# What is the unit vector that is normal to the plane containing  ( i +k)  and  ( i + 7 j + 4 k) ?

Aug 3, 2016

$\hat{v} = \frac{1}{\sqrt{107}} \cdot \left(\begin{matrix}7 \\ 3 \\ - 7\end{matrix}\right)$

#### Explanation:

first, you need to find the vector (cross) product vector, $\vec{v}$, of those 2 co-planar vectors, as $\vec{v}$ will be at right angles to both of these by definition:

$\vec{a} \times \vec{b} = \left\mid \vec{a} \right\mid \left\mid \vec{b} \right\mid \setminus \sin \theta \setminus {\hat{n}}_{\textcolor{red}{a b}}$

computationally, that vector is the determinant of this matrix, ie

$\vec{v} = \det \left(\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 1 \\ 1 & 7 & 4\end{matrix}\right)$

$= \hat{i} \left(- 7\right) - \hat{j} \left(3\right) + \hat{k} \left(7\right)$

$= \left(\begin{matrix}- 7 \\ - 3 \\ 7\end{matrix}\right)$ or as we are only interested in direction
$\vec{v} = \left(\begin{matrix}7 \\ 3 \\ - 7\end{matrix}\right)$

for the unit vector we have

$\hat{v} = \frac{\vec{v}}{\left\mid \vec{v} \right\mid} = \frac{1}{\sqrt{{7}^{2} + {3}^{3} + {\left(- 7\right)}^{2}}} \cdot \left(\begin{matrix}7 \\ 3 \\ - 7\end{matrix}\right)$

$= \frac{1}{\sqrt{107}} \cdot \left(\begin{matrix}7 \\ 3 \\ - 7\end{matrix}\right)$