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Question

Find the acute angle between the lines whose direction ratios are proportional to 2 : 3 : 6 and 1 : 2 : 2.

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Solution

Let a be a vector parallel to the vector with direction ratios 2, 3, 6. a=2i^+3j^+6k^. Let b be a vector parallel to the vector with direction ratios 1, 2, 2.b= i^ +2j^ +2k^Let θ be the angle between the the given vectors. Now,cos θ =a.ba b =2i^+3j^+6k^.i^ +2j^ +2k^2i^+3j^+6k^i^ +2j^ +2k^ = 2+6+12 4+9+36 1+4+4 = 2021 θ=cos-12021

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