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The vector equation of the plane passing through the points `hati+hatj-2hatk,2hati-hatj+hatk` and `hati+hatj+hatk`, isA. `vecr.(9hati+3hatj-hatk)=-14`B. `vecr.(9hati+3hatj-hatk)=14`C. `vecr.(3hati+9hatj-hatk)=14`D. none of these |
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Answer» Correct Answer - B Let A,B,C be the points with position vectors `hati+hatj-2hatk,2hati-hatj+hatk` and `hati+2hatj+hatk` respectively. Then, `vec(AB)=hati-2hatj+3hatk` and `vec(AB)=-hati+3hatj+0hatk` The vector normal to the plane containing points A,B and C is `vecn=vec(AB)xxvec(AC)=|(hati,hatj,hatk),(1,-2,3),(-1,3,0)|=-9hati-3hatj+hatk` The required plane passes through the point having position vector `veca=hati+hatj-2hatk` and is normal to the vector `vecn=-9hati-3hatj+hatk`. So its vector equation is `impliesvecr.vecn=veca.vecnimpliesveci.(9hati+3hatj-hatk)=14` |
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