# How do you find the unit vector that is orthogonal to the plane through the points P = (3, -3, 0), Q = (5, -1, 2), and R = (5, -1, 6)?

Jan 9, 2017

The reqd. unit normal $\frac{i}{\sqrt{2}} - \frac{j}{\sqrt{2}}$.

#### Explanation:

The given points $P , Q , \mathmr{and} , R$ are co-planer; therefore so are the

vectors $\vec{P Q} \mathmr{and} \vec{Q R} .$

Knowing that $\vec{n} = \vec{P Q} \times \vec{Q R}$ is orthogonal to both of

these vectors, $\vec{n}$ is the normal vector of the plane containing

the points P,Q, &, R.

Now, $\vec{P Q} = \left(5 , - 1 , 2\right) - \left(3 , - 3 , 0\right) = \left(2 , 2 , 2\right) = 2 \left(1 , 1 , 1\right) ,$

$\vec{Q R} = \left(0 , 0 , 4\right) = 4 \left(0 , 0 , 1\right)$

$\vec{n} = 8 | \left(i , j , k\right) , \left(1 , 1 , 1\right) , \left(0 , 0 , 1\right) | = 8 \left(i - j\right) \Rightarrow | | \vec{n} | | = 8 \sqrt{2.}$

Hence, the reqd. unit normal $\hat{n} = \frac{\vec{n}}{|} | \vec{n} | | = \frac{i}{\sqrt{2}} - \frac{j}{\sqrt{2}}$.

Enjoy Maths.!