# If Vector a= 2i - 4j + k and Vector b= -4i + j +2k how do you find a unit vector perpendicular to both vector a and b?

Nov 2, 2016

#### Explanation:

Given:

$\overline{a} = 2 \hat{i} - 4 \hat{j} + \hat{k} \mathmr{and} \overline{b} = - 4 \hat{i} + \hat{j} + 2 \hat{k}$

A vector perpendicular to any two vectors is found by computing the cross-product:

bara xx barb = | (hati, hatj, hatk, hati, hatj), (2,-4,1,2,-4), (-4, 1, 2, -4, 1) | =

$\hat{i} \left\{\left(- 4\right) \left(2\right) - \left(1\right) \left(1\right)\right\} + \hat{j} \left\{\left(1\right) \left(- 4\right) - \left(2\right) \left(2\right)\right\} + \hat{k} \left\{\left(2\right) \left(1\right) - \left(- 4\right) \left(- 4\right)\right\} =$
$- 9 \hat{i} - 8 \hat{j} - 14 \hat{k}$
We can multiply by the scalar -1 and still have a perpendicular vector:

$9 \hat{i} + 8 \hat{j} + 14 \hat{k}$

To obtain a unit vector, divide by the magnitude:

$| 9 \hat{i} + 8 \hat{j} + 14 \hat{k} | = \sqrt{{9}^{2} + {8}^{2} + {14}^{2}} = \sqrt{345}$

A unit vector perpendicular to both vectors is:

$9 \frac{\sqrt{345}}{345} \hat{i} + 8 \frac{\sqrt{345}}{345} \hat{j} + 14 \frac{\sqrt{345}}{345} \hat{k}$