Question

The projection of the line $\frac{x - 1}{2}=\frac{y + 1}{1}=\frac{z - 2}{3}$ on a plane $P$ is $\frac{x - 1}{1}=\frac{y + 1}{2}=\frac{z - 2}{1}$ , then the equation of the plane $P$ is

• A. $5x-8y+11z=35$
• B. $5x+8y-21z+45=0$
• C. $5x+8y+11z=35$
• D. $5x-8y+21z=45$

Answer 1

Answer: (A)

The required plane passes through $\left(1 , - 1,2\right)$
The required plane is parallel to $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & 2 & 1 \end{vmatrix}$
$=\hat{i}\left(- 5\right)-\hat{j}\left(- 1\right)+\hat{k}\left(3\right)$
$=-5\hat{i}+\hat{j}+3\hat{k}$
So, normal vector to the plane is $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & 3 \\ 1 & 2 & 1 \end{vmatrix}$
$=\hat{i}\left(- 5\right)-\hat{j}\left(- 8\right)+\hat{k}\left(- 11\right)$
$=-5\hat{i}+8\hat{j}-11\hat{k}$
Equation of the required plane is
$-5\left(x - 1\right)+8\left(y + 1\right)-11\left(z - 2\right)=0$
$-5x+5+8y+8-11z+22=0$
$-5x+8y-11z+35=0$
$5x-8y+11z-35=0$