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# Show that the points whose position vectors are as given below are collinear: (i) $2\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k},3\stackrel{^}{i}-2\stackrel{^}{j}+\stackrel{^}{k}\mathrm{and}\stackrel{^}{i}+4\stackrel{^}{j}-3\stackrel{^}{k}$ (ii) $3\stackrel{^}{i}-2\stackrel{^}{j}+4\stackrel{^}{k},\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\mathrm{and}-\stackrel{^}{i}+4\stackrel{^}{j}-2\stackrel{^}{k}$

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Solution

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