# How do you find a unit vector normal to the plane which passes through the following points (1,1,1),(2,0,2),(1,1,2)?

Aug 29, 2016

Reqd. unit normal is $\left(- \frac{1}{\sqrt{2}} , - \frac{1}{\sqrt{2}} , 0\right) , \mathmr{and} , \left(\frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} , 0\right)$.

#### Explanation:

Let $A \left(1 , 1 , 1\right) , B \left(2 , 0 , 2\right) , \mathmr{and} , C \left(1 , 1 , 2\right)$ be the pts. on the plane, say $\pi$.

We need a unit normal, call it $\hat{n}$, to $\pi$.

$\vec{A B} = \left(2 - 1 , 0 - 1 , 2 - 1\right) = \left(1 , - 1 , 1\right) \in \pi$, and,

$\vec{A C} = \left(1 - 1 , 1 - 1 , 2 - 1\right) = \left(0 , 0 , 1\right) \in \pi$.

$\therefore$ Fom Vector Geometry, $\vec{A B} \times \vec{A C}$ is a vector $\bot$

to $\vec{A B} \mathmr{and} \vec{A C}$, hence, is the normal to the plane $\pi$.

Now, $\vec{A B} \times \vec{A C} = \det | \left(\hat{i} , \hat{j} , \hat{k}\right) , \left(1 , - 1 , 1\right) , \left(0 , 0 , 1\right) |$

$= - 1 \hat{i} - 1 \hat{j} + 0 \hat{k} = \left(- 1 , - 1 , 0\right)$, so that,

$| | \vec{A B} \times \vec{A C} | | = \sqrt{2}$.

Therefore,

$\hat{n} = \frac{\vec{A B} \times \vec{A C}}{|} | \vec{A B} \times \vec{A C} | |$, i.e.,

$\hat{n} = \left(- \frac{1}{\sqrt{2}} , - \frac{1}{\sqrt{2}} , 0\right) , \mathmr{and} , \left(\frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} , 0\right)$

Enjoy maths.!