitude.
2 Prove directly from Definition 4.8 that $\mathbf{u}\cdot(\mathbf{u}\times\mathbf{v})=0$.
3 Verify Lagrange's identity $\|\mathbf{u}\times\mathbf{v}\|^2=\|\mathbf{u}\|^2\|\mathbf{v}\|^2-(\mathbf{u}\cdot\mathbf{v})^2$ for $\mathbf{u}=(1,2,-2)$, $\mathbf{v}=(3,0,1)$.
4 True or false: $\mathbf{u}\times(\mathbf{v}\times\mathbf{w})=(\mathbf{u}\times\mathbf{v})\times\mathbf{w}$ for all $\mathbf{u},\mathbf{v},\mathbf{w}$. Justify.
5Drones. A drone flies along $\mathbf{x}(t)=(0,0,10)+t(4,3,1)$ (metres, $t$ in seconds). A sloping roof lies in the plane $x+y+2z=100$. When and where does the drone reach the roof?
2 $\mathbf{u}\cdot(\mathbf{u}\times\mathbf{v})=u_1(u_2v_3-u_3v_2)+u_2(u_3v_1-u_1v_3)+u_3(u_1v_2-u_2v_1)$. Multiplying out, the six terms cancel in pairs: $u_1u_2v_3-u_2u_1v_3=0$, $-u_1u_3v_2+u_3u_1v_2=0$, $u_2u_3v_1-u_3u_2v_1=0$. $\blacksquare$ (Equivalently, it is a determinant with two equal rows.)
3 From Example 1, $\|\mathbf{u}\times\mathbf{v}\|^2=4+49+36=89$. Also $\|\mathbf{u}\|^2=9$, $\|\mathbf{v}\|^2=10$, $\mathbf{u}\cdot\mathbf{v}=3+0-2=1$, and $9\cdot10-1^2=89$ โ.
4False Counterexample: $\mathbf{i}\times(\mathbf{i}\times\mathbf{j})=\mathbf{i}\times\mathbf{k}=-\mathbf{j}$, but $(\mathbf{i}\times\mathbf{i})\times\mathbf{j}=\mathbf{0}\times\mathbf{j}=\mathbf{0}$.
5 $4t+3t+2(10+t)=9t+20=100$, so $t=\tfrac{80}{9}\approx8.89$ s, at $\left(\tfrac{320}{9},\tfrac{240}{9},\tfrac{170}{9}\right)\approx(35.56,\,26.67,\,18.89)$ m. Check: $\tfrac{320+240+340}{9}=100$ โ.
๐ฌ ResearchโTeaching NexusResearch-tutored ยท optional, about 30 minutes
Research question. How much power does a roof-mounted solar panel deliver at noon in winter and in summer, and how does the answer depend on the direction its surface faces?
Why it matters. The power a panel collects is proportional to $\cos$ of the angle between its normal and the direction of the sun. Installers in Oman, Spain, Kenya and Australia choose tilts and orientations using exactly this dot product โ and the normal comes from a cross product of two edges of the panel.
What to do.
Sample case: use axes east $=x$, north $=y$, up $=z$ (metres). A rectangular panel has corners $P(0,0,3)$, $Q(4,0,3)$ and $R(0,2,4.2)$ (the fourth corner is $Q+\overrightarrow{PR}$).
Compute $\mathbf{n}=\overrightarrow{PQ}\times\overrightarrow{PR}$. Explain why $\|\mathbf{n}\|$ is the panel's area, and find the unit normal and the tilt angle of the panel from the horizontal (the angle between $\mathbf{n}$ and $\mathbf{k}$). Which way does the panel face?
At a site near latitude $23.6^\circ$N the noon sun stands due south at elevation $\alpha\approx43^\circ$ in mid-winter and $\alpha\approx90^\circ$ in mid-summer. The unit vector towards the sun is $\mathbf{s}=(0,-\cos\alpha,\sin\alpha)$. Compute $\cos$ of the angle of incidence, $\dfrac{\mathbf{n}\cdot\mathbf{s}}{\|\mathbf{n}\|}$, in both seasons.
With sunlight of $1000$ W/mยฒ and panel efficiency $20\%$, estimate the electrical power in each season. What tilt would maximise the winter output?
Tools: Calculator or spreadsheet; optionally a free sun-position calculator to find $\alpha$ for your own town.
Hand in: A half-page note with $\mathbf{n}$, the area, the tilt, the two power estimates, and a recommendation for the tilt at your own latitude.
Going further. The same calculation (normal from a cross product, brightness from a dot product) is how 3-D games shade every triangle: look up "Lambertian shading". Or find out why panels near the equator are laid almost flat while panels in Norway stand nearly upright.
Instructor note: $\mathbf{n}=(4,0,0)\times(0,2,1.2)=(0,-4.8,8)$, area $\|\mathbf{n}\|=\sqrt{87.04}\approx9.33$ mยฒ, tilt $\cos^{-1}\tfrac{8}{9.33}\approx30.96^\circ$, facing south. Winter: $\cos i\approx0.961$, power $\approx1793$ W; summer ($\alpha=90^\circ$): $\cos i\approx0.857$, power $\approx1600$ W. Winter output is maximised when the panel faces the sun, tilt $=90^\circ-43^\circ=47^\circ$.
๐งช Self-check quiz
0/5 correct
Q1. $\mathbf{j}\times\mathbf{i}=$
Q2. $(1,2,0)\times(0,1,3)=$
Q3. The area of the parallelogram with sides $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^3$ is
Q4. A normal vector to the plane $2x-y+5z=3$ is
Q5. The distance from the origin to the plane $x+2y+2z=6$ is