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Question

# Prove that the points having position vectors $\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k},3\stackrel{^}{i}+4\stackrel{^}{j}+7\stackrel{^}{k},-3\stackrel{^}{i}-2\stackrel{^}{i}-5\stackrel{^}{k}$ are collinear.

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Solution

## Let $A,B,C$ be the points with position vectors $\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k},3\stackrel{^}{i}+4\stackrel{^}{j}+7\stackrel{^}{k},-3\stackrel{^}{i}-2\stackrel{^}{j}-5\stackrel{^}{k}.$ Then, $\stackrel{\to }{AB}=$ Position vector of B $-$ Position vector of A $=3\stackrel{^}{i}+4\stackrel{^}{j}+7\stackrel{^}{k}-\stackrel{^}{i}-2\stackrel{^}{j}-3\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}=2\stackrel{^}{i}+2\stackrel{^}{j}+4\stackrel{^}{k}$ $\stackrel{\to }{BC}=$Position vector of C $-$ Position vector of B $=-3\stackrel{^}{i}-2\stackrel{^}{j}-5\stackrel{^}{k}-3\stackrel{^}{i}-4\stackrel{^}{j}-7\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}=-6\stackrel{^}{i}-6\stackrel{^}{j}-12\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}=-3\left(2\stackrel{^}{i}+2\stackrel{^}{j}+4\stackrel{^}{k}\right)$ ∴ $\stackrel{\to }{BC}=-3\stackrel{\to }{AB}$ $\mathrm{So},\stackrel{\to }{AB}\mathrm{and}\stackrel{\to }{BC}$ are parallel vectors. But $B$ is a point common to them. So, $\stackrel{\to }{AB}$ and $\stackrel{\to }{BC}$ are collinear. Hence, $A,B,C$ are collinear.

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