# How would you find a unit vector with positive first coordinate that is orthogonal to the plane through the points P = (3, -4, 4), Q = (6, -1, 7), and R = (6, -1, 9)?

Oct 4, 2017

We need to create two vectors in the plane.

To create $\vec{P Q}$, we subtract each coordinate of point P from its respective coordinate of point Q:

$\vec{P Q} = < 6 - 3 , - 1 - \left(- 4\right) , 7 - 4 >$

$\vec{P Q} = < 3 , 3 , 3 >$

To create $\vec{Q R}$, we subtract each coordinate of point P from its respective coordinate of point R:

$\vec{P R} = < 6 - 3 , - 1 - \left(- 4\right) , 9 - 4 >$

$\vec{P R} = < 3 , 3 , 5 >$

A normal vector to the plane, $\vec{n}$, is the cross product of these two vectors:

$\vec{n} = \vec{P Q} \times \vec{P R}$

$\vec{n} = < 3 , 3 , 3 > \times < 3 , 3 , 5 >$

$\vec{n} = < 6 , - 6 , 0 >$

To make $\vec{n}$ a unit vector, we divide by the magnitude:

$| \vec{n} | = \sqrt{{6}^{2} + {\left(- 6\right)}^{2} + {0}^{2}}$

$| \vec{n} | = 6 \sqrt{2}$

$\hat{n} = \frac{1}{6 \sqrt{2}} < 6 , - 6 , 0 >$

$\hat{n} = \frac{\sqrt{2}}{12} < 6 , - 6 , 0 >$

$\hat{n} = < \frac{\sqrt{2}}{2} , - \frac{\sqrt{2}}{2} , 0 >$