InterviewSolution
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Equation of the plane that contains the lines vecr = (hati +hatj) + lambda(hati + 2hatj - hatk) and vecr = (hati + hatj) + mu(-hati + hatj-2hatk) is |
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Answer» `vecr(2hati + hatj - 3hatk)=-4` Therefore, the plane is NORMAL to the vector `vecn = vecb_(1) xx vecb_(2) = |{:(hati, hatj,hatk),(1,2,-1),(-1,1,-2):}|=-3hati + 3hatj + 3hatk` The required plane PASSES through `(hati + hatj)` and is normal to the vector `vecn`. Therefore, its equation is `vecr.vecn = veca. vecn` `rArr vecr.(-3hati + 3hatj + 3hatk) = (hati + hatj).(-3hati + 3hatj + 3hatk)` `rArr vecr.(-3hati + 3hatj + 3hatk) = -3+3` `rArr vecr.(-hati + hatj +hatk)=0` |
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