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Question Number 56711 by Tawa1 last updated on 22/Mar/19

$$\mathrm{Find}\mathrm{the}\mathrm{shotest}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{line}$$$$\frac{\mathrm{x}-8}{3}=\frac{\mathrm{y}-2}{4}=\frac{\mathrm{z}+1}{1},\frac{\mathrm{x}-3}{3}=\frac{\mathrm{y}+4}{5}=\frac{\mathrm{z}-2}{2}$$

Answered by mr W last updated on 23/Mar/19

$$usingvectormethod:$$$$$$$$L1:(8,2,-1)+s(3,4,1)$$$$L2:(3,-4,2)+t(3,5,2)$$$$normaltoL1andL2:$$$$(3,4,1)\times (3,5,2)=(3,-3,3)$$$$unitnormal:$$$$(3,-3,3)/\sqrt{3^2+{(-3)}^2+3^2}=(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$$$$d=\mid (8-3,2+4,-1-2)\bullet (\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})\mid$$$$d=\mid (5,6,-3)\bullet (1,-1,1)\mid /\sqrt{3}$$$$d=\mid -4\mid /\sqrt{3}$$$$\Rightarrow d=\frac{4}{\sqrt{3}}\approx 2.309$$

Commented by Tawa1 last updated on 26/Mar/19

$$\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{effort}\mathrm{sir}$$

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