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Question

# Decompose the vector $6\stackrel{^}{i}-3\stackrel{^}{j}-6\stackrel{^}{k}$ into vectors which are parallel and perpendicular to the vector $\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}.$

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Solution

## $\text{Let}\stackrel{\to }{a}\text{=6}\stackrel{^}{i}-3\stackrel{^}{j}-6\stackrel{^}{k}\text{and}\stackrel{\to }{b}\text{=}\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\mathrm{and}\stackrel{\to }{x}\text{and}\stackrel{\to }{y}\text{be such that}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{a}=\stackrel{\to }{x}+\stackrel{\to }{y}\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{y}=\stackrel{\to }{a}-\stackrel{\to }{x}...\left(1\right)\phantom{\rule{0ex}{0ex}}\text{Since}\stackrel{\to }{x}\text{is parallel to}\stackrel{\to }{b}\text{,}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{x}=t\stackrel{\to }{b}\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{x}=t\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=t\stackrel{^}{i}+t\stackrel{^}{j}+t\stackrel{^}{k}\text{...(2)}\phantom{\rule{0ex}{0ex}}\text{Substituting the values of}\stackrel{\to }{x}\text{and}\stackrel{\to }{a}\text{in (1), we get}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{y}=6\stackrel{^}{i}-3\stackrel{^}{j}-6\stackrel{^}{k}-\left(t\stackrel{^}{i}+t\stackrel{^}{j}+t\stackrel{^}{k}\right)=\left(6-t\right)\stackrel{^}{i}+\left(-3-t\right)\stackrel{^}{j}+\left(-6-t\right)\stackrel{^}{k}...\left(3\right)\phantom{\rule{0ex}{0ex}}\text{Since}\stackrel{\to }{y}\text{is perpendicular to}\stackrel{\to }{b}\text{,}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{y}.\stackrel{\to }{b}=0\phantom{\rule{0ex}{0ex}}⇒\left[\left(6-t\right)\stackrel{^}{i}+\left(-3-t\right)\stackrel{^}{j}+\left(-6-t\right)\stackrel{^}{k}\right].\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=0\phantom{\rule{0ex}{0ex}}⇒1\left(6-t\right)+1\left(-3-t\right)+1\left(-6-t\right)=0\phantom{\rule{0ex}{0ex}}⇒-3-3t=0\phantom{\rule{0ex}{0ex}}⇒t=-1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{From (2) and (3), we get}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{x}=-\stackrel{^}{i}-\stackrel{^}{j}-\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{y}=7\stackrel{^}{i}-2\stackrel{^}{j}-5\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{So,}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{a}=\stackrel{\to }{x}+\stackrel{\to }{y}=\left(-\stackrel{^}{i}-\stackrel{^}{j}-\stackrel{^}{k}\right)+\left(7\stackrel{^}{i}-2\stackrel{^}{j}-5\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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