# What is the unit vector that is orthogonal to the plane containing  (3i – 4j + 4k)  and  (2i+j+2k) ?

Mar 9, 2017

$\vec{u} = < \frac{- 12 \sqrt{269}}{269} , \frac{2 \sqrt{269}}{269} , \frac{11 \sqrt{269}}{269} >$

#### Explanation:

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

First, write each vector in vector form:

$\vec{a} = < 3 , - 4 , 4 >$ and $\vec{b} = < 2 , 1 , 2 >$

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

$\vec{a} \times \vec{b} = \left\mid \begin{matrix}\vec{i} & \vec{j} & \vec{k} \\ 3 & - 4 & 4 \\ 2 & 1 & 2\end{matrix} \right\mid$

For the i component, we have:

$\left(- 4 \cdot 2\right) - \left(4 \cdot 1\right) = \left(- 8\right) - \left(4\right) = - 12$

For the j component, we have:

$- \left[\left(3 \cdot 2\right) - \left(4 \cdot 2\right)\right] = - \left[6 - 8\right] = 2$

For the k component, we have:

$\left(3 \cdot 1\right) - \left(- 4 \cdot 2\right) = 3 - \left(- 8\right) = 11$

Therefore, $\vec{n} = < - 12 , 2 , 11 >$

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

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

$| \vec{n} | = \sqrt{{\left(- 12\right)}^{2} + {\left(2\right)}^{2} + {\left(11\right)}^{2}}$

$| \vec{n} | = \sqrt{144 + 4 + 121} = \sqrt{269}$

The unit vector is then given by:

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

$\vec{u} = \frac{< - 12 , 2 , 11 >}{\sqrt{269}}$

$\vec{u} = < - \frac{12}{\sqrt{269}} , \frac{2}{\sqrt{269}} , \frac{11}{\sqrt{269}} >$

We can rationalize the denominators to get:

$\vec{u} = < \frac{- 12 \sqrt{269}}{269} , \frac{2 \sqrt{269}}{269} , \frac{11 \sqrt{269}}{269} >$