Problem on finding a vector perpendicular to two vectors
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Find a unit vector perpendicular to $\langle 1, 1, 1\rangle$ and $\langle 1, 0, 1 \rangle.$
Solution
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 $\bfx = \langle 1, 1, 1\rangle$ and $\bfy = \langle 1, 0, 1 \rangle$.
Computing the cross product,
Determinant of a 3x3 matrix
\begin{align}\mathbf{x} \times \mathbf{y} &= \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix}\\ \ \\ &= \begin{vmatrix}1 & 1 \\ 0 & 1 \end{vmatrix} \ \mathbf{i} - \begin{vmatrix}1 & 1 \\ 1 & 1 \end{vmatrix} \ \mathbf{j} + \begin{vmatrix}1 & 1 \\ 1 & 0 \end{vmatrix} \ \mathbf{k} \end{align}
Determinant of a 2x2 matrix
Evaluating the $2 \times 2$ determinants, $$\bfx \times \bfy= \bfi - \bfk.$$
Hence, $\bfi - \bfk = \langle 1, 0, -1\rangle$ is a vector that is perpendicular to both $\langle 1, 1, 1\rangle$ and $\langle 1, 0, 0\rangle.$
Let's determine if this vector has unit length.
Recall that
Definition of vector length
The length of the vector $\mathbf{z} = \langle z_1, z_2, z_3 \rangle$ is $|\mathbf{z}| = \sqrt{z_1^2 + z_2^2 + z_3^2}.$
Hence, we observe $\left | \langle 1, 0, -1 \rangle \right | = \sqrt{2}$.
Recall that
Finding the unit vector in the same direction as another vector
The find the unit vector in the same direction as $\bfz$, divide by the length of $\bfz$.
Hence, a unit vector that is perpendicular to both $\langle 1, 1, 1\rangle$ and $\langle 1, 0, 0\rangle$ is $$ \frac{\langle 1, 0, -1\rangle}{\sqrt{2}} = \left \langle \frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}} \right \rangle.$$
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