# What is the unit vector that is orthogonal to the plane containing  (3i + 2j - 6k)  and  (3i – 4j + 4k) ?

Feb 17, 2016

${u}_{n} = \frac{- 16 i - 30 j - 18 k}{38.5}$
Notice in the picture I actually drew the unit vector in the opposite direction, i.e.: ${u}_{n} = \frac{16 i + 30 j + 18 k}{38.5}$
It does matter it depend what you're are rotating to what as you apply the Right Hand Rule...

#### Explanation:

As you can see you vectors - let's call them
${v}_{red} = 3 i + 2 j - 6 k$ and ${v}_{b l u e} = 3 i - 4 j + 4 k$
This two vector constitute a plane see the figure.
The vector formed by their x-product => ${v}_{n} = {v}_{red} \times {v}_{b l u e}$
is an orthogonal vector. The unit vector is obtained by normalizing the ${u}_{n} = {v}_{n} / | {v}_{n} |$

Now let's sub and calculate our orthonormal vector ${u}_{n}$
${v}_{n} = \left[\begin{matrix}i & j & k \\ 3 & 2 & - 6 \\ 3 & - 4 & 4\end{matrix}\right]$
${v}_{n} = i \left[\begin{matrix}2 & - 6 \\ - 4 & 4\end{matrix}\right] - j \left[\begin{matrix}3 & - 6 \\ 3 & 4\end{matrix}\right] + k \left[\begin{matrix}3 & 2 \\ 3 & - 4\end{matrix}\right]$
${v}_{n} = \left(\left(2 \cdot 4\right) - \left(- 4 \cdot - 6\right)\right) i - \left(\left(3 \cdot 4\right) - \left(3 \cdot - 6\right)\right) j + \left(\left(3 \cdot - 4\right) - \left(3 \cdot 2\right)\right) k$
${v}_{n} = \left(8 - 24\right) i - \left(12 + 18\right) j + \left(- 12 - 6\right) = - 16 i - 30 j - 18 k$
$| {v}_{n} | = \sqrt{{16}^{2} + {30}^{2} + {18}^{2}} = \sqrt{256 + 900 + 324} \approx 38.5$
${u}_{n} = \frac{- 16 i - 30 j - 18 k}{38.5}$