Question

If the equation of the plane passing through the line of intersection of the planes $ax+by+cz+d=0,$ $a_{1}x+b_{1}y+c_{1}z+d_1=0$ and perpendicular to the $xy$-plane is $px+qy+rz+s=0$, then $s=$

• A. $d{c}_1-d_{1}c$
• B. $d{c}_1+d_{1}c$
• C. $d{d}_1+c{c}_1$
• D. $a{a}_1+b{b}_1++c{c}_1$.

Answer 1

The correct option is A $d{c}_1-d_{1}c$

Equation of plane passing through the line of intersection of plane
$ax+by+cz+d=0$ and $a_{1}x+b_{1}y+c_{1}z+d_1=0$ is given by,
$ax+by+cz+d+\lambda (a_{1}x+b_{1}y+c_{1}z+d_1)=0$ ......$\left(1\right)$
$(a+\lambda a_1)x+(b+\lambda b_1)y+(c+\lambda c_1)z+(d+\lambda d_1)=0$
Also given this plane is perpendicular to $xy$ plane and we know equation of $xy$ plane is $z=k$, ($k$ is any constant)
$\Rightarrow (c+\lambda c_1).1=0\Rightarrow \lambda =\frac{-c}{c_1}$
Finally comparing equation $\left(1\right)$ and given plane equation, we get
$s=c_{1}d-c{d}_1$