# How do you find all unit vectors orthogonal to v=i+j+k?

Jan 22, 2017

A generalised unit vector is:

$m a t h b f u = \frac{1}{\sqrt{2 \left({\alpha}_{2}^{2} + {\alpha}_{3}^{2} + {\alpha}_{2} {\alpha}_{3}\right)}} \left(\begin{matrix}- {\alpha}_{2} - {\alpha}_{3} \\ \setminus {\alpha}_{2} \\ {\alpha}_{3}\end{matrix}\right)$

#### Explanation:

There are an infinite number of vectors thare are orthogonal to $m a t h b f v = \left(\begin{matrix}1 \\ 1 \\ 1\end{matrix}\right)$.

If $m a t h b f \alpha$ is one such vector, we know from the dot product that $m a t h b f v \cdot m a t h b f \alpha = 0 \implies {\alpha}_{1} + {\alpha}_{2} + {\alpha}_{3} = 0$

A generalised vector is therefore:

$m a t h b f \alpha = \left(\begin{matrix}- {\alpha}_{2} - {\alpha}_{3} \\ \setminus {\alpha}_{2} \\ {\alpha}_{3}\end{matrix}\right)$

A generalised unit vector is:

$m a t h b f u = \frac{1}{\sqrt{2 \left({\alpha}_{2}^{2} + {\alpha}_{3}^{2} + {\alpha}_{2} {\alpha}_{3}\right)}} \left(\begin{matrix}- {\alpha}_{2} - {\alpha}_{3} \\ \setminus {\alpha}_{2} \\ {\alpha}_{3}\end{matrix}\right)$