# What is the cross product of <2 , 5 ,-3 > and <3 ,5 ,-2 >?

Jan 28, 2016

To find the cross product of two vectors, $\left[{a}_{1} , {a}_{2} , {a}_{3}\right]$ and $\left[{b}_{1} , {b}_{2} , {b}_{3}\right]$, which we will call the new vector $\left[{c}_{1} , {c}_{2} , {c}_{3}\right]$, we multiply and subtract as follows (careful, it's a little tricky!):

To give the first element of the cross product, ignore the first elements of each of the vectors being multiplied and multiply and subtract the remaining elements as follows:

${c}_{1} = {a}_{2} \cdot {b}_{3} - {a}_{3} \cdot {b}_{2}$

The second element is the trickiest. Ignore the second elements of each vector, and multiply and subtract as shown:

${c}_{2} = {a}_{3} \cdot {b}_{1} - {a}_{1} \cdot {b}_{3}$

Carefully note the order.

Finally, to find the third element of the resultant vector, ignore the third element of the two component vectors, and multiply and subtract as shown:

${c}_{3} = {a}_{1} \cdot {b}_{2} - {a}_{2} \cdot {b}_{1}$

Putting it all together:

$\left[{a}_{1} , {a}_{2} , {a}_{3}\right] \times \left[{b}_{1} , {b}_{2} , {b}_{3}\right]$
= $\left[\left({a}_{2} \cdot {b}_{3} - {a}_{3} \cdot {b}_{2}\right) + \left({a}_{3} \cdot {b}_{1} - {a}_{1} \cdot {b}_{3}\right) + \left({a}_{1} \cdot {b}_{2} - {a}_{2} \cdot {b}_{1}\right)\right]$
= $\left[{c}_{1} , {c}_{2} , {c}_{3}\right]$

[2, 5, −3] xx [3, 5, −2]
$= \left[\left(5 \cdot \left(- 2\right) - \left(- 3\right) \cdot 5\right) + \left(\left(- 3\right) \cdot 3 - 2 \cdot \left(- 2\right)\right) + \left(2 \cdot 5 - 5 \cdot 3\right)\right]$
= $\left[\left(\left(- 10\right) - \left(- 15\right)\right) + \left(\left(- 9\right) - \left(- 4\right)\right) + \left(10 - 15\right)\right]$
= $\left[5 , - 5 , - 5\right]$