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Question

# The vector $\stackrel{\to }{b}=3\stackrel{^}{i}+4\stackrel{^}{k}$ is to be written as the sum of a vector $\stackrel{\to }{\mathrm{\alpha }}$parallel to $\stackrel{\to }{a}=\stackrel{^}{i}+\stackrel{^}{j}$ and a vector $\stackrel{\to }{\mathrm{\beta }}$perpendicular to $\stackrel{\to }{a}$. Then $\stackrel{\to }{\mathrm{\alpha }}=$ (a) $\frac{3}{2}\left(\stackrel{^}{i}+\stackrel{^}{j}\right)$ (b) $\frac{2}{3}\left(\stackrel{^}{i}+\stackrel{^}{j}\right)$ (c) $\frac{1}{2}\left(\stackrel{^}{i}+\stackrel{^}{j}\right)$ (d) $\frac{1}{3}\left(\stackrel{^}{i}+\stackrel{^}{j}\right)$

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Solution

## (a) $\frac{3}{2}\left(\stackrel{^}{i}+\stackrel{^}{j}\right)$ $\text{Let:}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\alpha }\text{=}{a}_{1}\stackrel{^}{i}+{a}_{2}\stackrel{^}{j}+{a}_{3}\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\beta }={b}_{1}\stackrel{^}{i}+{b}_{2}\stackrel{^}{j}+{b}_{3}\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{Now,}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{b}\text{=3}\stackrel{^}{i}\text{+4}\stackrel{^}{k}\text{=}\stackrel{\to }{\alpha }\text{+}\stackrel{\to }{\beta }\text{(Given)}\phantom{\rule{0ex}{0ex}}⇒3\stackrel{^}{i}+0\stackrel{^}{j}+4\stackrel{^}{k}=\left({a}_{1}+{b}_{1}\right)\stackrel{^}{i}+\left({a}_{2}+{b}_{2}\right)\stackrel{^}{j}+\left({a}_{3}+{b}_{3}\right)\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}⇒{a}_{1}+{b}_{1}=3;{a}_{2}+{b}_{2}=0;{a}_{3}+{b}_{3}=4\phantom{\rule{0ex}{0ex}}⇒{a}_{1}+{b}_{1}=3;{a}_{2}=-{b}_{2};{a}_{3}+{b}_{3}=4...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{a}\text{=}\stackrel{^}{i}\text{+}\stackrel{^}{j}\text{}\left(\mathrm{Given}\right)\phantom{\rule{0ex}{0ex}}\text{Also,}\stackrel{\to }{\alpha }\text{is parallel to}\stackrel{\to }{a}\text{.}\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{\alpha }×\stackrel{\to }{a}=\stackrel{\to }{0}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\left|\begin{array}{ccc}\stackrel{^}{i}& \stackrel{^}{j}& \stackrel{^}{k}\\ {a}_{1}& {a}_{2}& {a}_{3}\\ 1& 1& 0\end{array}\right|=\stackrel{\to }{0}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒-{a}_{3}\stackrel{^}{i}+{a}_{3}\stackrel{^}{j}+\left({a}_{1}-{a}_{2}\right)\stackrel{^}{k}=0\stackrel{^}{i}+0\stackrel{^}{j}+0\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}⇒{a}_{3}=0;{a}_{1}-{a}_{2}=0\phantom{\rule{0ex}{0ex}}⇒{a}_{3}=0;{a}_{1}={a}_{2}...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{Since}\stackrel{\to }{\beta }\text{is perpendicular to}\stackrel{\to }{a}\text{, we get}\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{\beta }.\stackrel{\to }{a}=0\phantom{\rule{0ex}{0ex}}⇒\left({b}_{1}\stackrel{^}{i}+{b}_{2}\stackrel{^}{j}+{b}_{3}\stackrel{^}{k}\right).\left(\stackrel{^}{i}\text{+}\stackrel{^}{j}\right)=0\phantom{\rule{0ex}{0ex}}⇒{b}_{1}+{b}_{2}=0\phantom{\rule{0ex}{0ex}}⇒{b}_{1}=-{b}_{2}...\left(3\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{Solving (1), (2) and (3), we get}\phantom{\rule{0ex}{0ex}}{a}_{1}=\frac{3}{2};{a}_{2}=\frac{3}{2};{a}_{3}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{∴}\stackrel{\to }{\alpha }\text{=}{a}_{1}\stackrel{^}{i}+{a}_{2}\stackrel{^}{j}+{a}_{3}\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}=\frac{3}{2}\stackrel{^}{i}+\frac{3}{2}\stackrel{^}{j}+0\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}=\frac{3}{2}\left(\stackrel{^}{i}+\stackrel{^}{j}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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