LinAlg4.2.5

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Find the vector equation of a plane which is at a distance of 5units from the origin and which has 2{\bar  {i}}+3{\bar  {j}}+6{\bar  {k}}\, as a normal vector.

Given p=5 and let {\bar  {n}}=2{\bar  {i}}+3{\bar  {j}}+6{\bar  {k}},|n|={\sqrt  {4+9+36}}={\sqrt  {49}}=7\,

Therefore,unit vector of n is {\frac  {{\bar  {n}}}{|n|}}={\frac  {1}{7}}(2{\bar  {i}}+3{\bar  {j}}+6{\bar  {k}})\,

Hence the required equation of the plane is {\bar  {r}}\cdot {\frac  {1}{7}}(2{\bar  {i}}+3{\bar  {j}}+6{\bar  {k}})=5\,

{\bar  {r}}\cdot (2{\bar  {i}}+3{\bar  {j}}+6{\bar  {k}})=35\,

Cartesian equation of the plane is (x{\bar  {i}}+y{\bar  {j}}+z{\bar  {k}})\cdot (2{\bar  {i}}+3{\bar  {j}}+6{\bar  {k}})=35\,

2x+3y+6z=35\,

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