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Find the vector equation of the plane in which the lines `vecr=hati+hatj+lambda(hati+2hatj-hatk)` and `vecr=(hati+hatj)+mu(-hati+hatj-2hatk)` lie.A. ` vecr.(2hati+hatj-3hatk)=-4`B. `vecrxx(-hati+hatj+hatk)=vec0`C. `vecr.(-hati+hatj+hatk)=0`D. none of these |
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Answer» Correct Answer - C The lines are parallel to the vector `vecb_(1)=hati+2hatj-hatk` and `vecb_(2)=-hati+hatj-hatk`. Therefore, the plane is normal to the vector `vecn=vecb_(1)xxvecb_(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` `impliesvecr.(-3hati+3hatj+3hatk)=(hati+hatj).(-3hati+hatj+3hatk)` `impliesvecr.(-3hati+3hatj+3hatk)=3+3impliesvecr.(-hati+hatj+hatk)=0` |
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