Question

The line $L_1:\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies in the plane $2x-4y+z=7$, then the equation of the plane containing the line $L_1$ and perpendicular to the plane $2x-4y+z=7,$ is:

• A. $x-2y-3z=0$
• B. $3x+y-2z=0$
• C. $-2x+3y-z=0$
• D. $2x+3y-z=0$

Answer 1

The correct option is B $3x+y-2z=0$

For the given line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ to lie in the plane $2x-4y+z=7$, the fixed point of the line $(4,2,k)$ must lie on the plane.
Hence, $2\left(4\right)-4\left(2\right)+k=7$
$\Rightarrow k=7$
Now, $D.r^{\prime }s$ of line $=(a_1,b_1,c_1)=(1,1,2)$
and plane $P_1:2x-4y+z=7;D.r^{\prime }s=(a,b,c)=(2,-4,1)$
So, equation of plane containing the line $L_1$ and is perpendicular to plane $P_1$ is given by,
$\left|\begin{array}{ccc}x-4& y-2& z-7\\ a& b& c\\ a_1& b_1& c_1\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}x-4& y-2& z-7\\ 2& -4& 1\\ 1& 1& 2\end{array}\right|=0\Rightarrow 3x+y-2z=0$