## Problem on finding the unit vectors perpendicular to two vectors Find all the vectors in 3d that have unit length and are perpendicular to $$\Bigl(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \Bigr) \ \text{ and } \ \Bigl(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \Bigr).$$
• ## Solution Let's start by figuring out how many vectors satisfy these properties. In three dimensions, if two vectors are not parallel, there is only one pair of (opposite) directions that can be perpendicular to both. Hence, there are two vectors of unit length that are perpendicular to the two provided vectors, and these unit vectors are negatives of each other. Recall that We identify $\bfx = \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\rangle$ and $\bfy = \langle 0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle$. Computing the cross product, Hence, $\frac{1}{2} \bfi - \frac{1}{2} \bfj + \frac{1}{2} \bfk = \langle \frac{1}{2}, -\frac{1}{2}, \frac{1}{2} \rangle$ is a vector that is perpendicular to both $\langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\rangle$ and $\langle 0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\rangle.$ This vector does not have unit length, so we recall Hence a unit vector in the same direction as $\frac{1}{2} \bfi - \frac{1}{2} \bfj + \frac{1}{2} \bfk$ is $$\frac{\frac{1}{2} \bfi - \frac{1}{2} \bfj + \frac{1}{2} \bfk}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} \bfi - \frac{1}{\sqrt{3}} \bfj + \frac{1}{\sqrt{3}} \bfk$$ Hence, the two unit vectors that are perpendicular to $\Bigl(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \Bigr)$ and $\Bigl(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \Bigr)$ are $$\frac{1}{\sqrt{3}} \bfi - \frac{1}{\sqrt{3}} \bfj + \frac{1}{\sqrt{3}} \bfk \ \text{ and } \ -\frac{1}{\sqrt{3}} \bfi + \frac{1}{\sqrt{3}} \bfj - \frac{1}{\sqrt{3}} \bfk.$$