# What is the unit vector that is normal to the plane containing 3i+7j-2k and 8i+2j+9k?

The unit vector normal to the plane is
$\left(\frac{1}{94.01}\right) \left(67 \hat{i} - 43 \hat{j} + 50 \hat{k}\right)$.

#### Explanation:

Let us consider $\vec{A} = 3 \hat{i} + 7 \hat{j} - 2 \hat{k} , \vec{B} = 8 \hat{i} + 2 \hat{j} + 9 \hat{k}$
The normal to the plane $\vec{A} , \vec{B}$ is nothing but the vector perpendicular i.e., cross product of $\vec{A} , \vec{B}$.
=>vecAxxvecB= hati(63+4)-hatj(27+16)+hatk(6-56) =67hati-43hatj+50hatk.
The unit vector normal to the plane is
$\pm \left[\vec{A} \times \vec{B} / \left(| \vec{A} \times \vec{B} |\right)\right]$
So$| \vec{A} \times \vec{B} | = \sqrt{{\left(67\right)}^{2} + {\left(- 43\right)}^{2} + {\left(50\right)}^{2}} = \sqrt{8838} = 94.01 \approx 94$
Now substitute all in above equation, we get unit vector =$\pm \left\{\left[\frac{1}{\sqrt{8838}}\right] \left[67 \hat{i} - 43 \hat{j} + 50 \hat{k}\right]\right\}$.