# What is the unit vector that is normal to the plane containing (- 3 i + j -k) and ( i +k )?

Nov 29, 2016

The unit vector is:

$\hat{C} = \frac{\sqrt{6}}{6} \hat{i} + \frac{\sqrt{6}}{3} \hat{j} - \frac{\sqrt{6}}{6} \hat{k}$

#### Explanation:

Let $\overline{A} = - 3 \hat{i} + \hat{j} - \hat{k}$
Let $\overline{B} = \hat{i} + \hat{k}$
Let $\overline{C} =$ the vector perpendicular to both $\overline{A} \mathmr{and} \overline{B}$ such that the following is true:

$\overline{C} = \overline{A} \times \overline{B}$

barC = | (hati, hatj, hatk, hati, hatj), (-3,1,-1,-3,1), (1,0,1,1,0) |#

$\overline{C} = \left\{1 \left(1\right) - \left(- 1\right) \left(0\right)\right\} \hat{i} + \left\{- 1 \left(1\right) - \left(- 3\right) \left(1\right)\right\} \hat{j} + \left\{- 3 \left(0\right) - \left(1\right) \left(1\right)\right\} \hat{k}$

$\overline{C} = \hat{i} + 2 \hat{j} - \hat{k}$

Check:

$\overline{C} \cdot \overline{A} = \left(1\right) \left(- 3\right) + \left(2\right) \left(1\right) + \left(- 1\right) \left(- 1\right) = 0$
$\overline{C} \cdot \overline{B} = \left(1\right) \left(1\right) + \left(2\right) \left(0\right) + \left(- 1\right) \left(1\right) = 0$

This checks but $\overline{C}$ is not a unit vector.

To make it a unit vector we must divide $\overline{C}$ by its magnitude.

$| \overline{C} | = \sqrt{{1}^{2} + {2}^{2} + {\left(- 1\right)}^{2}}$

$| \overline{C} | = \sqrt{6}$

Dividing by $\sqrt{6}$ is the same as multiplying by $\frac{\sqrt{6}}{6}$

The unit vector is:

$\hat{C} = \frac{\sqrt{6}}{6} \hat{i} + \frac{\sqrt{6}}{3} \hat{j} - \frac{\sqrt{6}}{6} \hat{k}$