10.4 The Vector (Cross) Product

An engineer installing a solar panel on a roof in Jodhpur knows two edges of the panel as vectors — one along the ridge of the roof, one running down the slope. Two questions decide how much electricity it will make: how large is the panel? and which way does it face? The second question asks for a direction perpendicular to both edges — the panel's normal. The dot product cannot give it: it turns two vectors into a number.

The vector product (or cross product) $\vec a\times\vec b$ answers both questions at once. It is a vector perpendicular to $\vec a$ and $\vec b$, pointing the way your right thumb points when your fingers curl from $\vec a$ to $\vec b$, and its length $|\vec a|\,|\vec b|\sin\theta$ is exactly the area of the parallelogram with sides $\vec a$ and $\vec b$. In components it is a $3\times3$ determinant of the kind you expanded in Section 4.1. With it you can find areas of triangles in space, unit vectors perpendicular to two given vectors, and test points for collinearity — and in Chapter 11 — Three Dimensional Geometry it gives the direction of the shortest distance between two lines.

The Vector Product

Three non-coplanar vectors $\vec a,\vec b,\vec c$ drawn from one point form a right-handed system if, when the fingers of your right hand curl from $\vec a$ towards $\vec b$ (through the angle less than $180^\circ$), your thumb points to the side of the plane of $\vec a,\vec b$ where $\vec c$ lies. The axes $\hat i,\hat j,\hat k$ form a right-handed system.

Why. (a) is built into the definition. (b) Curling the fingers from $\vec b$ to $\vec a$ instead of from $\vec a$ to $\vec b$ turns the thumb round, so $\hat n$ becomes $-\hat n$ while the length $|\vec a|\,|\vec b|\sin\theta$ is unchanged. (c) $\sin\theta=0$ exactly when $\theta=0$ or $\pi$. (d) For $\lambda\gt0$ only the length changes, by the factor $\lambda$; for $\lambda\lt0$ the vector $\lambda\vec a$ points the other way, the right-hand rule reverses the normal, and the length is multiplied by $|\lambda|$ — net factor $\lambda$. (f) $\hat i,\hat j$ are perpendicular unit vectors, so $|\hat i\times\hat j|=1\cdot1\cdot\sin90^\circ=1$, and the right-hand rule from $\hat i$ to $\hat j$ points along $\hat k$; the other two follow by going round the cycle $\hat i\to\hat j\to\hat k\to\hat i$, and reversing the order changes the sign by (b). $\square$

Memory aid for (f): write $\hat i,\hat j,\hat k$ round a circle clockwise. The product of two neighbours taken with the arrows is the third; taken against the arrows it is minus the third.

The Cross Product in Components

Proof. Expand $(a_1\hat i+a_2\hat j+a_3\hat k)\times(b_1\hat i+b_2\hat j+b_3\hat k)$ by Theorem 10.12(d), (e) into nine terms. The three terms $a_1b_1(\hat i\times\hat i)$, … vanish. The others are $a_1b_2\hat k-a_1b_3\hat j-a_2b_1\hat k+a_2b_3\hat i+a_3b_1\hat j-a_3b_2\hat i$ by the table (f). Collecting: $(a_2b_3-a_3b_2)\hat i+(a_3b_1-a_1b_3)\hat j$ $+(a_1b_2-a_2b_1)\hat k$, which is the expansion of the determinant along its first row (signs $+,-,+$ as in Section 4.1). $\square$

Two consequences are worth stating. The angle between $\vec a$ and $\vec b$ can now be found from a sine as well as a cosine, $\sin\theta=\dfrac{|\vec a\times\vec b|}{|\vec a|\,|\vec b|}$; and since $\sin^2\theta+\cos^2\theta=1$, the two products are tied together by Lagrange's identity $$|\vec a\times\vec b|^2+(\vec a\cdot\vec b)^2=|\vec a|^2\,|\vec b|^2.$$

Areas of Parallelograms and Triangles

Proof. (a) Take $\vec a$ as the base. The height is the distance of the tip of $\vec b$ from the line of $\vec a$, which is $|\vec b|\sin\theta$. So the area is $|\vec a|\,|\vec b|\sin\theta=|\vec a\times\vec b|$. (b) If the sides are $\vec a,\vec b$, the diagonals are $\vec d_1=\vec a+\vec b$ and $\vec d_2=\vec b-\vec a$. Then $\vec d_1\times\vec d_2=\vec a\times\vec b-\vec a\times\vec a+\vec b\times\vec b-\vec b\times\vec a$ $=2\,\vec a\times\vec b$, so $|\vec a\times\vec b|=\tfrac12|\vec d_1\times\vec d_2|$. (c) The diagonal $BC$ cuts the parallelogram on $\overrightarrow{AB},\overrightarrow{AC}$ into two congruent triangles. (d) The points are collinear exactly when the triangle is flat, i.e. has area $0$ (or, by Theorem 10.12(c), when $\overrightarrow{AB}\parallel\overrightarrow{AC}$). $\square$

For points in the $xy$-plane, (c) gives $\tfrac12|(x_2-x_1)(y_3-y_1)-(x_3-x_1)(y_2-y_1)|$ — the same number as the determinant formula for the area of a triangle in Section 4.1. The cross product extends that formula to triangles anywhere in space.

Exercises