4 Using determinants to evaluate a vector product

Evaluation of a vector product using the formula in Key Point 20 is very cumbersome. A more convenient and easily remembered method is to use determinants. Recall from Workbook 7 that, for a $3\times 3$ determinant,

$\left|\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill c\hfill \\ \hfill d\hfill & \hfill e\hfill & \hfill f\hfill \\ \hfill g\hfill & \hfill h\hfill & \hfill i\hfill \end{array}\right|=a\left|\begin{array}{cc}\hfill e\hfill & \hfill f\hfill \\ \hfill h\hfill & \hfill i\hfill \\ \hfill \hfill \end{array}\right|-b\left|\begin{array}{cc}\hfill d\hfill & \hfill f\hfill \\ \hfill g\hfill & \hfill i\hfill \\ \hfill \hfill \end{array}\right|+c\left|\begin{array}{cc}\hfill d\hfill & \hfill e\hfill \\ \hfill g\hfill & \hfill h\hfill \\ \hfill \hfill \end{array}\right|$

The vector product of two vectors $\underset{̲}{a}=a_1\underset{̲}{i}+a_2\underset{̲}{j}+a_3\underset{̲}{k}$ and $\underset{̲}{b}=b_1\underset{̲}{i}+b_2\underset{̲}{j}+b_3\underset{̲}{k}$ can be found by evaluating the determinant:

$\underset{̲}{a}\times \underset{̲}{b}=\left|\begin{array}{ccc}\hfill \underset{̲}{i}\hfill & \hfill \underset{̲}{j}\hfill & \hfill \underset{̲}{k}\hfill \\ \hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill b_3\hfill \\ \hfill \hfill \end{array}\right|$

in which $\underset{̲}{i}$ , $\underset{̲}{j}$ and $\underset{̲}{k}$ are (temporarily) treated as if they were scalars.

Example 18

Find the vector product of $\underset{̲}{a}=3\underset{̲}{i}-4\underset{̲}{j}+2\underset{̲}{k}$ and $\underset{̲}{b}=9\underset{̲}{i}-6\underset{̲}{j}+2\underset{̲}{k}$ .

Solution

We have $\underset{̲}{a}\times \underset{̲}{b}=\left|\begin{array}{ccc}\hfill \underset{̲}{i}\hfill & \hfill \underset{̲}{j}\hfill & \hfill \underset{̲}{k}\hfill \\ \hfill 3\hfill & \hfill -4\hfill & \hfill 2\hfill \\ \hfill 9\hfill & \hfill -6\hfill & \hfill 2\hfill \\ \hfill \hfill \end{array}\right|$ which, when evaluated, gives

$\underset{̲}{a}\times \underset{̲}{b}=\underset{̲}{i}\left(-8-\left(-12\right)\right)-\underset{̲}{j}\left(6-18\right)+\underset{̲}{k}\left(-18-\left(-36\right)\right)=4\underset{̲}{i}+12\underset{̲}{j}+18\underset{̲}{k}$

Example 19

The area $A_T$ of the triangle shown in Figure 43 is given by the formula

$A_T=\frac{1}{2}bcsin\alpha$ . Show that an equivalent formula is $A_T=\frac{1}{2}\left|\overrightarrow{AB}\times \overrightarrow{AC}\right|$ .

Figure 43

Solution

We use the definition of the vector product $|\overrightarrow{AB}\times \overrightarrow{AC}|=\left|\overrightarrow{AB}\right|\left|\overrightarrow{AC}\right|sin\alpha .$

Since $\alpha$ is the angle between $\overrightarrow{AB}$ and $\overrightarrow{AC}$ , and $\left|\overrightarrow{AB}\right|=c$ and $\left|\overrightarrow{AC}\right|=b$ , the required result follows immediately:

$\frac{1}{2}\left|\overrightarrow{AB}\times \overrightarrow{AC}\right|=\frac{1}{2}c\cdot b\cdot sin\alpha$ .

4.1 Moments

The moment (or torque ) of the force $\underset{̲}{F}$ about a point $O$ is defined as

${\underset{̲}{M}}_o=\underset{̲}{r}\times \underset{̲}{F}$

where $\underset{̲}{r}$ is a position vector from $O$ to any point on the line of action of $\underset{̲}{F}$ as shown in Figure 44.

Figure 44

It may seem strange that any point on the line of action may be taken but it is easy to show that exactly the same vector ${\underset{̲}{M}}_o$ is always obtained.

By the properties of the cross product the direction of ${\underset{̲}{M}}_o$ is perpendicular to the plane containing $\underset{̲}{r}$ and $\underset{̲}{F}$ (i.e. out of the paper). The magnitude of the moment is

$\left|{\underset{̲}{M}}_0\right|=\left|\underset{̲}{r}\right|\left|\underset{̲}{F}\right|sin\theta$ .

From Figure 32, $\left|\underset{̲}{r}\right|sin\theta =D$ . Hence $\left|{\underset{̲}{M}}_o\right|=D\left|\underset{̲}{F}\right|$ . This would be the same no matter which point on the line of action of $\underset{̲}{F}$ was chosen.

Example 20

Find the moment of the force given by $\underset{̲}{F}=3\underset{̲}{i}+4\underset{̲}{j}+5\underset{̲}{k}$ (N) acting at the point $\left(14,-3,6\right)$ about the point $P\left(2,-2,1\right)$ .

Figure 45

Solution

The vector $\underset{̲}{r}$ can be any vector from the point $P$ to any point on the line of action of $\underset{̲}{F}$ . Choosing $\underset{̲}{r}$ to be the vector connecting $P$ to $\left(14,-3,6\right)$ (and measuring distances in metres) we have:

$\underset{̲}{r}=\left(14-2\right)\underset{̲}{i}+\left(-3-\left(-2\right)\right)\underset{̲}{j}+\left(6-1\right)\underset{̲}{k}=12\underset{̲}{i}-\underset{̲}{j}+5\underset{̲}{k}$ .

The moment is $\underset{̲}{M}=\underset{̲}{r}\times \underset{̲}{F}=\left|\begin{array}{ccc}\hfill \underset{̲}{i}\hfill & \hfill \underset{̲}{j}\hfill & \hfill \underset{̲}{k}\hfill \\ \hfill 12\hfill & \hfill -1\hfill & \hfill 5\hfill \\ \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill \end{array}\right|=-25\underset{̲}{i}-45\underset{̲}{j}+51\underset{̲}{k}$  (N m)

Exercises

1. Show that if $\underset{̲}{a}$ and $\underset{̲}{b}$ are parallel vectors then their vector product is the zero vector.
2. Find the vector product of $\underset{̲}{p}=-2\underset{̲}{i}-3\underset{̲}{j}$ and $\underset{̲}{q}=4\underset{̲}{i}+7\underset{̲}{j}$ .
3. If $\underset{̲}{a}=\underset{̲}{i}+2\underset{̲}{j}+3\underset{̲}{k}$ and $\underset{̲}{b}=4\underset{̲}{i}+3\underset{̲}{j}+2\underset{̲}{k}$ find $\underset{̲}{a}\times \underset{̲}{b}$ . Show that $\underset{̲}{a}\times \underset{̲}{b}\ne \underset{̲}{b}\times \underset{̲}{a}$ .
4. Points $A$ , $B$ and $C$ have coordinates $\left(9,1,-2\right)$ , (3,1,3), and $\left(1,0,-1\right)$ respectively. Find the vector product $\overrightarrow{AB}\times \overrightarrow{AC}$ .
5. Find a vector which is perpendicular to both of the vectors $\underset{̲}{a}=\underset{̲}{i}+2\underset{̲}{j}+7\underset{̲}{k}$ and $\underset{̲}{b}=\underset{̲}{i}+\underset{̲}{j}-2\underset{̲}{k}$ . Hence find a unit vector which is perpendicular to both $\underset{̲}{a}$ and $\underset{̲}{b}$ .
6. Find a vector which is perpendicular to the plane containing $6\underset{̲}{i}+\underset{̲}{k}$ and $2\underset{̲}{i}+\underset{̲}{j}$ .
7. For the vectors $\underset{̲}{a}=4\underset{̲}{i}+2\underset{̲}{j}+\underset{̲}{k}$ , $\underset{̲}{b}=\underset{̲}{i}-2\underset{̲}{j}+\underset{̲}{k}$ , and $\underset{̲}{c}=3\underset{̲}{i}-3\underset{̲}{j}+4\underset{̲}{k}$ , evaluate both $\underset{̲}{a}\times \left(\underset{̲}{b}\times \underset{̲}{c}\right)$ and $\left(\underset{̲}{a}\times \underset{̲}{b}\right)\times \underset{̲}{c}$ . Deduce that, in general, the vector product is not associative.
8. Find the area of the triangle with vertices at the points with coordinates $\left(1,2,3\right)$ , $\left(4,-3,2\right)$ and $\left(8,1,1\right)$ .
9. For the vectors $\underset{̲}{r}=\underset{̲}{i}+2\underset{̲}{j}+3\underset{̲}{k}$ , $\underset{̲}{s}=2\underset{̲}{i}-2\underset{̲}{j}-5\underset{̲}{k}$ , and $\underset{̲}{t}=\underset{̲}{i}-3\underset{̲}{j}-\underset{̲}{k}$ , evaluate
   1. $\left(\underset{̲}{r}\cdot \underset{̲}{t}\right)\underset{̲}{s}-\left(\underset{̲}{s}\cdot \underset{̲}{t}\right)\underset{̲}{r}$ .
   2. $\left(\underset{̲}{r}\times \underset{̲}{s}\right)\times \underset{̲}{t}$ . Deduce that $\left(\underset{̲}{r}\cdot \underset{̲}{t}\right)\underset{̲}{s}-\left(\underset{̲}{s}\cdot \underset{̲}{t}\right)\underset{̲}{r}=\left(\underset{̲}{r}\times \underset{̲}{s}\right)\times \underset{̲}{t}$ .

Answer