Find the vector equation of a plane, which is at a distance of 5 units from
the origin and whose normal vector is 2\hat{i}-\hat{j}+2\hat{k}\cdot

 

 

 

 
 
 
 
 

Answers (1)

\vec{n}= 2\hat{i}-\hat{j}+2\hat{k}
we know that  \vec{n}=\frac{\vec{n}}{\left | \vec{n} \right |}= \frac{ 2\hat{i}-\hat{j}+2\hat{k}}{\sqrt{\left ( 2 \right )^{2}+\left ( -1 \right )^{2}+\left ( 2 \right )^{2}}}
                      = \frac{ 2\hat{i}-\hat{j}+2\hat{k}}{3}= \frac{2}{3}\hat{i}-\frac{1}{3}\hat{j}+\frac{2}{3}\hat{k}
r\cdot \vec{n}= p  where p is the distance from origin
\Rightarrow r\cdot \left [\frac{2}{3}\hat{i} -\frac{1}{3}\hat{j} +\frac{2}{3}\hat{k}\right ]= 5
\Rightarrow r \left [2\hat{i}-\hat{j}+2\hat{k}]= 15

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