# What is the unit vector that is orthogonal to the plane containing  ( i - 2 j + 3 k)  and  ( - 4 i - 5 j + 2 k) ?

Nov 4, 2016

The unit vector is $\left(\frac{11 \vec{i}}{\sqrt{486}} - \frac{14 \vec{j}}{\sqrt{486}} - \frac{13 \vec{k}}{\sqrt{486}}\right)$

#### Explanation:

Firstly, we need the vector perpendicular to other two vectros:
For this we do the cross product of the vectors:
Let vecu=〈1,-2,3〉 and vecv=〈-4,-5,2〉
The cross product $\vec{u}$x$\vec{v}$ $=$the determinant
∣((veci,vecj,veck),(1,-2,3),(-4,-5,2))∣
=veci∣((-2,3),(-5,2))∣-vecj∣((1,3),(-4,2))∣+veck∣((1,-2),(-5,-5))∣
$= 11 \vec{i} - 14 \vec{j} - 13 \vec{k}$
So vecw=〈11,-14,-13〉
We can check that they are perpendicular by doing the dot prodct.
$\vec{u} . \vec{w} = 11 + 28 - 39 = 0$
$\vec{v} . \vec{w} = - 44 + 70 - 26 = 0$
The unit vector hatw=vecw/(∥vecw∥)
The modulus of $\vec{w} = \sqrt{121 + 196 + 169} = \sqrt{486}$
So the unit vector is $\left(\frac{11 \vec{i}}{\sqrt{486}} - \frac{14 \vec{j}}{\sqrt{486}} - \frac{13 \vec{k}}{\sqrt{486}}\right)$