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 $(x\bar{i}+y\bar{j}+z\bar{k})\cdot(2\bar{i}+3\bar{j}+6\bar{k})=35\,$

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

Main Page:Linear Algebra

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