# What is the unit vector that is orthogonal to the plane containing  (2i + 3j – 7k)  and  (3i + 2j - 6k) ?

Dec 28, 2016

$\vec{u} = < - \frac{4}{\sqrt{122}} , - \frac{9}{\sqrt{122}} , - \frac{5}{\sqrt{122}} >$

#### Explanation:

A vector which is orthogonal 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 $\vec{a} = < 2 , 3 , - 7 >$ and $\vec{b} = < 3 , 2 , - 6 >$, $\vec{a} \times \vec{b}$ is found by For the $i$ component, we have

$\left(3 \cdot - 6\right) - \left(2 \cdot - 7\right) = - 18 + 14 = - 4$

For the $j$ component, we have

$- \left[\left(2 \cdot - 6\right) - \left(3 \cdot - 7\right)\right] = - \left[- 12 + 21\right] = - 9$

For the $k$ component, we have

$\left(2 \cdot 2\right) - \left(3 \cdot 3\right) = 4 - 9 = - 5$

Our vector is $\vec{n} = < - 4 , - 9 , - 5 >$

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

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

$| \vec{n} | = \sqrt{{\left(- 4\right)}^{2} + {\left(- 9\right)}^{2} + {\left(- 5\right)}^{2}}$

$| \vec{n} | = \sqrt{16 + 81 + 25} = \sqrt{122}$

The unit vector is then given by:

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

$\vec{u} = \frac{< - 4 , - 9 , - 5 >}{\sqrt{122}}$

$\vec{u} = \frac{1}{\sqrt{122}} < - 4 , - 9 , - 5 >$

or equivalently,

$\vec{u} = < - \frac{4}{\sqrt{122}} , - \frac{9}{\sqrt{122}} , - \frac{5}{\sqrt{122}} >$

You may also choose to rationalize the denominator.