InterviewSolution
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InterviewSolution
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The distance of the point `(1,-2,4)` from the plane passing through the point `(1,2,2)` perpendicular to the planes `x-y+2z=3` and `2x-2y+z+12=0` isA. `1/(sqrt(2))`B. `2`C. `sqrt(2)`D. `2sqrt(2)` |
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Answer» Correct Answer - D Let `vecn` be a vector normal to the required plane. Then, its equation is `(vecr-(hati+2hatj+2hatk)).vecn=0`…………i This plane is perpendicular to the plane `x-y+2z=3` and `2x-2y+z+12=0` whose normals are `vecn_(1)=hati-hatj+2hatk` and `vecn_(2)=2hati-2hatj+hatk` `:.vecn=vecn_(1)xxvecn_(2)=|(hati,hatj,hatk),(1,-1,2),(2,-2,1)|=3hati+3hatj+0hatk` Substituting `vecn=3hati+3hatj` in i we obtain `vecr.(3hati+3hatj)=9` or `vecr.(hati+hatj)=3`............ii The distance of this plane from `veca=hati-2hatj+4hatk` is given by `d=|((hati-2hatj+4hatk).(hati+hatj)-3)/(sqrt(1+1))|=2sqrt(2)` |
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