Question

The vector equation of the plane which is at a distance of $\frac{3}{\sqrt{14}}$ from the origin and the normal from the origin is $2\hat{i}-3\hat{j}+\hat{k}$ is

• A. $\overrightarrow{r}.(2\hat{i}-3\hat{j}+\hat{k})=3$
• B. $\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k})=9$
• C. $\overrightarrow{r}.(\hat{i}+2\hat{j})=3$
• D. $\overrightarrow{r}.(2\hat{i}+\hat{k})=3$

Answer 1

Answer: (A)

$\overrightarrow{r}.(2\hat{i}-3\hat{j}+\hat{k})=3$

The unit vector along the normal of the plane is given by
$\overrightarrow{n}=\frac{2i-3j+k}{\sqrt{4+9+1}}=\frac{2i-3j+k}{\sqrt{14}}$
Therefore, the vector equation of the plane with the above normal is
$\overrightarrow{r}.\left(\frac{2i-3j+k}{\sqrt{14}}\right)=d$ where 'd' is the distance of the plane from the origin.
Now it is given that $d=\frac{3}{\sqrt{14}}$, hence the vector equation of the plane becomes
$\overrightarrow{r}.\left(\frac{2i-3j+k}{\sqrt{14}}\right)=\frac{3}{\sqrt{14}}$ or
$\overrightarrow{r}.(2i-3j+k)=3$.