# How do you find the vector perpendicular to the plane containing (0,-2,2), (1,2,-3), and (4,0,-1)?

Oct 20, 2016

Make two vectors from the 3 points and then take the cross product:

$\overline{v} \times \overline{u} = 2 \hat{i} + 17 \hat{j} + 14 \hat{k}$

#### Explanation:

Use the 3 points to create two vectors in the plane:

let $\overline{u} = \left(1 - 0\right) \hat{i} + \left(2 - - 2\right) \hat{j} + \left(- 3 - 2\right) \hat{k}$

$\overline{u} = 1 \hat{i} + 4 \hat{j} - 5 \hat{k}$

let $\overline{v} = \left(4 - 0\right) \hat{i} + \left(0 - - 2\right) \hat{j} + \left(- 1 - 2\right) \hat{k}$

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

Use the cross product to find a vector perpendicular to these two vectors:

$\overline{u} \times \overline{v} = | \left(\hat{i} , \hat{j} , \hat{k} , \hat{i} , \hat{j}\right) , \left(1 , 4 , - 5 , 1 , 4\right) , \left(4 , 2 , - 3 , 4 , 2\right) |$

$\overline{u} \times \overline{v} = \hat{i} \left\{\left(4\right) \left(- 3\right) - \left(- 5\right) \left(- 2\right)\right\} + \hat{j} \left\{\left(- 5\right) \left(4\right) - \left(1\right) \left(- 3\right)\right\} + \hat{k} \left\{\left(1\right) \left(2\right) - \left(4\right) \left(4\right)\right\}$

$\overline{u} \times \overline{v} = - 2 \hat{i} - 17 \hat{j} - 14 \hat{k}$

The negative of this is also valid:

$\overline{v} \times \overline{u} = 2 \hat{i} + 17 \hat{j} + 14 \hat{k}$