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

Jun 1, 2016

$\left\{- 4 \sqrt{\frac{2}{61}} , \frac{7}{\sqrt{122}} , - \frac{3}{\sqrt{122}}\right\}$

#### Explanation:

Given two non aligned vectors $\vec{u}$ and $\vec{v}$ the cross product given by $\vec{w} = \vec{u} \times \vec{v}$ is orthogonal to $\vec{u}$ and $\vec{v}$

Their cross product is calculated by the determinant rule, expanding the subdeterminants headed by $\vec{i} , \vec{j} , \vec{k}$

$\vec{w} = \vec{u} \times \vec{v} = \det \left(\begin{matrix}\vec{i} & \vec{j} & \vec{k} \\ {u}_{x} & {u}_{y} & {u}_{z} \\ {v}_{x} & {v}_{y} & {v}_{z}\end{matrix}\right)$

$\vec{u} \times \vec{v} = \left({u}_{y} {v}_{z} - {u}_{z} {v}_{y}\right) \vec{i} - \left({u}_{x} {v}_{z} - {u}_{z} {v}_{x}\right) \vec{j} + \left({u}_{x} {v}_{y} - {u}_{y} {v}_{x}\right) \vec{k}$

so
$\vec{w} = \det \left(\begin{matrix}\vec{i} & \vec{j} & \vec{k} \\ 1 & 2 & 2 \\ 2 & 1 & - 3\end{matrix}\right) = - 8 \vec{i} + 7 \vec{j} - 3 \vec{k}$

Then the unit vector is $\frac{\vec{w}}{\left\lVert \vec{w} \right\rVert} = \left\{- 4 \sqrt{\frac{2}{61}} , \frac{7}{\sqrt{122}} , - \frac{3}{\sqrt{122}}\right\}$