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# If P (3, 2, −4), Q (5, 4, −6) and R (9, 8, −10) are collinear, then R divides PQ in the ratio (a) 3 : 2 internally (b) 3 : 2 externally (c) 2 : 1 internally (d) 2 : 1 externally

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Solution

## $\left(\mathrm{b}\right)3:2\mathrm{externally}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Suppose}\mathrm{the}\mathrm{point}R\mathrm{divides}PQ\mathrm{in}\mathrm{the}\mathrm{ratio}\lambda :1.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Coordinates}\mathrm{of}R\mathrm{are}\left(\frac{5\lambda +3}{\lambda +1},\frac{4\lambda +2}{\lambda +1},\frac{-6\lambda -4}{\lambda +1}\right).\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{But}\mathrm{the}\mathrm{coordinates}\mathrm{of}R\mathrm{are}\left(9,8,-10\right).\phantom{\rule{0ex}{0ex}}\therefore \frac{5\lambda +3}{\lambda +1}=9,\frac{4\lambda +2}{\lambda +1}=8\mathrm{and}\frac{-6\lambda -4}{\lambda +1}=-10\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{From}\mathrm{each}\mathrm{of}\mathrm{these}\mathrm{equation}s,\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\lambda =-\frac{3}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore R\mathrm{divides}PQ\mathrm{in}\mathrm{the}\mathrm{ratio}3:2\mathrm{externally}.$

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