# What is the unit vector that is orthogonal to the plane containing  (8i + 12j + 14k)  and  (3i – 4j + 4k) ?

Jan 20, 2016

Use the cross product to find the vector ${v}_{1} \times {v}_{2} \bot = {v}_{3}$
${v}_{3} = 104 i + 10 j - 68 k$
$\vec{{u}_{3}} = \frac{1}{\sqrt{{104}^{2} + {10}^{2} + {68}^{2}}} \left(104 i + 10 j - 68 k\right)$

#### Explanation:

let $\vec{{v}_{1}} = \left(8 , 12 , 14\right) \mathmr{and} \vec{{v}_{2}} = \left(3 , - 4 , 4\right)$
cross product vec(v_1)xx vec(v_2) = [(12xx4) +(14xx4)] i + [(14xx3) -(8xx4)] j + [(8xx-4 - 12xx3)]k

Simplify to get the Orthogonal vector.

$\vec{{v}_{3}} = 104 i + 10 j - 68 k$
To calculate the unit vector find:
$\vec{{u}_{3}} = \frac{1}{|} \vec{{v}_{3}} | \vec{{v}_{3}}$
where: $| \vec{{v}_{3}} | = m a g n i t u \mathrm{de}$
$| \vec{{v}_{3}} | = \sqrt{{104}^{2} + {10}^{2} + {68}^{2}}$