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

Feb 20, 2017

The unit vector is $= < \frac{5}{\sqrt{42}} , \frac{4}{\sqrt{42}} , \frac{1}{\sqrt{42}} >$

#### Explanation:

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

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

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

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

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

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

$= < 5 , 4 , 1 >$

Verification

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

$\vec{b} . \vec{c} = < 1 , - 2 , 3 > . < 5 , 4 , 1 \ge 5 - 8 + 3 = 0$

The modulus of $\vec{c} = | | \vec{c} | | = | | < 5 , 4 , 1 > | | = \sqrt{25 + 16 + 1} = \sqrt{42}$

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

$= \frac{1}{\sqrt{42}} < 5 , 4 , 1 >$