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VECTOR ALGEBRA — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSVECTOR ALGEBRA2 Marks
Question
Find a unit vector perpendicular to each of the vector $\vec{a}+\vec{b}\ \text{and}\ \vec{a}-\vec{b},\ \text{where}$  $\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}\ \text{and}\ \vec{b}=\hat{i}+2\hat{j}-2\hat{k}.$
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Answer

$\text{Given:}\ \ \vec{a}=3\hat{i}+2\hat{j}+2\hat{k}\ \text{and}\ \vec{b}=\hat{i}+2\hat{j}-2\hat{k}$
$\text{On Adding}\ \vec{c}=\vec{a}+\vec{b}=3\hat{i}+2\hat{j+2\hat{k}}\ +\ \hat{i}+2\hat{j}-2\hat{k}$ $=4\hat{i}+4\hat{j}+0\hat{k}$
$\text{On Subtracting}\ \ \ \vec{d}=\vec{a}-\vec{b}=3\hat{i}+2\hat{j}+2\hat{k}\ - \ \hat{i}-2\hat{j}+2\hat{k}$ $=2\hat{i}+0\hat{j}+4\hat{k}$
$\text{Therefore,}\ \ \ \vec{n}=\vec{c}\times\vec{d}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\4&4&0\\2&0&4 \end{vmatrix}$
Expanding along first row $=\hat{i}(16-0)-\hat{i}(16-0)+\hat{k}(0-8)$
$\Rightarrow\ \ \vec{n}=16\hat{i}-16\hat{j}-8\hat{k}$
$\therefore\ \big|\vec{n}\big|=\sqrt{(16)^2+(-16)^2+(-8)^2}$ $\sqrt{256+256+64}=\sqrt{576}=24$
Therefore, a unit vector perpendicular to both $\vec{a}\ \text{and}\ \vec{b}\ \text{is}$
$\hat{n}=\pm\frac{\vec{n}}{|\vec{n}|}=\pm\frac{(16\hat{i}-16\hat{j}-8\hat{k})}{24}$ 
$=\pm\Big(\frac{16}{24}\hat{i}-\frac{16}{24}\hat{j}-\frac{8}{24}\hat{k}\Big)=\pm\Big(\frac{2}{3}\hat{i}-\frac{2}{3} \hat{j}-\frac{1}{3}\hat{k}\Big)$

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