Question

Equation of the plane passing through a point with position vector $3\hat{i}-3\hat{j}+\hat{k}$ & normal to the line joining the points with position vectors $3\hat{i}+4\hat{j}-\hat{k}$ & $2\hat{i}-\hat{j}=5\hat{k}$ is

• A. $\overline{r}.(-\hat{i}-5\hat{j}+6\hat{k})+18=0$
• B. $\overline{r}.(\hat{i}-5\hat{j}+6\hat{k})=22$
• C. $\overline{r}.(\hat{i}+5\hat{j}-6\hat{k})+18=0$
• D. $\overline{r}.(-\hat{i}+5\hat{j}+6\hat{k})+12=0$

Answer 1

The correct option is C $\overline{r}.(\hat{i}+5\hat{j}-6\hat{k})+18=0$

$(\overline{r}-\overline{a})\cdot \overline{n}=0$
$\Rightarrow \overline{r}\cdot \overline{n}=\overline{a}\cdot \overline{n}$ where $\overline{n}=AB=-\hat{i}-5\hat{j}+6\hat{k}$
$\Rightarrow \overline{r}\cdot (-\hat{i}-5\hat{j}+6\hat{k})=(3\hat{i}-3\hat{j}+\hat{k})\cdot (-\hat{i}-5\hat{j}+6\hat{k})$
$\Rightarrow \overline{r}\cdot (\hat{i}+5\hat{j}-6\hat{k})+18=0$
Hence the correct option becomes c.