# How do you find a unit vector perpendicular to both (2,1,1) and the x-axis?

Nov 23, 2016

Please see the explanation for steps leading to: $\hat{C} = \frac{\sqrt{2}}{2} \hat{j} - \frac{\sqrt{2}}{2} \hat{k}$

#### Explanation:

Given: $\overline{A} = 2 \hat{i} + \hat{j} + \hat{k}$

Let $\overline{B} = \text{a vector along the x axis} = \hat{i}$

The vector $\overline{C} = \overline{A} \times \overline{B}$ will be perpendicular to both but is will not be a unit vector.

barC =barA xx barB = | (hati, hatj, hatk, hati, hatj), (2,1,1,2,1), (1,0,0,1,0) |=

$\left\{1 \left(0\right) - 1 \left(0\right)\right\} \hat{i} + \left\{1 \left(1\right) - 2 \left(0\right)\right\} \hat{j} + \left\{2 \left(0\right) - \left(1\right) \left(1\right)\right\} \hat{k}$

$\overline{C} = \hat{j} - \hat{k}$

Compute the magnitude of $\overline{C}$

$| \overline{C} | = \sqrt{{1}^{2} + {1}^{2}} = \sqrt{2}$

The unit vector perpendicular to both is:

$\hat{C} = \frac{\overline{C}}{|} \overline{C} |$

$\hat{C} = \frac{\hat{j} - \hat{k}}{\sqrt{2}}$

$\hat{C} = \frac{\sqrt{2}}{2} \hat{j} - \frac{\sqrt{2}}{2} \hat{k}$