# How do you find the cross product and state whether the resulting vectors are perpendicular to the given vectors <3,0,4>times<-1,5,2>?

Jan 13, 2017

I use a 5 column determinant to compute the cross-product:
a xx b =| (hati,hatj,hatk,hati,hatj), (a_x,a_y,a_z,a_x,a_y), (b_x,b_y,b_z,b_x,b_y) |

#### Explanation:

Make the 3 dimensional determinant with 5 columns so that it is easier to visualize what is being computed:

<3,0,4>xx<-1,5,2> =| (hati,hatj,hatk,hati,hatj), (3,0,4,3,0), (-1,5,2,-1,5) |

Please observe that columns 4 and 5 are copies of columns 1 and 2 respectively.

I will make the elements of the determinant that are used in each computation turn $\textcolor{red}{red}$

Compute the major diagonal for the $\hat{i}$ component:

| (color(red)hati,hatj,hatk,hati,hatj), (3,color(red)0,4,3,0), (-1,5,color(red)2,-1,5) | = color(red)((0)(2)hati)

Subtract the minor diagonal for the $\hat{i}$ component:

| (hati,hatj,hatk,color(red)hati,hatj), (3,0,color(red)4,3,0), (-1,color(red)5,2,-1,5) | = {(0)(2) color(red)(- (4)(5))}color(red)hati

Add the major diagonal for the $\hat{j}$ component:

| (hati,color(red)hatj,hatk,hati,hatj), (3,0,color(red)4,3,0), (-1,5,2,color(red)(-1),5) | = {(0)(2) - (4)(5)}hati + color(red)((4)(-1)hatj)

Subtract the minor diagonal for the $\hat{j}$ component:

| (hati,hatj,hatk,hati,color(red)hatj), (3,0,4,color(red)3,0), (-1,5,color(red)2,-1,5) | = {(0)(2) - (4)(5)}hati + {(4)(-1)-color(red)((3)(2))}color(red)hatj

Add the major diagonal for the $\hat{k}$ component:

| (hati,hatj,color(red)hatk,hati,hatj), (3,0,4,color(red)3,0), (-1,5,2,-1,color(red)5) | = {(0)(2) - (4)(5)}hati + {(4)(-1)-(3)(2)}hatj + color(red)((3)(5)hatk)

Subtract the minor diagonal for the $\hat{k}$ component:

| (hati,hatj,color(red)hatk,hati,hatj), (3,color(red)0,4,3,0), (color(red)(-1),5,2,-1,5) | = {(0)(2) - (4)(5)}hati + {(4)(-1)-(3)(2)}hatj + {(3)(5) color(red)(- (0)(-1))}color(red)hatk

Simplify:

| (hati,hatj,hatk,hati,hatj), (3,0,4,3,0), (-1,5,2,-1,5) | = -20hati -10hatj + 15hatk

You may prefer the form $< - 20 , - 10 , 15 >$

This vector will be perpendicular to both given vectors but let's compute the dot-product and verify that both dot products are zero:

$< - 20 , - 10 , 15 > \cdot < 3 , 0 , 4 > = \left(- 20\right) \left(3\right) + \left(- 10\right) \left(0\right) + \left(15\right) \left(4\right) = 0$

$< - 20 , - 10 , 15 > \cdot < - 1 , 5 , 2 > = \left(- 20\right) \left(- 1\right) + \left(- 10\right) \left(5\right) + \left(15\right) \left(2\right) = 0$