# What is the unit vector that is orthogonal to the plane containing  <3, -6, 2>  and  <3, 2, 1> ?

Feb 21, 2016

$\left(- \frac{10}{\sqrt{685}} \cdot i + \frac{3}{\sqrt{685}} \cdot j + \frac{24}{\sqrt{685.}} k\right)$

#### Explanation:

$\text{step 1:Find cross product of two vector}$
$\text{the cross product of two vector in the same plane is perpendicular to the plane}$
$\text{let } \vec{C} = \vec{A} X \vec{B}$
$\vec{C} = i \left({a}_{y} \cdot {b}_{z} - {a}_{z} \cdot {b}_{y}\right) - j \left({a}_{x} \cdot {b}_{z} - {a}_{z} \cdot {b}_{x}\right) + k \left({a}_{x} \cdot {b}_{y} - {a}_{y} \cdot {b}_{x}\right)$
$\vec{c} = i \left(- 10\right) - j \left(- 3\right) + k \left(24\right)$
$\vec{C} = - 10 i + 3 j + 24 k$
$\text{step 2:Find magnitude of the vector of } \vec{C}$
$| | c | | = \sqrt{{\left(- 10\right)}^{2} + {3}^{2} + {24}^{2}}$
$| | c | | = \sqrt{100 + 9 + 576}$
$| | c | | = \sqrt{685}$
step 3:" use : vec C /||c||
$\frac{- 10 i + 3 j + 24 k}{\sqrt{685}}$
$\left(- \frac{10}{\sqrt{685}} \cdot i + \frac{3}{\sqrt{685}} \cdot j + \frac{24}{\sqrt{685.}} k\right)$