Cross product, lines and planes

This week we learn the second product of vector algebra and the geometry built with it. The dot product gave a number and measured angle and perpendicularity. The cross product, on the other hand, produces a new vector perpendicular to both vectors. Its length gives area as well. It is the shortest way to write the equation of a plane, find the area of a triangle, and compute the distance from a point to a line. The exam on this topic usually has one “find the plane through three points” and one distance or intersection question.

Golden rules:

1. The cross product is defined only in three dimensions and its result is a vector.
2. In the determinant expansion, the middle $\mathbf j$ term has a minus in front.
3. A line needs a point + a direction vector, a plane needs a point + a normal vector.
4. A plane from three points: $\mathbf n = \overrightarrow{PQ}\times\overrightarrow{PR}$.
5. Check every result: the cross product must be perpendicular to both vectors, and the plane equation must be satisfied by all the given points.

The cross product

Definition.

$\mathbf u\times\mathbf v = \begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\u_1&u_2&u_3\\v_1&v_2&v_3\end{vmatrix} = (u_2v_3-u_3v_2)\,\mathbf i-(u_1v_3-u_3v_1)\,\mathbf j+(u_1v_2-u_2v_1)\,\mathbf k.$

Geometry.

• $\mathbf u\times\mathbf v$ is perpendicular to both $\mathbf u$ and $\mathbf v$.
• Its direction is found with the right-hand rule: as the fingers of your right hand curl from $\mathbf u$ to $\mathbf v$, the thumb points in the direction of $\mathbf u\times\mathbf v$.
• $\lvert\mathbf u\times\mathbf v\rvert = \lvert\mathbf u\rvert\lvert\mathbf v\rvert\sin\theta$, which is the area of the parallelogram formed by $\mathbf u$ and $\mathbf v$.

Properties. $\mathbf v\times\mathbf u = -(\mathbf u\times\mathbf v)$ (NOT commutative). $\mathbf u\times\mathbf u = \mathbf 0$. The cross product of parallel vectors is $\mathbf 0$. $\mathbf i\times\mathbf j = \mathbf k$, $\mathbf j\times\mathbf k = \mathbf i$, $\mathbf k\times\mathbf i = \mathbf j$ (cyclic order; minus in the reverse order).

Example 1. $\langle2,1,-1\rangle\times\langle1,3,2\rangle$.

i component:   1·2 - (-1)·3 = 2 + 3 = 5
j component: -(2·2 - (-1)·1) = -(4 + 1) = -5
k component:   2·3 - 1·1 = 5
result: <5, -5, 5>
check: <5,-5,5>·<2,1,-1> = 10 - 5 - 5 = 0   ✓
       <5,-5,5>·<1,3,2>  = 5 - 15 + 10 = 0   ✓

Area and volume

Example 2 (area of a triangle). $P(1,1,0)$, $Q(3,1,2)$, $R(1,4,1)$. $\overrightarrow{PQ} = \langle2,0,2\rangle$, $\overrightarrow{PR} = \langle0,3,1\rangle$.

$\overrightarrow{PQ}\times\overrightarrow{PR} = \langle0\cdot1-2\cdot3,\;-(2\cdot1-2\cdot0),\;2\cdot3-0\rangle = \langle-6,-2,6\rangle,\qquad\text{Area} = \frac{\sqrt{76}}{2} = \sqrt{19}.$

Triple scalar product. $\mathbf u\cdot(\mathbf v\times\mathbf w)$ is the $3\times3$ determinant whose rows are $\mathbf u,\mathbf v,\mathbf w$. Its absolute value is the volume of the parallelepiped formed by these three vectors. If it is zero, the three vectors are in the same plane (coplanar).

Example 3. $\langle1,2,0\rangle,\langle0,1,3\rangle,\langle2,0,1\rangle$: $1(1-0)-2(0-6)+0 = 1+12 = 13$. The volume is $13$.

Lines in space

A line is determined by a point $P_0(x_0,y_0,z_0)$ on it and a direction vector $\mathbf v = \langle a,b,c\rangle$.

$\mathbf r(t) = \mathbf r_0+t\mathbf v\qquad\Longleftrightarrow\qquad x = x_0+at,\quad y = y_0+bt,\quad z = z_0+ct.$

The line through two points: the direction is $\overrightarrow{P_0P_1}$. The line through $(2,0,-1)$ and $(4,3,1)$: $\mathbf v = \langle2,3,2\rangle$, $x = 2+2t$, $y = 3t$, $z = -1+2t$. $t=0$ gives the first point, $t=1$ the second.

The relationship of two lines. If the directions are parallel, the lines are either parallel or coincident. If they are not parallel, they either intersect or are skew: neither parallel nor intersecting. To find the intersection, write the equalities of $x$, $y$, $z$ ($t$ and $s$ are different parameters!), solve $t,s$ from two of them, and test in the third.

Example 4 (skew lines). $L_1: (1+t,\,t,\,0)$ and $L_2: (0,\,1+s,\,3+s)$. From $z$, $0 = 3+s$, $s=-3$; from $y$, $t = 1+s = -2$; in $x$, $1+t = -1\neq0$. They do not intersect; the directions $\langle1,1,0\rangle$ and $\langle0,1,1\rangle$ are not parallel. Skew. The distance between them is $\dfrac{\lvert\overrightarrow{P_1P_2}\cdot(\mathbf v_1\times\mathbf v_2)\rvert}{\lvert\mathbf v_1\times\mathbf v_2\rvert} = \dfrac{\lvert\langle-1,1,3\rangle\cdot\langle1,-1,1\rangle\rvert}{\sqrt3} = \dfrac{1}{\sqrt3}$.

Planes in space

A plane is determined by a point $P_0$ on it and a normal vector $\mathbf n = \langle a,b,c\rangle$ perpendicular to the plane. For every $P$ in the plane, $\overrightarrow{P_0P}\perp\mathbf n$:

$a(x-x_0)+b(y-y_0)+c(z-z_0) = 0\qquad\Longleftrightarrow\qquad ax+by+cz = d.$

The coefficients of $x$, $y$, $z$ directly give the normal vector. The normal of the plane $2x-y+3z = 7$ is $\langle2,-1,3\rangle$.

Example 5 (a plane from three points). For $P$, $Q$, $R$ of Example 2: the normal is $\langle-6,-2,6\rangle\parallel\langle3,1,-3\rangle$. With $P$: $3(x-1)+(y-1)-3z = 0$, that is, $3x+y-3z = 4$. $Q$: $9+1-6 = 4$ ✓. $R$: $3+4-3 = 4$ ✓.

Computations involving planes

Example 6. The distance from the point $(1,2,3)$ to the plane $2x+y-2z = 5$: $\dfrac{\lvert2+2-6-5\rvert}{3} = \dfrac73$.

Example 7 (parallel planes). Between $x+2y+2z = 3$ and $x+2y+2z = 12$: take a point from the first plane, $(3,0,0)$; its distance to the second is $\dfrac{\lvert3-12\rvert}{3} = 3$. Shortcut: $\dfrac{\lvert d_1-d_2\rvert}{\lvert\mathbf n\rvert}$ (if the normals are written the same).

PITFALL · line or plane? $x+2y+2z = 3$ is a single equation, a plane in space. A line in space cannot be written with a single linear equation; it is written either parametrically or as the intersection of two planes. In two dimensions the “equation of a line” $ax+by = c$ is, in three dimensions, a plane parallel to the $z$ axis.

Cylinders: an introduction to surfaces

If one of the variables is missing from the equation, the surface is a cylinder extending along that axis. $x^2+y^2 = 4$ is a circular cylinder of radius $2$ parallel to the $z$ axis in space; at every height $z$ it gives the same circle. $z = y^2$ is a parabolic cylinder extending along the $x$ axis. Next week we will move on to second-degree surfaces (quadric surfaces) in which all three variables appear.

Common mistakes