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

Jul 1, 2016

$= \frac{- 2 \hat{i} + \hat{j} + 7 \hat{k}}{3 \sqrt{6}}$

#### Explanation:

you'll do this by calculating the vector cross product of these 2 vectors to get the normal vector

so $\vec{n} = \left(- 3 i + j - k\right) \times \left(2 i - 3 j + k\right)$

$= \det \left[\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ - 3 & 1 & - 1 \\ 2 & - 3 & 1\end{matrix}\right]$

= hat i (1*1 -( -3*-1)) - hat j(-3*1 - (-1*2) ) + hat k (-3*-3 - 2*1))

$= - 2 \hat{i} + \hat{j} + 7 \hat{k}$

the unit normal is $\hat{n} = \frac{- 2 \hat{i} + \hat{j} + 7 \hat{k}}{\sqrt{{\left(- 2\right)}^{2} + {1}^{2} + {7}^{2}}}$

$= \frac{- 2 \hat{i} + \hat{j} + 7 \hat{k}}{3 \sqrt{6}}$

you could check this by doing a scalar dot product between the normal and each of the original vectors, should get zero as they are orthogonal.

so for example

${\vec{v}}_{1} \cdot \vec{n}$

$= \left(- 3 i + j - k\right) \cdot \left(- 2 i + j + 7 k\right)$

$= 6 + 1 - 7 = 0$