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

Jan 7, 2017

$\vec{u} = < \frac{\sqrt{3}}{3} , - \frac{\sqrt{3}}{3} , - \frac{\sqrt{3}}{3} >$

#### Explanation:

A vector which is normal (orthogonal, perpendicular) to a plane containing two vectors is also normal to both of the given vectors. We can find the normal vector by taking the cross product of the two given vectors. We can then find a unit vector in the same direction as that vector.

First, write each vector in vector form:

$\vec{a} = < 1 , 0 , 1 >$
$\vec{b} = < 1 , - 2 , 3 >$

The cross product, $\vec{a} \times \vec{b}$ is found by:

$\vec{a} \times \vec{b} = \left\mid \begin{matrix}\vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & 1 \\ 1 & - 2 & 3\end{matrix} \right\mid$

For the i component, we have:

$\left(0 \cdot 3\right) - \left(- 2 \cdot 1\right) = 0 - \left(- 2\right) = 2$

For the j component, we have:

$- \left[\left(1 \cdot 3\right) - \left(1 \cdot 1\right)\right] = - \left[3 - 1\right] = - 2$

For the k component, we have:

$\left(1 \cdot - 2\right) - \left(0 \cdot 1\right) = - 2 - 0 = - 2$

Therefore, $\vec{n} = < 2 , - 2 , - 2 >$

Now, to make this a unit vector, we divide the vector by its magnitude. The magnitude is given by:

$| \vec{n} | = \sqrt{{\left({n}_{x}\right)}^{2} + {\left({n}_{y}\right)}^{2} + {\left({n}_{z}\right)}^{2}}$

$| \vec{n} | = \sqrt{{\left(2\right)}^{2} + {\left(- 2\right)}^{2} + {\left(- 2\right)}^{2}}$

$| \vec{n} | = \sqrt{4 + 4 + 4} = \sqrt{12} = 2 \sqrt{3}$

The unit vector is then given by:

$\vec{u} = \frac{\vec{a} \times \vec{b}}{| \vec{a} \times \vec{b} |} = \frac{\vec{n}}{| \vec{n} |}$

$\vec{u} = \frac{< 2 , - 2 , - 2 >}{2 \sqrt{3}}$

$\vec{u} = < \frac{2}{2 \sqrt{3}} , - \frac{2}{2 \sqrt{3}} , - \frac{2}{2 \sqrt{3}} >$

$\vec{u} = < \frac{1}{\sqrt{3}} , - \frac{1}{\sqrt{3}} , - \frac{1}{\sqrt{3}} >$

By rationalizing the denominator, we get:

$\vec{u} = < \frac{\sqrt{3}}{3} , - \frac{\sqrt{3}}{3} , - \frac{\sqrt{3}}{3} >$