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

Feb 15, 2017

The unit vector is $= < - \frac{10}{\sqrt{117}} , \frac{1}{\sqrt{117}} , \frac{4}{\sqrt{117}} >$

#### Explanation:

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

Let $\vec{a} = < 1 , - 2 , 3 >$

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

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

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

$= \hat{i} \left(- 10\right) - \hat{j} \left(- 1\right) + \hat{k} \left(4\right)$

$= < - 10 , 1 , 4 >$

Verification

$\vec{a} . \vec{c} = < 1 , - 2 , 3 > . < - 10 , 1 , 4 \ge - 10 - 2 + 12 = 0$

$\vec{b} . \vec{c} = < 1 , 2 , 2 > . < - 10 , 1 , 4 \ge - 10 + 2 + 8 = 0$

The modulus of $\vec{c} = | | \vec{c} | = | | < - 10 , 1 , 4 > | | = \sqrt{100 + 1 + 16} = \sqrt{117}$

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

$= \frac{1}{\sqrt{117}} < - 10 , 1 , 4 >$