# How do you find the cross product and state whether the resulting vectors are perpendicular to the given vectors <-1,-3,2>times<6,-1,-2>?

Dec 10, 2016

The cross product of two vectors will always result in a third vector that is perpendicular to both. You can check that is it perpendicular by verifying that the dot-product with either vector is zero.

#### Explanation:

I use a 5 column determinant to compute the cross-product:

| (hati, hatj, hatk, hati, hatj), (-1,-3,2,-1,-3), (6,-1,-2,6,-1) | =

Add the product of the diagonal descending to the right and subtract the product of the diagonal descending to the left.

$\left\{- 3 \left(- 2\right) - \left(2\right) \left(- 1\right)\right\} \hat{i} + \left\{2 \left(6\right) - \left(- 1\right) \left(- 2\right)\right\} \hat{j} + \left\{- 1 \left(- 1\right) - \left(- 3\right) \left(6\right)\right\} \hat{k} =$

Simplify:

$8 \hat{i} + 10 \hat{j} + 19 \hat{k}$

Convert the vector notation: $< 8 , 10 , 19 >$

Verify that it is perpendicular to the first vector:

<8,10,19>•<-1, -3, 2> = 8(-1) + 10(-3) + 19(2) = 0

The dot-product is 0, therefore, the two vectors are perpendicular.

Verify that it is perpendicular to the second vector:

<8,10,19>•<6, -1, -2> = 8(6) + 10(-1) + 19(-2) = 0

The dot-product is 0, therefore, the two vectors are perpendicular.