Question

Equation of the plane passing through a point with position vector $\hat{i}+\hat{j}-\hat{k}$ and $\perp$er to the planes $\overline{r}.(\hat{i}+2\hat{j}+3\hat{k})=7$ & $\overline{r}.(2\hat{i}-3\hat{j}+4\hat{k})=0$ is

• A. $\overline{r}.(17\hat{i}+2\hat{j}-7\hat{k})=26$
• B. $\overline{r}.(3\hat{i}-2\hat{j}+3\hat{k})+2=0$
• C. $\overline{r}.(6\hat{i}+3\hat{j}-4\hat{k})=13$
• D. $\overline{r}.(17\hat{i}+2\hat{j}+7\hat{k})=12$

Answer 1

Answer: (A)

$\overline{r}.(17\hat{i}+2\hat{j}-7\hat{k})=26$

The required plane passes through the point having position vector $\overline{a}=\hat{i}+\hat{j}-\hat{k}$
Let normal vector to the plane be $\overline{n}$, then $\overline{n}$ is $\perp$ler to the normal to the given planes
$\overline{r}\cdot (\hat{i}+2\hat{j}+3\hat{k})=7and\overline{r}\cdot (2\hat{i}-3\hat{j}+4\hat{k})=0$
$\therefore \overline{n}={\overline{n}}_1\times {\overline{n}}_2=\left|\begin{array}{ccc}i& j& k\\ 1& 2& 3\\ 2& -3& 4\end{array}\right|=17\hat{i}+2\hat{j}-7\hat{k}$
$\therefore (\overline{r}-\overline{a}).\overline{n}=0$
$\Rightarrow \overline{r}\cdot \overline{n}=\overline{a}\cdot \overline{n}$
$\overline{r}\cdot (17\hat{i}+2\hat{j}-7\hat{k})=(\hat{i}+\hat{j}-\hat{k})\cdot (17\hat{i}+2\hat{j}-7\hat{k})$
$\overline{r}\cdot (17\hat{i}+2\hat{j}-7\hat{k})=17+2+7=26$
Hence choice (A) is correct.