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