A formula for finding the vector product

3 A formula for finding the vector product

We can use the results in Key Point 19 to develop a formula for finding the vector product of two vectors given in Cartesian form: Suppose $\underset{̲}{a} = a_{1} \underset{̲}{i} + a_{2} \underset{̲}{j} + a_{3} \underset{̲}{k}$ and $\underset{̲}{b} = b_{1} \underset{̲}{i} + b_{2} \underset{̲}{j} + b_{3} \underset{̲}{k}$ then

\begin{array}{rcl} \underset{̲}{a} \times \underset{̲}{b} & = & (a_{1} \underset{̲}{i} + a_{2} \underset{̲}{j} + a_{3} \underset{̲}{k}) \times (b_{1} \underset{̲}{i} + b_{2} \underset{̲}{j} + b_{3} \underset{̲}{k}) \\ = & a_{1} \underset{̲}{i} \times (b_{1} \underset{̲}{i} + b_{2} \underset{̲}{j} + b_{3} \underset{̲}{k}) \\ + a_{2} \underset{̲}{j} \times (b_{1} \underset{̲}{i} + b_{2} \underset{̲}{j} + b_{3} \underset{̲}{k}) \\ + a_{3} \underset{̲}{k} \times (b_{1} \underset{̲}{i} + b_{2} \underset{̲}{j} + b_{3} \underset{̲}{k}) \\ = & a_{1} b_{1} (\underset{̲}{i} \times \underset{̲}{i}) + a_{1} b_{2} (\underset{̲}{i} \times \underset{̲}{j}) + a_{1} b_{3} (\underset{̲}{i} \times \underset{̲}{k}) \\ + a_{2} b_{1} (\underset{̲}{j} \times \underset{̲}{i}) + a_{2} b_{2} (\underset{̲}{j} \times \underset{̲}{j}) + a_{2} b_{3} (\underset{̲}{j} \times \underset{̲}{k}) \\ + a_{3} b_{1} (\underset{̲}{k} \times \underset{̲}{i}) + a_{3} b_{2} (\underset{̲}{k} \times \underset{̲}{j}) + a_{3} b_{3} (\underset{̲}{k} \times \underset{̲}{k}) \end{array}

Using Key Point 19, this expression simplifies to

$\underset{̲}{a} \times \underset{̲}{b} = (a_{2} b_{3} - a_{3} b_{2}) \underset{̲}{i} - (a_{1} b_{3} - a_{3} b_{1}) \underset{̲}{j} + (a_{1} b_{2} - a_{2} b_{1}) \underset{̲}{k}$

This gives us Key Point 20:

Key Point 20

If $\underset{̲}{a} = a_{1} \underset{̲}{i} + a_{2} \underset{̲}{j} + a_{3} \underset{̲}{k}$ and $\underset{̲}{b} = b_{1} \underset{̲}{i} + b_{2} \underset{̲}{j} + b_{3} \underset{̲}{k}$ then

\underset{̲}{a} \times \underset{̲}{b} = (a_{2} b_{3} - a_{3} b_{2}) \underset{̲}{i} - (a_{1} b_{3} - a_{3} b_{1}) \underset{̲}{j} + (a_{1} b_{2} - a_{2} b_{1}) \underset{̲}{k}

Example 17

Evaluate the vector product $\underset{̲}{a} \times \underset{̲}{b}$ if $\underset{̲}{a} = 3 \underset{̲}{i} - 2 \underset{̲}{j} + 5 \underset{̲}{k}$ and $\underset{̲}{b} = 7 \underset{̲}{i} + 4 \underset{̲}{j} - 8 \underset{̲}{k}$ .

Solution

Identifying $a_{1} = 3$ , $a_{2} = - 2$ , $a_{3} = 5$ , $b_{1} = 7$ , $b_{2} = 4$ , $b_{3} = - 8$ we find

\begin{array}{rcl} \underset{̲}{a} \times \underset{̲}{b} & = & ((- 2) (- 8) - (5) (4)) \underset{̲}{i} - ((3) (- 8) - (5) (7)) \underset{̲}{j} + ((3) (4) - (- 2) (7)) \underset{̲}{k} \\ = & - 4 \underset{̲}{i} + 59 \underset{̲}{j} + 26 \underset{̲}{k} \end{array}

Task!

Use Key Point 20 to find the vector product of

$\underset{̲}{p} = 3 \underset{̲}{i} + 5 \underset{̲}{j}$ and $\underset{̲}{q} = 2 \underset{̲}{i} - \underset{̲}{j}$ .

Note that in this case there are no $\underset{̲}{k}$ components so $a_{3}$ and $b_{3}$ are both zero:

$- 13 \underset{̲}{k}$

3 A formula for finding the vector product

Key Point 20

Example 17

Solution

Task!

Answer