Question

The equation of the plane through the line of intersection of the planes $ax + by + cz + d = 0$ and $\alpha x + \beta y + \gamma z + e = 0$ and perpendicular to $xy$ plane is
(a)$\left( {a\gamma - c\alpha } \right)x + \left( {b\gamma - c\beta } \right)y + \left( {d\gamma - ce} \right) = 0$
(b)$\left( {a\gamma + c\alpha } \right)x + \left( {b\gamma - c\beta } \right)y + e = 0$
(c)$\left( {a\gamma - c\alpha } \right)x + \left( {b\gamma - c\beta } \right)y + d = 0$
(d)None of these

Answer 1

Here, we will use the formula of the equation of the plane passing through the intersection of two planes. Then we will substitute the value of a point lying on this plane using the fact that it is perpendicular to the$xy$ plane. From this we will get the value of $\lambda$, and substituting its value, we will get the required equation of the plane.

Formula Used:
Equation of passing through intersection of two plane $= \left( {\,{A_1}x + {B_1}y + {C_1}z - {d_1}} \right) + \lambda \left( {{A_2}x + {B_2}y + {C_2}z - {d_2}} \right)$

Complete step-by-step answer:
We know that, equation of passing through intersection of two planes which are in the form of ${A_1}x + {B_1}y + {C_1}z = {d_1}$ and ${A_2}x + {B_2}y + {C_2}z = {d_2}$ is given as:
$\left( {\,{A_1}x + {B_1}y + {C_1}z - {d_1}} \right) + \lambda \left( {{A_2}x + {B_2}y + {C_2}z - {d_2}} \right)$…………………………………$\left( 1 \right)$
Comparing the general equation of two planes by the given planes, we get,
${A_1}x + {B_1}y + {C_1}z - {d_1} = ax + by + cz + d$
and
${A_2}x + {B_2}y + {C_2}z - {d_2} = \alpha x + \beta y + \gamma z + e$
Hence, from equation $\left( 1 \right)$, we get
Equation of the planes passing through intersection of given planes $= \left( {ax + by + cz + d} \right) + \lambda \left( {\alpha x + \beta y + \gamma z + e} \right)$
Opening the brackets and taking the like variables common, we get,
$\Rightarrow$ Equation of the plane $= \left( {a + \lambda \alpha } \right)x + \left( {b + \lambda \beta } \right)y + \left( {c + \lambda \gamma } \right)z + d + \lambda e = 0$……………….$\left( 2 \right)$
Also, according to the question, this plane is also perpendicular to $xy$ plane.
Hence, if a plane is perpendicular to $xy$ plane then it means that it lies on $z$ plane.
Then, its coordinates are of the form $\left( {0,0,1} \right)$, where $x$ and $y$ coordinates are 0 and $z$ coordinate is 1.
Hence, substituting $x = 0$, $y = 0$ and $z = 1$ in $\left( {a + \lambda \alpha } \right)x + \left( {b + \lambda \beta } \right)y + \left( {c + \lambda \gamma } \right)z$, we get
$\left( {a + \lambda \alpha } \right)\left( 0 \right) + \left( {b + \lambda \beta } \right)\left( 0 \right) + \left( {c + \lambda \gamma } \right)\left( 1 \right) = 0$
$\Rightarrow 0 + 0 + \left( {c + \lambda \gamma } \right) = 0$
Now, rewriting this as:
$\Rightarrow \lambda = \dfrac{{ - c}}{\gamma }$
Now, substituting $\lambda = \dfrac{{ - c}}{\gamma }$ in equation $\left( 2 \right)$, we get
$\left( {a + \left( {\dfrac{{ - c}}{\gamma }} \right)\alpha } \right)x + \left( {b + \left( {\dfrac{{ - c}}{\gamma }} \right)\beta } \right)y + \left( {c + \left( {\dfrac{{ - c}}{\gamma }} \right)\gamma } \right)z + d + \left( {\dfrac{{ - c}}{\gamma }} \right)e = 0$
Taking LCM and solving further, we get
$\Rightarrow \left( {a\gamma - c\alpha } \right)x + \left( {b\gamma - c\beta } \right)y + \left( {c\gamma - c\gamma } \right)z + d\gamma - ce = 0$
$\Rightarrow \left( {a\gamma - c\alpha } \right)x + \left( {b\gamma - c\beta } \right)y + d\gamma - ce = 0$
Hence, the equation of the plane through the line of intersection of the planes is $\left( {a\gamma - c\alpha } \right)x + \left( {b\gamma - c\beta } \right)y + \left( {d\gamma - ce} \right) = 0$.
Therefore, option A is the correct answer.

Note: A plane is a two dimensional flat surface in which if any two random points are chosen then, the straight line joining those points completely lie on that plane. A plane is defined by three points unless they are forming a straight line. In a scalar equation, when a plane passes through the intersection of two planes, then we combine their equations to form one single equation of the required plane.