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
Cross product gives a perpendicular vector

The cross product $\bfx \times \bfy$ is a vector perpendicular to both $\bfx$ and $\bfy$.
Cross product definition

The cross product of $\mathbf{x} = \langle x_1, x_2, x_3\rangle$ and $\mathbf{y} = \langle y_1, y_2, y_3\rangle$ is \begin{align}\mathbf{x} \times \mathbf{y} &= \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{vmatrix} \end{align}
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,
Determinant of a 3x3 matrix

\begin{align}\mathbf{x} \times \mathbf{y} &= \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{vmatrix}\\ \ \\ &= \begin{vmatrix}\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{vmatrix} \ \mathbf{i} - \begin{vmatrix}\frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} \end{vmatrix} \ \mathbf{j} + \begin{vmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} \end{vmatrix} \ \mathbf{k} \end{align}
Determinant of a 2x2 matrix

Evaluating the $2 \times 2$ determinants, $$\bfx \times \bfy= \frac{1}{2} \bfi - \frac{1}{2} \bfj + \frac{1}{2} \bfk.$$
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
Finding the unit vector in the same direction as another vector

The find the unit vector in the same direction as $\bfz$, divide $\bfz$ by its length.
Definition of vector length

We can compute that the length of $ \frac{1}{2} \bfi - \frac{1}{2} \bfj + \frac{1}{2} \bfk$ is $\frac{\sqrt{3}}{2}.$
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 $$\frac{1}{\sqrt{3}} \bfi - \frac{1}{\sqrt{3}} \bfj + \frac{1}{\sqrt{3}} \bfk \ \text{ and } \ -\frac{1}{\sqrt{3}} \bfi + \frac{1}{\sqrt{3}} \bfj - \frac{1}{\sqrt{3}} \bfk.$$

Problem on finding the unit vectors perpendicular to two vectors

Solution

Related topics