Question

The equation to the plane through the line of intersection of $2x+y+3z=2$ and $x-y+z+4=0$ which is parallel to the z-axis is

• A. $x-4y+14=0$
• B. $4x+y-2=0$
• C. $x+y+9=0$
• D. $3x-2y+7=0$

Answer 1

The correct option is A $x-4y+14=0$

The general equation of plane passing through the intersection of two lines $l_1$ and $l_2$ is given by

$l_1+\lambda \left(l_2\right)=0$ where $\lambda \in R$

So,

The general equation of plane passing through intersection of given lines is

$(2x+y+3z-2)+\lambda (x-y+z+4)=0$

$\to (\lambda +2)x+(1-\lambda )y+(3+\lambda )z-(2-4\lambda )=0$

For the plane to be parallel to z-axis, it must be of form, ax + by = c.

So,

$\to 3+\lambda =0$

$\to \lambda =-3$

So, the required equation of plane is,

$-x+4y-14=0$

$\to x-4y+14=0$

Hence, (A) is the correct answer.