A unit vector perpendicular to the vectors $\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $\vec{b}=3 \vec{j}+2…
A unit vector perpendicular to the vectors $\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $\vec{b}=3 \vec{j}+2 \vec{k}$ is
- $\frac{3 \hat{i}+2 \hat{j}-2 \hat{k}}{\sqrt{22}}$
- $\frac{3 \hat{i}+2 \hat{j}-3 \hat{k}}{\sqrt{22}}$
- $\frac{3 \hat{i}-2 \hat{j}+3 \hat{k}}{\sqrt{22}}$
- $\frac{3 \hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{22}}$
Solution
$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & 3 & 2\end{array}\right|=-6 \hat{i}-4 \hat{j}+6 \hat{k}$
$\therefore$ Unit vector perpendicular to $\vec{a}$ and $\vec{b}$
$= \pm \frac{3 \hat{i}+2 \hat{j}-3 \hat{k}}{\sqrt{22}}$
Asked in: AP EAMCET 2024 (23 May Shift 1)
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