Question

Equation of the line $L=0$ is given by $2x-y+z-2=0=x+2y-z-3$, then which of the following is/are true?

• A. Equation of the plane containing the line $L=0$ and passing through the point $(3,2,1)$ is $x-3y+2z+1=0$
• B. Equation of the plane containing the line $L=0$ and passing through the point $(3,2,1)$ is $2x-3y+z-1=0$
• C. Vector equation of the plane which passes through the point $(-1,3,2)$ and is perpendicular to the line $L=0$ is $\overrightarrow{r}.(\hat{i}-3\hat{j}-5\hat{k})+20=0$
• D. Vector equation of the plane which passes through the point $(-1,3,2)$ and is perpendicular to the line $L=0$ is $\overrightarrow{r}.(\hat{i}+3\hat{j}+5\hat{k})-18=0$

Answer 1

Answer: (A), (C)

The correct options are
A Equation of the plane containing the line $L=0$ and passing through the point $(3,2,1)$ is $x-3y+2z+1=0$
C Vector equation of the plane which passes through the point $(-1,3,2)$ and is perpendicular to the line $L=0$ is $\overrightarrow{r}.(\hat{i}-3\hat{j}-5\hat{k})+20=0$

Equation of a plane passes through $(-1,3,2)$ is $a(x+1)+b(y-3)+c(z-2)=0\dots \dots \left(1\right)$

$\left(1\right)$ is perpendicular to $L=0$

dr's of $L$ are $(2\hat{i}-\hat{j}+\hat{k})\times (\hat{i}+2\hat{j}-\hat{k})=(-1,3,5)$

we get $\frac{a}{-1}=\frac{b}{3}=\frac{c}{5}$,
So equation of plane is $(\overrightarrow{r}-(-\hat{i}+3\hat{j}+2\hat{k}\left)\right).(\hat{i}-3\hat{j}-5\hat{k})=0$
$\Rightarrow \overrightarrow{r}.(\hat{i}-3\hat{j}-5\hat{k})+20=0$

Equation of a plane containing the line formed by intersection of two planes $P_1$ and $P_2$ is given by:
$P_1+\lambda P_2=0$
$\Rightarrow (2x-y+z-2)+\lambda (x+2y-z-3)=0$

Now, the above plane passes through $(3,2,1)$

$\Rightarrow \lambda =-1$

So required plane is $x-3y+2z+1=0$