Applications of the Cross Product

The cross product gives a vector that is perpendicular to two given vectors, and its length is $|\vec{a}||\vec{b}|\sin\theta$. Those two facts make it surprisingly useful. In this lesson you’ll use them to find areas and volumes in 3-space, to measure the turning effect of a force (torque), and to find a direction at right angles to two others, which is exactly what you’ll need for the equations of planes.

Key ideas

Area of a parallelogram

The parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ has base $|\vec{a}|$ and height $|\vec{b}|\sin\theta$, so

$$\text{area of parallelogram} = |\vec{a}|\,|\vec{b}|\sin\theta = |\vec{a} \times \vec{b}|$$

In 3-space this is usually the easiest way to find an area: you don’t need to find any angles or heights first.

Area of a triangle

A triangle with sides $\vec{a}$ and $\vec{b}$ from one vertex is half of that parallelogram:

$$\text{area of triangle} = \frac{1}{2}|\vec{a} \times \vec{b}|$$

For a triangle with vertices $A$, $B$, $C$, use two sides from the same vertex: area $= \dfrac{1}{2}\left|\overrightarrow{AB} \times \overrightarrow{AC}\right|$.

Volume of a parallelepiped: the scalar triple product

A parallelepiped is a “slanted box”: six faces, each a parallelogram, with opposite faces parallel. If its edges from one corner are $\vec{a}$, $\vec{b}$, and $\vec{c}$, its volume is

$$V = \left|\vec{a} \cdot (\vec{b} \times \vec{c})\right|$$

The expression $\vec{a} \cdot (\vec{b} \times \vec{c})$ is called the scalar triple product. It’s a number (a dot product), and you take its absolute value because a volume can’t be negative.

Why it works. The base is the parallelogram formed by $\vec{b}$ and $\vec{c}$, so the base area is $|\vec{b} \times \vec{c}|$. The vector $\vec{b} \times \vec{c}$ is perpendicular to the base, so the height $h$ is the size of the scalar projection of $\vec{a}$ onto $\vec{b} \times \vec{c}$. Then

$$V = (\text{base area})(h) = |\vec{b} \times \vec{c}| \cdot \frac{\left|\vec{a} \cdot (\vec{b} \times \vec{c})\right|}{|\vec{b} \times \vec{c}|} = \left|\vec{a} \cdot (\vec{b} \times \vec{c})\right|$$

Base area $|\vec{b} \times \vec{c}|$ times height $h$ gives $V = |\vec{a} \cdot (\vec{b} \times \vec{c})|$.

Coplanar vectors. If $\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$, the “box” is flat and has no volume. That happens exactly when $\vec{a}$, $\vec{b}$, and $\vec{c}$ lie in the same plane (they are coplanar). So the scalar triple product is also a test for coplanarity.

Torque

When you push on a wrench, the turning effect on the bolt is called torque. If $\vec{r}$ is the vector from the centre of the bolt to the point where you push, and $\vec{F}$ is the force, the torque is

$$\vec{\tau} = \vec{r} \times \vec{F}, \qquad |\vec{\tau}| = |\vec{r}|\,|\vec{F}|\sin\theta$$

where $\theta$ is the angle between the handle direction $\vec{r}$ and the force. With $\vec{r}$ in metres and $\vec{F}$ in newtons, torque is in newton-metres (N·m).

Only the part of $\vec{F}$ perpendicular to the handle, $|\vec{F}|\sin\theta$, turns the bolt.

What this tells you about using a wrench:

• Push at right angles to the handle. $\sin\theta$ is largest, $1$, at $\theta = 90^\circ$. Pushing along the handle ($\theta = 0^\circ$) does nothing at all.
• Use a longer handle, or push farther from the bolt. Doubling $|\vec{r}|$ doubles the torque.
• Ratchets help. In a tight space you can only swing the handle a little. A ratchet lets you swing back without removing the wrench, so you can keep pushing in the best position, near $90^\circ$ to the handle, on every stroke.
• Direction. $\vec{\tau}$ points along the axis of the bolt (right-hand rule). For an ordinary bolt, turning it counterclockwise as you look at its head gives a torque pointing toward you, and the bolt loosens and moves toward you.

A vector normal to two given vectors

A vector perpendicular to a surface or to a pair of directions is called a normal vector. The cross product gives one directly:

$$\vec{n} = \vec{a} \times \vec{b} \text{ is perpendicular to both } \vec{a} \text{ and } \vec{b}$$

Any non-zero multiple of $\vec{n}$ is also normal, so it’s fine (and tidier) to divide out a common factor. For a unit normal, divide by the length: $\dfrac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$, or its negative. In the next unit, a normal vector is the key ingredient of the equation of a plane.

Worked examples

Example 1: Normal vector and parallelogram area

Let $\vec{a} = [3, 1, 2]$ and $\vec{b} = [1, -2, 4]$.

(a) Find a vector perpendicular to both, and a unit vector perpendicular to both.

(b) Find the area of the parallelogram with sides $\vec{a}$ and $\vec{b}$.

Solution.

(a)

$$\begin{aligned} \vec{a} \times \vec{b} &= [\,1(4) - 2(-2),\ \ 2(1) - 3(4),\ \ 3(-2) - 1(1)\,] \\ &= [4 + 4,\ 2 - 12,\ -6 - 1] \\ &= [8, -10, -7] \end{aligned}$$

Check: $[8, -10, -7] \cdot [3, 1, 2] = 24 - 10 - 14 = 0$ ✓ and $[8, -10, -7] \cdot [1, -2, 4] = 8 + 20 - 28 = 0$ ✓

Its length is $\sqrt{64 + 100 + 49} = \sqrt{213}$, so a unit normal is $\dfrac{1}{\sqrt{213}}[8, -10, -7]$ (its negative works too).

(b) Area $= |\vec{a} \times \vec{b}| = \sqrt{213} \approx 14.59$ square units.

Example 2: Area of a triangle from its vertices

Find the area of the triangle with vertices $P(1, 0, 2)$, $Q(3, 1, 1)$, and $R(0, 2, 4)$.

Solution. Use two sides from $P$:

$$\overrightarrow{PQ} = [2, 1, -1], \qquad \overrightarrow{PR} = [-1, 2, 2]$$ $$\overrightarrow{PQ} \times \overrightarrow{PR} = [\,1(2) - (-1)(2),\ \ (-1)(-1) - 2(2),\ \ 2(2) - 1(-1)\,] = [4, -3, 5]$$ $$\text{area} = \frac{1}{2}\sqrt{16 + 9 + 25} = \frac{1}{2}\sqrt{50} = \frac{5\sqrt{2}}{2} \approx 3.54 \text{ square units}$$

Example 3: Volume, and testing for coplanarity

(a) Find the volume of the parallelepiped with edges $\vec{a} = [2, 0, 1]$, $\vec{b} = [1, 3, 0]$, and $\vec{c} = [0, 1, 4]$.

(b) Are $\vec{u} = [1, 2, 3]$, $\vec{v} = [2, 1, 0]$, and $\vec{w} = [4, 5, 6]$ coplanar?

Solution.

(a) First the cross product, then the dot product:

$$\vec{b} \times \vec{c} = [\,3(4) - 0(1),\ \ 0(0) - 1(4),\ \ 1(1) - 3(0)\,] = [12, -4, 1]$$ $$\vec{a} \cdot (\vec{b} \times \vec{c}) = 2(12) + 0(-4) + 1(1) = 25$$

The volume is $|25| = 25$ cubic units.

(b)

$$\vec{v} \times \vec{w} = [\,1(6) - 0(5),\ \ 0(4) - 2(6),\ \ 2(5) - 1(4)\,] = [6, -12, 6]$$ $$\vec{u} \cdot (\vec{v} \times \vec{w}) = 6 - 24 + 18 = 0$$

The scalar triple product is $0$, so the vectors are coplanar. In fact $\vec{w} = 2\vec{u} + \vec{v}$: check $2[1, 2, 3] + [2, 1, 0] = [4, 5, 6]$. ✓

Example 4: Torque on a wrench

You push with a force of $80$ N on the end of a wrench $25$ cm long.

(a) Find the magnitude of the torque if the force makes an angle of $70^\circ$ with the handle, to two decimal places.

(b) What is the greatest torque you can get with this force, and how?

Solution. Convert to metres: $|\vec{r}| = 0.25$ m.

(a)

$$|\vec{\tau}| = |\vec{r}||\vec{F}|\sin\theta = (0.25)(80)\sin 70^\circ = 20\sin 70^\circ \approx 18.79 \text{ N·m}$$

(b) The torque is largest when $\sin\theta = 1$, that is, when you push at $90^\circ$ to the handle. Then $|\vec{\tau}| = (0.25)(80)(1) = 20$ N·m. Pushing at $70^\circ$ already gets you about $94\%$ of the maximum, so you don’t have to be exact.

Common mistakes

Forgetting the one-half for a triangle. $|\vec{a} \times \vec{b}|$ is the area of the parallelogram. A triangle with the same two sides has half that area.

Using position vectors instead of side vectors. The vertices $P(1, 0, 2)$ and $Q(3, 1, 1)$ are points. Crossing $[1, 0, 2]$ and $[3, 1, 1]$ finds the area of a parallelogram with a corner at the origin, which is not your triangle. Subtract first to get the sides.

Leaving a volume negative. The scalar triple product can be negative (it depends on the order of the vectors), but volume is $|\vec{a} \cdot (\vec{b} \times \vec{c})|$. Take the absolute value.

Computing in the wrong order in the triple product. Do the cross product first, then the dot product. $(\vec{a} \cdot \vec{b}) \times \vec{c}$ makes no sense, because $\vec{a} \cdot \vec{b}$ is a number.

Mixing units in torque. A $25$ cm wrench is $0.25$ m. Using $25$ gives an answer $100$ times too big. Torque is in N·m.

Practice

1. (Warm-up) Find the area of the parallelogram with sides $\vec{a} = [1, 2, 0]$ and $\vec{b} = [3, 1, 0]$.

Solution

$$\vec{a} \times \vec{b} = [\,2(0) - 0(1),\ \ 0(3) - 1(0),\ \ 1(1) - 2(3)\,] = [0, 0, -5]$$

Area $= |[0, 0, -5]| = 5$ square units. (Both vectors lie in the $xy$-plane, so their cross product points along the $z$-axis.)

2. (Warm-up) A $40$ N force is applied to the end of a wrench $0.30$ m long. Find the magnitude of the torque if the force is (a) perpendicular to the handle and (b) at $30^\circ$ to the handle.

Solution

(a) $|\vec{\tau}| = (0.30)(40)\sin 90^\circ = 12$ N·m

(b) $|\vec{\tau}| = (0.30)(40)\sin 30^\circ = 12 \times \dfrac{1}{2} = 6$ N·m

3. (Core) Find the area of the triangle with vertices $A(2, -1, 0)$, $B(4, 1, 1)$, and $C(1, 3, 2)$.

Solution

$\overrightarrow{AB} = [2, 2, 1]$ and $\overrightarrow{AC} = [-1, 4, 2]$.

$$\overrightarrow{AB} \times \overrightarrow{AC} = [\,2(2) - 1(4),\ \ 1(-1) - 2(2),\ \ 2(4) - 2(-1)\,] = [0, -5, 10]$$$$\text{area} = \frac{1}{2}\sqrt{0 + 25 + 100} = \frac{1}{2}\sqrt{125} = \frac{5\sqrt{5}}{2} \approx 5.59 \text{ square units}$$

4. (Core) Find the two unit vectors that are perpendicular to both $[1, -1, 2]$ and $[3, 0, 1]$.

Solution

$$[1, -1, 2] \times [3, 0, 1] = [\,(-1)(1) - 2(0),\ \ 2(3) - 1(1),\ \ 1(0) - (-1)(3)\,] = [-1, 5, 3]$$

Check: $[-1, 5, 3] \cdot [1, -1, 2] = -1 - 5 + 6 = 0$ ✓ and $[-1, 5, 3] \cdot [3, 0, 1] = -3 + 0 + 3 = 0$ ✓

Its length is $\sqrt{1 + 25 + 9} = \sqrt{35}$, so the unit vectors are

$$\frac{1}{\sqrt{35}}[-1, 5, 3] \qquad\text{and}\qquad \frac{1}{\sqrt{35}}[1, -5, -3]$$

5. (Core) Find the volume of the parallelepiped with edges $\vec{a} = [1, 1, 0]$, $\vec{b} = [0, 2, 1]$, and $\vec{c} = [3, 0, 2]$.

Solution

$$\vec{b} \times \vec{c} = [\,2(2) - 1(0),\ \ 1(3) - 0(2),\ \ 0(0) - 2(3)\,] = [4, 3, -6]$$$$\vec{a} \cdot (\vec{b} \times \vec{c}) = 4 + 3 + 0 = 7$$

The volume is $7$ cubic units.

6. (Core) Find the value of $k$ for which $[1, 2, k]$, $[2, 0, 1]$, and $[3, 1, 2]$ are coplanar.

Solution

The vectors are coplanar when the scalar triple product is $0$.

$$[2, 0, 1] \times [3, 1, 2] = [\,0(2) - 1(1),\ \ 1(3) - 2(2),\ \ 2(1) - 0(3)\,] = [-1, -1, 2]$$$$[1, 2, k] \cdot [-1, -1, 2] = -1 - 2 + 2k = 0 \quad\Rightarrow\quad k = \frac{3}{2}$$

7. (Core) A wrench lies along the $x$-axis, with $\vec{r} = [0.2, 0, 0]$ m from the bolt to your hand. You push with force $\vec{F} = [0, 30, -40]$ N. Find the torque vector $\vec{\tau} = \vec{r} \times \vec{F}$ and its magnitude. Check the magnitude with $|\vec{r}||\vec{F}|\sin\theta$.

Solution

$$\vec{\tau} = [\,0(-40) - 0(30),\ \ 0(0) - 0.2(-40),\ \ 0.2(30) - 0(0)\,] = [0, 8, 6]$$

$|\vec{\tau}| = \sqrt{0 + 64 + 36} = 10$ N·m.

Check: $|\vec{r}| = 0.2$ and $|\vec{F}| = \sqrt{900 + 1600} = 50$. Since $\vec{r} \cdot \vec{F} = 0$, the force is perpendicular to the handle, so $\sin\theta = 1$ and $|\vec{\tau}| = (0.2)(50)(1) = 10$ N·m. ✓

8. (Challenge) Parallelogram $ABCD$ has vertices $A(1, 1, 1)$, $B(3, 2, 1)$, and $D(2, 1, 3)$, with $C$ opposite $A$. Find the coordinates of $C$ and the area of the parallelogram.

Solution

In a parallelogram, $\overrightarrow{BC} = \overrightarrow{AD}$. Since $\overrightarrow{AD} = [1, 0, 2]$,

$$C = B + \overrightarrow{AD} = (3 + 1,\ 2 + 0,\ 1 + 2) = (4, 2, 3)$$

The sides from $A$ are $\overrightarrow{AB} = [2, 1, 0]$ and $\overrightarrow{AD} = [1, 0, 2]$:

$$\overrightarrow{AB} \times \overrightarrow{AD} = [\,1(2) - 0(0),\ \ 0(1) - 2(2),\ \ 2(0) - 1(1)\,] = [2, -4, -1]$$$$\text{area} = \sqrt{4 + 16 + 1} = \sqrt{21} \approx 4.58 \text{ square units}$$

9. (Challenge) A rusty bolt needs a torque of at least $30$ N·m to loosen. You use a wrench $20$ cm long and can push with a force of $180$ N. For which angles between the force and the handle will the bolt loosen? Give the angles to one decimal place.

Solution

You need $|\vec{r}||\vec{F}|\sin\theta \ge 30$:

$$(0.20)(180)\sin\theta \ge 30 \quad\Rightarrow\quad 36\sin\theta \ge 30 \quad\Rightarrow\quad \sin\theta \ge \frac{5}{6}$$

$\sin^{-1}\left(\dfrac{5}{6}\right) \approx 56.4^\circ$, and the sine is also $\dfrac{5}{6}$ at $180^\circ - 56.4^\circ = 123.6^\circ$. Between those angles the sine is larger, so the bolt loosens for angles from about $56.4^\circ$ to $123.6^\circ$. (Even at the best angle, $90^\circ$, you only get $36$ N·m, so there’s not much room for error. A longer wrench would help.)