Question

Find the vector equation of a plane which is at a distance of 5 units from the origin and which is normal to the vector  $$\hat{i}-\text{2 }\hat{j}-\text{2 }\hat{k}.$$

 

Answer 1

${\displaystyle \text{It is given that the normal vector,}\overrightarrow{n}=\widehat{i }-2 \widehat{j }-2 \widehat{k }}$

${\displaystyle \text{Now,} \hat{n}=\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}=\frac{\hat{i}-2 \widehat{j }-2\widehat{k }}{\sqrt{1+4+4}}=\frac{\hat{i}-2\hat{j}-2\hat{k}}{3}=\frac{1}{3}\widehat{i }-\frac{2}{3}\widehat{j }-\frac{2}{3} \widehat{k }}$

The equation of a plane in normal form is  

${\displaystyle \overrightarrow{r}.\hat{n}=\text{ d (where d is the distance of the plane from the origin)} }$

${\displaystyle \text{Substituting} \hat{n}=\frac{1}{3}\widehat{i }-\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k}\text{and}d=5}$ 

Here, 

${\displaystyle \overrightarrow{r}.(\frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k}=5}$