# What is the unit vector that is normal to the plane containing (- 3 i + j -k) and #(- 2i - j - k)?

Feb 19, 2017

The unit vector is $= < - \frac{2}{\sqrt{30}} , - \frac{1}{\sqrt{30}} , \frac{5}{\sqrt{30}} >$

#### Explanation:

We calculate the vector that is perpendicular to the other 2 vectors by doing a cross product,

Let $\vec{a} = < - 3 , 1 , - 1 >$

$\vec{b} = < - 2 , - 1 , - 1 >$

$\vec{c} = | \left(\hat{i} , \hat{j} , \hat{k}\right) , \left(- 3 , 1 , - 1\right) , \left(- 2 , - 1 , - 1\right) |$

$= \hat{i} | \left(1 , - 1\right) , \left(- 1 , - 1\right) | - \hat{j} | \left(- 3 , - 1\right) , \left(- 2 , - 1\right) | + \hat{k} | \left(- 3 , 1\right) , \left(- 2 , - 1\right) |$

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

$= < - 2 , - 1 , 5 >$

Verification

$\vec{a} . \vec{c} = < - 3 , 1 , - 1 > . < - 2 , - 1 , 5 \ge 6 - 1 - 5 = 0$

$\vec{b} . \vec{c} = < - 2 , - 1 , - 1 > . < - 2 , - 1 , 5 \ge 4 + 1 - 5 = 0$

The modulus of $\vec{c} = | | \vec{c} | | = | | < - 2 , - 1 , 5 > | | = \sqrt{4 + 1 + 25} = \sqrt{30}$

The unit vector $= \frac{\vec{c}}{| | \vec{c} | |}$

$= \frac{1}{\sqrt{30}} < - 2 , - 1 , 5 >$