Find a unit vector orthogonal to the given vectors

For each of the following pairs of vectors find an orthogonal vector of unit length.

$A = \mathbf{i} + \mathbf{j} + \mathbf{k}, \qquad B = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k}$ ;
$A = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}, \qquad B = -\mathbf{i} + 5\mathbf{j} + 7\mathbf{k}$ ;
$A = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k}, \qquad B = -3\mathbf{i} + 2\mathbf{j} - \mathbf{k}$ .

First, we have
$\begin{align*} A \times B &= \begin{vmatrix*} \mathbf{i} & \mathbf{j} &\mathbf{k} \\ 1 & 1 & 1 \\ 2 & 3 & -1 \end{vmatrix*} \\[9pt] &= \begin{vmatrix*} 1 & 1 \\ 3 & -1 \end{vmatrix*} \mathbf{i} - \begin{vmatrix*} 1 & 1 \\ 2 & -1 \end{vmatrix*} \mathbf{j} + \begin{vmatrix*} 1 & 1 \\ 2 & 3 \end{vmatrix*}\mathbf{k} \\[9pt] &= -4\mathbf{i} +3 \mathbf{j} + \mathbf{k} \end{align*}$

Therefore, the vectors orthogonal to both $A$ and $B$ are of the form $N = (-4c, 3c, c)$ for $c \neq 0$ . Setting the length of this equal to 1 we have

$\begin{align*} \lVert N \rVert = 1 && \implies && 16c^2 + 9c^2 + c^2 &= 1 \\ && \implies && c^2 &= \frac{1}{26} \\ && \implies && c &= \pm \frac{1}{\sqrt{26}}. \end{align*}$

Therefore, the vectors orthogonal to $A$ and $B$ with unit length are

$N = \pm \frac{1}{\sqrt{26}} (-4,3,1).$
First, we have
$\begin{align*} A \times B &= \begin{vmatrix*} \mathbf{i} & \mathbf{j} &\mathbf{k} \\ 2 & -3 & 4 \\ -1 & 5 & 7 \end{vmatrix*} \\[9pt] &= \begin{vmatrix*} -3 & 4 \\ 5 & 7 \end{vmatrix*} \mathbf{i} - \begin{vmatrix*} 2 & 4 \\ -1 & 7 \end{vmatrix*} \mathbf{j} + \begin{vmatrix*} 2 & -3 \\ -1 & 5 \end{vmatrix*}\mathbf{k} \\[9pt] &= -41\mathbf{i} -18 \mathbf{j} + 7 \mathbf{k} \end{align*}$

Therefore, the vectors orthogonal to both $A$ and $B$ are of the form $N = (-41c, -18c, 7c)$ for $c \neq 0$ . Setting the length of this equal to 1 we have

$\begin{align*} \lVert N \rVert = 1 && \implies && 1681c^2 + 324c^2 + 49c^2 &= 1 \\ && \implies && c^2 &= \frac{1}{2054} \\ && \implies && c &= \pm \frac{1}{\sqrt{2054}}. \end{align*}$

Therefore, the vectors orthogonal to $A$ and $B$ with unit length are

$N = \pm \frac{1}{\sqrt{2054}} (-41,-18,7).$
First, we have
$\begin{align*} A \times B &= \begin{vmatrix*} \mathbf{i} & \mathbf{j} &\mathbf{k} \\ 1 & -2 & 3 \\ -3 & 2 & -1 \end{vmatrix*} \\[9pt] &= \begin{vmatrix*} -2 & 3 \\ 2 & -1 \end{vmatrix*} \mathbf{i} - \begin{vmatrix*} 1 & 3 \\ -3 & -1 \end{vmatrix*} \mathbf{j} + \begin{vmatrix*} 1 & -2 \\ -3 & 2 \end{vmatrix*}\mathbf{k} \\[9pt] &= -4 \mathbf{i} -8 \mathbf{j} - 4\mathbf{k} \end{align*}$

Therefore, the vectors orthogonal to both $A$ and $B$ are of the form $N = (-4c', -8c', -4c') = -4(c',2c',c') = (c,2c,c)$ for $c \neq 0$ . Setting the length of this equal to 1 we have

$\begin{align*} \lVert N \rVert = 1 && \implies && c^2 + 4c^2 + c^2 &= 1 \\ && \implies && c^2 &= \frac{1}{6} \\ && \implies && c &= \pm \frac{1}{\sqrt{6}}. \end{align*}$

Therefore, the vectors orthogonal to $A$ and $B$ with unit length are

$N = \pm \frac{1}{\sqrt{6}} (1,2,1).$