Question

The equation of a plane containing the line of intersection of the plane $2x-y-4=0\And y+2z-4=0$ and passing through the point $\left( 1,1,0 \right)$ is:
(a)$x+3y+z=4$
(b)$x-y-z=0$
(c)$x-3y-2z=-2$
(d)$2x-z=2$

Answer 1

Hint: We have given the equations of two planes $2x-y-4=0\And y+2z-4=0$. Now, the family of planes passing through the intersection of these two planes is given by $2x-y-4+\lambda \left( y+2z-4 \right)=0$. Now, it is given that the equation of plane that we have to find passes through the line of intersection of given two planes and also passes through the point $\left( 1,1,0 \right)$ so we have to satisfy this point in the family of planes passing through the intersection of two planes and then get the value of $\lambda $. After that, substitute the value of this $\lambda $ in the family of planes passing through the line of intersection of two planes.

Complete step by step answer:
We have given two equations of planes that are:
$2x-y-4=0\And y+2z-4=0$
Now, we have to find the equation of the plane passing through the line of intersection of these two planes and also passing through the point $\left( 1,1,0 \right)$.
The family of planes passing through the line of intersection of two given planes is equal to:
$2x-y-4+\lambda \left( y+2z-4 \right)=0$
Now, the equation of plane that we desire for is contained in the above family of planes so to get our plane we have to satisfy the point $\left( 1,1,0 \right)$ in the above equation we get,
$\begin{align}
  & 2\left( 1 \right)-\left( 1 \right)-4+\lambda \left( \left( 1 \right)+2\left( 0 \right)-4 \right)=0 \\
 & \Rightarrow 2-1-4+\lambda \left( -3 \right)=0 \\
 & \Rightarrow 2-5-3\lambda =0 \\
 & \Rightarrow -3-3\lambda =0 \\
\end{align}$
 Adding 3 on both the sides of the above equation we get,
$-3\lambda =3$
Dividing 3 on both the sides we get,
$\begin{align}
  & -\dfrac{3\lambda }{3}=\dfrac{3}{3} \\
 & \Rightarrow \lambda =-1 \\
\end{align}$
Now, substituting the above value of $\lambda $ in the family of planes passing through the line of intersection of two given planes we get,
$\begin{align}
  & 2x-y-4+\left( -1 \right)\left( y+2z-4 \right)=0 \\
 & \Rightarrow 2x-y-4-y-2z+4=0 \\
 & \Rightarrow 2x-2y-2z=0 \\
\end{align}$
Dividing 2 on both the sides of the above equation we get,
$\begin{align}
  & 2\dfrac{\left( x-y-z \right)}{2}=0 \\
 & \Rightarrow x-y-z=0 \\
\end{align}$
From the above solution, we have got the equation of plane containing the line of intersection of the plane $2x-y-4=0\And y+2z-4=0$ and passing through the point $\left( 1,1,0 \right)$ is $x-y-z=0$.
Hence, the correct option is (b).

Note:
The mistake that could happen in the above problem is in putting the point $\left( 1,1,0 \right)$ in the family of planes passing through the line of intersection of two given planes.
The family of planes which we have shown in the above solution is:
$2x-y-4+\lambda \left( y+2z-4 \right)=0$
Now, while substituting the point $\left( 1,1,0 \right)$ in the above equation, when substituting the values of x, y and z in the second equation written after $\lambda $ you might substitute the value of y as the value of x because generally, the equations we deal with have first term as x so the possibility of substituting the value in place of y as considering as x is pretty high. But here, you are lucky because x and y coordinates are the same so you could not get the wrong answer but this mistake can repeat in other problems so make sure you carefully substitute the values of x, y and z.