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VECTOR ALGEBRA — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA4 Marks
Question
Find a unit vector perpendicular to the plane containing the vectors $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}.}$
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Answer

A vector perpendicular to the plane containing the vector $\vec{\text{a}}$ and $\vec{\text{b}}$ is given by $\vec{\text{a}}\times\vec{\text{b}}=\pm\vec{\text{c}}$ (say)
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2 \end{vmatrix}$
$\vec{\text{c}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&1&1\\1&2&1 \end{vmatrix}$
$\vec{\text{c}}=\hat{\text{i}}(1-2)-\hat{\text{j}}(2-1)+\hat{\text{k}}(4-1)$
$\vec{\text{c}}=-\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$
$\hat{\text{c}}=\frac{\vec{\text{c}}}{|\vec{\text{c}}|}$
$=\frac{-\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}}{\sqrt{(-1)^2+(-1)^2+(3)^2}}$
$=\frac{-\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}}{\sqrt{1+1+9}}$
$=\frac{1}{\sqrt{11}}\big(-\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)$
Unit vector perpendicular to the plane of $\vec{\text{a}}$ and $\vec{\text{b}}=\pm\frac{1}{\sqrt{11}}\big(-\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big).$

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