Question

Equation of a plane passing through $\hat{i}+\hat{j}+\hat{k}$ parallel to both $\hat{i}+\hat{j}and\hat{j}+\hat{k}$ can be given by

• A. $\overline{r}.(\hat{i}+\hat{j}+\hat{k})=1$
• B. $\overline{r}.(\hat{i}-\hat{j}+\hat{k})=1$
• C. $\overline{r}.(\hat{i}+\hat{j}-\hat{k})=1$
• D. $\overline{r}.(\hat{i}-\hat{j}-\hat{k})=1$

Answer 1

The correct option is B

$\overline{r}.(\hat{i}-\hat{j}+\hat{k})=1$

We know equation of plane passing through $\overline{a}$ perpendicular to $\hat{n}$is $(r-\overline{a}).\hat{n}=0$
Here, $a=\hat{i}+\hat{j}+\hat{k}$
How to get $\hat{n}$?

We can see $\hat{n}$ is perpendicular to both $\hat{i}+\hat{j}and\hat{j}+\hat{k}$ i.e., along $(\hat{i}+\hat{j})\times (\hat{j}+\hat{k})$.

$\therefore (\hat{i}+\hat{j})\times (\hat{j}+\hat{k})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 0\\ 0& 1& 1\end{array}\right|=\hat{i}\left(1\right)-\hat{j}\left(1\right)+\hat{k}\left(1\right)=\hat{i}-\hat{j}+\hat{k}$

$\Rightarrow \hat{n}=\frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{1^2+1^2+1^2}}=\frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}\therefore equationofplane\Rightarrow (\overline{r}-(\hat{i}+\hat{j}+\hat{k}\left)\right).\hat{n}=0$

$\overline{r}.\left(\frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}\right)=\frac{1^2-1^2+1^2}{\sqrt{3}}=\frac{1}{\sqrt{3}}\therefore \overline{r}.(\hat{i}-\hat{j}+\hat{k})=1$