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Question

(i) Find a unit vector perpendicular to both the vectors $4\stackrel{^}{i}-\stackrel{^}{j}+3\stackrel{^}{k}\mathrm{and}-2\stackrel{^}{i}+\stackrel{^}{j}-2\stackrel{^}{k}.$ (ii) Find a unit vector perpendicular to the plane containing the vectors $\stackrel{\to }{a}=2\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\mathrm{and}\stackrel{\to }{b}=\stackrel{^}{i}+2\stackrel{^}{j}+\stackrel{^}{k}.$

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Solution

$\left(\mathrm{i}\right)\mathrm{Given}:\phantom{\rule{0ex}{0ex}}\stackrel{\to }{a}=4\stackrel{^}{i}-\stackrel{^}{j}+3\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{b}=-2\stackrel{^}{i}+\stackrel{^}{j}-2\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore \stackrel{\to }{a}×\stackrel{\to }{b}=\left|\begin{array}{ccc}\stackrel{^}{i}& \stackrel{^}{j}& \stackrel{^}{k}\\ 4& -1& 3\\ -2& 1& -2\end{array}\right|\phantom{\rule{0ex}{0ex}}=\left(2-3\right)\stackrel{^}{i}-\left(-8+6\right)\stackrel{^}{j}+\left(4-2\right)\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}=-\stackrel{^}{i}+2\stackrel{^}{j}+2\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\left|\stackrel{\to }{a}×\stackrel{\to }{b}\right|=\sqrt{1+{2}^{2}+{2}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{9}\phantom{\rule{0ex}{0ex}}=3\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{Unit vector perpendicular to}\stackrel{\to }{a}\text{and}\stackrel{\to }{b}\text{=}\frac{\stackrel{\to }{a}×\stackrel{\to }{b}}{\left|\stackrel{\to }{a}×\stackrel{\to }{b}\right|}=\frac{-\stackrel{^}{i}+2\stackrel{^}{j}+2\stackrel{^}{k}}{3}\phantom{\rule{0ex}{0ex}}$ $\left(\mathrm{ii}\right)\mathrm{Given}:\phantom{\rule{0ex}{0ex}}\stackrel{\to }{a}=2\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{b}=\stackrel{^}{i}+2\stackrel{^}{j}+\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\stackrel{}{\therefore \stackrel{\to }{a}}×\stackrel{}{\stackrel{\to }{b}}=\left|\begin{array}{ccc}\stackrel{^}{i}& \stackrel{^}{j}& \stackrel{^}{k}\\ 2& 1& 1\\ 1& 2& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}=\left(1-2\right)\stackrel{^}{i}-\left(2-1\right)\stackrel{^}{j}+\stackrel{}{\left(4-1\right)\stackrel{⏜}{k}}\phantom{\rule{0ex}{0ex}}=-\stackrel{^}{i}-\stackrel{^}{j}+3\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\left|\stackrel{\to }{a}×\stackrel{\to }{b}\right|=\sqrt{1+1+9}\phantom{\rule{0ex}{0ex}}=\sqrt{11}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{Unit vector perpendicular to the plane containing vectors}\stackrel{\to }{a}\text{and}\stackrel{\to }{b}=±\frac{\stackrel{\to }{a}×\stackrel{\to }{b}}{\left|\stackrel{\to }{a}×\stackrel{\to }{b}\right|}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{Unit vector perpendicular to the plane containing vectors}\stackrel{\to }{a}\text{and}\stackrel{\to }{b}=±\frac{1}{\sqrt{11}}\left(-\stackrel{^}{i}-\stackrel{^}{j}+3\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}$

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