Question

The equation of a plane passing through $\overline{r}.(\hat{i}+\hat{j}+\hat{k})=1$ and $\overline{r}.(2\hat{i}+\hat{k})=2$ can be given by

• A. $\overline{r}.\left(\right(1+2\lambda )\hat{i}+\hat{j}+(1+\lambda \left)\hat{k}\right)=1+2\lambda$
• B. $\overline{r}.\left(\right(\lambda +2)\hat{i}+\lambda \hat{j}+(\lambda +1\left)\hat{k}\right)=\lambda +2$
• C. $\overline{r}.\left(\right(3\lambda \hat{i}+2\lambda \hat{j}+2\lambda \hat{k})=3)=3\lambda$
• D. $\overline{r}.(\hat{i}+\hat{j}+\hat{k})=1$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\overline{r}.\left(\right(1+2\lambda )\hat{i}+\hat{j}+(1+\lambda \left)\hat{k}\right)=1+2\lambda$
B $\overline{r}.\left(\right(\lambda +2)\hat{i}+\lambda \hat{j}+(\lambda +1\left)\hat{k}\right)=\lambda +2$
C $\overline{r}.\left(\right(3\lambda \hat{i}+2\lambda \hat{j}+2\lambda \hat{k})=3)=3\lambda$
D $\overline{r}.(\hat{i}+\hat{j}+\hat{k})=1$

We already know equation of any plane passing through intersection of $\overline{r}.\widehat{n_1}=d_1$ and $\overline{r}.\widehat{n_2}=d_2$ will be $\overline{r}(\widehat{n_1}+\lambda \widehat{n_2})=d_1+\lambda d_2$
Therefore, $\overline{r}.(\hat{i}+\hat{j}+\hat{k}+\lambda (2\hat{i}+\hat{k}\left)\right)=1+2\lambda$
$\overline{r}.\left(\right(1+2\lambda )\hat{i}+\hat{j}+(1+\lambda \left)\hat{k}\right)=1+2\lambda$
$\therefore$ Option a:- Satisfies.

But we can do it the other way too i.e,

$\overline{r}.(\lambda \widehat{n_1}+\widehat{n_2})=\lambda d_1+d_2$ which gives,
$\overline{r}.\left(\right(\lambda +2\left)\hat{i}\right)+\lambda \hat{j}+(\lambda +1)\hat{k})=\lambda +2$

Option C is nothing but $\overline{r}.\lambda (3\hat{i}+2\hat{j}+2\hat{k})=3\lambda$
For $\lambda \ne 0$ this can be written as,
$\lambda (\overline{r}.((\hat{i}+\hat{j}+\hat{k}+1(2\hat{i}+\hat{k}\left)\right))=(1+2)\lambda$
Which is a linear combination of 2 planes.
Option d can be arrived easily by substituting $\lambda =0$ in $r.(n_1+\lambda \widehat{n_2})=d_2+\lambda d_2$
$\therefore$ Answer is a,b,c,d