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The vector equation of the plane which is at a distance of 8 units from the origin which is normal to the vector `2hati+hatj+2hatk` isA. `vecr.(2hati+hatj+2hatk)=8`B. `vecr.(2hati+hatj+2hatk)=24`C. `vecr.(2hati+hatj+2hatk)=4`D. none of these |
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Answer» Correct Answer - B Here `d=8` and `vecn=2hati+hatj+2hatk` `:. hatn=(vecn)/(|vecn|)=(2hati+hatj+2hatk)/(sqrt(4+1+4))=2/3hati+1/3hatj+2/3hatk` Hence the required equation of the plane is `vecr.(2/3hati+1/3hatj+2/3hatk)=8impliesvecr.(2hati+hatj+2hatk)=24` |
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