Consider the points \(A(1,-1,2)\text{,}\) \(B(2,0,-1)\text{,}\) and \(C(0,-2,3)\text{.}\) Find a unit vector which is orthogonal to both \(\vec{AB}\) and \(\vec{AC}\text{.}\)
Section 4.3 Computing a Vector Perpendicular to Two Other Vectors
Example 4.13. Computing a vector perpendicular to two other vectors.
Solution.
We first compute the two vectors:
\begin{align*}
\vec{AB} \amp= \langle 1,1,-3\rangle\\
\vec{AC} \amp= \langle -1,-1,1\rangle.
\end{align*}
The cross product \(\vec{AB} \times \vec{AC}\) is orthogonal to both vectors \(\vec{AB}\) and \(\vec{AC}\text{:}\)
\begin{align*}
\vec{AB} \times \vec{AC} \amp=
\begin{vmatrix}
\mathbf i \amp \mathbf j \amp \mathbf k \\
1 \amp 1 \amp -3 \\
-1 \amp -1 \amp 1
\end{vmatrix}
=
\begin{vmatrix} 1 \amp -3 \\ -1 \amp 1 \end{vmatrix}\mathbf i
- \begin{vmatrix} 1 \amp -3 \\ -1 \amp 1 \end{vmatrix}\mathbf j
+ \begin{vmatrix} 1 \amp 1 \\ -1 \amp -1 \end{vmatrix}\mathbf k\\
\amp= (1-3)\mathbf i - (1-3)\mathbf j + (-1+1)\mathbf k
= -2\mathbf i + 2\mathbf j
= \langle -2,2,0\rangle.
\end{align*}
The unit vector in question is then computed as follows:
\begin{equation*}
\begin{aligned}
\mathbf n \amp = \frac{\vec{AB} \times \vec{AC}}{\left|\vec{AB} \times \vec{AC}\right|}\\
\amp = \frac{1}{\sqrt 8}\langle -2,2,0\rangle
= \left\langle -\frac{1}{\sqrt 2},\frac{1}{\sqrt 2},0\right\rangle
= -\frac{1}{\sqrt 2}\mathbf i + \frac{1}{\sqrt 2}\mathbf j.
\end{aligned}
\end{equation*}
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