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

Sep 8, 2016

$\frac{\sqrt{2}}{2} \left\{1 , - 1 , 0\right\}$

Explanation:

If $P = \left(3 , - 3 , 0\right) , Q = \left(5 , - 1 , 2\right)$, and $R = \left(5 , - 1 , 6\right)$ pertain to the plane $\Pi$

then $\left(P - Q\right) \times \left(R - Q\right)$ is normal to $\Pi$

so

$\vec{n} = \left(P - Q\right) \times \left(R - Q\right) = \left\{- 8 , 8 , 0\right\}$. Any vector proportional to $\vec{n}$ like $\lambda \vec{n}$ with $\lambda \in \mathbb{R} , \lambda \ne 0$ is also normal to $\Pi$ so we choose

$\lambda = - \frac{1}{\left\lVert \vec{n} \right\rVert} = - \frac{1}{8 \sqrt{2}}$. The sought unit vector is then

$\lambda \vec{n} = \frac{\sqrt{2}}{2} \left\{1 , - 1 , 0\right\}$