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>?

1 Answer
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 color(red)(red)

Compute the major diagonal for the hati 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 hati 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 hatj 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 hatj 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 hatk 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 hatk 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>*<3, 0, 4> = (-20)(3)+(-10)(0) + (15)(4) = 0

< -20, -10, 15>*<-1, 5, 2> = (-20)(-1)+(-10)(5) + (15)(2) = 0