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Question

Find the vector equation of a line passing through the point (2, 3, 2) and parallel to the line r=-2i^+3j^ +λ2i^-3j^+6k^ . Also find the distance between these lines.

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Solution

Vector equation of a line passing through (2, 3, 2) and parallel to the line r=-2i^+3j^+λ2i^-3j^+6k^ has direction ratios (2, −3, 6)
Hence, its equation is

r=2i^+3j^+2k^+λ2i^-3j^+6k^
Hence, distance between r=a1+λb1 and r=a2+μb1=a2+μb1 is d=a1-a2×bb
Here, a1=-2i^+3j^, a2=2i^+3j^+2k^ and b=2i^-3j^+6k^i.e. d=-4i^-2k^×2i^-3j^+6k^22+-32+62a1-a2×b=i^j^k^-40-22-36 =i^-6-j^-24+4+k^12i.e a1-a2×b=-6i^+20j^+12k^ d=-6i^+20j^+12k^4+9+36 =-6i^+20j^+12k^7 =2-3i^+10j^+6k^7 =279+100+36 =27×145




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