Question

The vector equation of the plane which is perpendicular to $2i+3j+k$ and at the distance of $5$ units from the origin is $(EAM-1991)$

• A. $\overline{r}.(2\overline{i}-3\overline{j}+\overline{k})=5\sqrt{14}$
• B. $\overline{r}.(2\overline{i}-3\overline{j}+\overline{k})=5$
• C. $\overline{r}.\frac{(2\overline{i}-3\overline{j}+\overline{k})}{\sqrt{14}}$
• D. $\frac{\overline{r}.(2\overline{i}-3\overline{j}+\overline{k})}{\sqrt{14}}$

Answer 1

The correct option is A $\overline{r}.(2\overline{i}-3\overline{j}+\overline{k})=5\sqrt{14}$

Let the required equation of the plane be

$\overrightarrow{r}.\hat{n}=d$

$\overrightarrow{r}.\hat{n}=5$

$\overrightarrow{r}.(l\hat{i}+m\hat{j}+n\hat{k})=5$

Since the plane is perpendicular to $2\hat{i}+3\hat{j}+\hat{k}$

So, \dfrac{l}{2}=\frac{m}{3}=\dfrac{n}{1}=k$

$l=2k,m=3k,n=k$

and, $l^2+m^2+n^2=1$

$4k^2+9k^2+k^2=1$

$k=\frac{1}{\sqrt{14}}$

So, $l=\frac{2}{\sqrt{14}},m=\frac{3}{\sqrt{14}},n=\frac{1}{\sqrt{14}}$

Hence required equation of the plane is

$\overrightarrow{r}.\frac{(2\hat{i}+3\hat{j}+\hat{k})}{\sqrt{14}}=5$

$\overrightarrow{r}.(2\hat{i}+3\hat{j}+\hat{k})=5\sqrt{14}$