# How do you find the unit vectors perpendicular to the plane determined by the three points (1, 3, 5), (3, -1, 2), and (4, 0, 1)?

Oct 31, 2016

Please see the explanation.

#### Explanation:

Let $\overline{A} =$ the vector from the first point to the second

$\overline{A} = \left(3 - 1\right) \hat{i} + \left(- 1 - 3\right) \hat{j} + \left(2 - 5\right) \hat{k}$

$\overline{A} = 2 \hat{i} - 4 \hat{j} - 3 \hat{k}$

Let $\overline{B} =$ the vector from the first point to the third

$\overline{B} = \left(4 - 1\right) \hat{i} + \left(0 - 3\right) \hat{j} + \left(1 - 5\right) \hat{k}$

$\overline{B} = 3 \hat{i} - 3 \hat{j} - 4 \hat{k}$

A vector perpendicular to the plane can be found by computing the cross product of these two vectors:

AxxB = |(hati, hatj, hatk, hati, hatj),(2, - 4, - 3, 2, -4) ,(3, -3, -4, 3, -3) | =

$\hat{i} \left\{\left(- 4\right) \left(- 4\right) - \left(- 3\right) \left(- 3\right)\right\} = 7 \hat{i}$
$\hat{j} \left\{\left(- 3\right) \left(3\right) - \left(2\right) \left(- 4\right)\right\} = - 1 \hat{j}$
$\hat{k} \left\{\left(2\right) \left(- 3\right) - \left(- 4\right) \left(3\right)\right\} = 6 \hat{k}$

A vector $\overline{C}$ perpendicular to the plane is:

$\overline{C} = 7 \hat{i} - \hat{j} + 6 \hat{k}$

However, you want the unit vector $\hat{C}$

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

$| \overline{C} | = \sqrt{{7}^{2} + {\left(- 1\right)}^{2} + {6}^{2}}$

$| \overline{C} | = \sqrt{86}$

$\hat{C} = 7 \frac{\sqrt{86}}{86} \hat{i} - \frac{\sqrt{86}}{86} \hat{j} + 6 \frac{\sqrt{86}}{86} \hat{k}$