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Find the vector equation of the plane which is at a distance of 5 units from the origin and which is normal to the vector `2hati + hatj + 2hatk`. |
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Answer» Let `vecn = 2hati+ hatj+2 hatk` be a normal vector to the plane. `therefore` Unit vector along the normal to the plane `hatn =(veca)/(|veca|)` `=(2hati + hatj + 2hatk)/(sqrt(2^(2)+1^(2)+2^(2)))` `=(2hati+hatj+2hatk)/(3)` `therefore `Equation of the required plane is `vecr . hatn = p` `rArr" "vecr.((2hati + hatj + 2hatk)/(3))=5` `rArr" "vecr. (2hati+hatj+2hatk)=15` |
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