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VECTOR ALGEBRA — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA3 Marks
Question
Find a unit vector perpendicular to both the vectors $4\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ and $-2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}.}$
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Answer

A vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}=\vec{\text{a}}\times\vec{\text{b}}.$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2 \end{vmatrix}$
$\vec{\text{c}}(\text{say})=\begin{bmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\4&-1&3\\-2&1&-2 \end{bmatrix}$
$\vec{\text{c}}=\hat{\text{i}}(2-3)-\hat{\text{j}}(-8+6)+\hat{\text{k}}(4-2)$
$\vec{\text{c}}=-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{c}}$ is a vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}.$
$\hat{\text{c}}=\frac{\vec{\text{c}}}{|\vec{\text{c}}|}$
$=\frac{-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}}{\sqrt{(-1)^2+(2)^2+(2)^2}}$
$=\frac{-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}}{\sqrt{1+4+4}}$
$=\frac{1}{3}\big(-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)$
So, unit vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}=\frac{1}{3}\big(-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big).$

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