# What is the unit vector that is orthogonal to the plane containing  (8i + 12j + 14k)  and  (2i + 3j – 7k) ?

Dec 30, 2016

$\vec{u} = < \frac{- 3 \sqrt{13}}{13} , \frac{2 \sqrt{13}}{13} , 0 >$

#### 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.

Given $\vec{a} = < 8 , 12 , 14 >$ and $\vec{b} = < 2 , 3 , - 7 >$, $\vec{a} \times \vec{b}$is found by

For the $i$ component, we have

$\left(12 \cdot - 7\right) - \left(14 \cdot 3\right) = - 84 - 42 = - 126$

For the $j$ component, we have

$- \left[\left(8 \cdot - 7\right) - \left(2 \cdot 14\right)\right] = - \left[- 56 - 28\right] = 84$

For the $k$ component, we have

$\left(8 \cdot 3\right) - \left(12 \cdot 2\right) = 24 - 24 = 0$

Our normal vector is $\vec{n} = < - 126 , 84 , 0 >$

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

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

$| \vec{n} | = \sqrt{{\left(- 126\right)}^{2} + {\left(84\right)}^{2} + {\left(0\right)}^{2}}$

$| \vec{n} | = \sqrt{15878 + 7056 + 0} = \sqrt{22932} = 42 \sqrt{13}$

The unit vector is then given by:

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

$\vec{u} = \frac{< - 126 , 84 , 0 >}{42 \sqrt{13}}$

$\vec{u} = \frac{1}{42 \sqrt{13}} < - 126 , 84 , 0 >$

or equivalently,

$\vec{u} = < - \frac{3}{\sqrt{13}} , \frac{2}{\sqrt{13}} , 0 >$

You may also choose to rationalize the denominator:

$\vec{u} = < \frac{- 3 \sqrt{13}}{13} , \frac{2 \sqrt{13}}{13} , 0 >$