Question

If the plane $2x-y+2z+3 = 0$ has the distance of $\dfrac{1}{3}$ from the plane $4x-2y+4z+\lambda =0$ and a distance of $\dfrac{2}{3}$ from the plane $2x-y+2z+\mu =0$, then the maximum value of $\lambda +\mu $ is
[a] 15
[b] 5
[c] 13
[d] 9

Answer 1

Hint: Use the fact that the distance between two parallel planes ax+by+cz+d = 0 and ax+by+cz+e = 0 is given by $\dfrac{\left| d-e \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$. Hence find the possible value of $\lambda $ and $\mu $ and hence find the maximum value of $\lambda +\mu $. Hence determine which o the options is correct.

Complete step-by-step solution:
We have
$\begin{align}
  & {{\pi }_{1}}:2x-y+2z+3=0 \\
 & {{\pi }_{2}}:4x-2y+4z+\lambda =0 \\
\end{align}$
Multiplying equation of ${{\pi }_{1}}$ by 2, we get
$\begin{align}
  & {{\pi }_{1}}:4x-2y+4z+6=0 \\
 & {{\pi }_{2}}:4x-2y+4z+\lambda =0 \\
\end{align}$
We know that the distance between two parallel planes ax+by+cz+d = 0 and ax+by+cz+e = 0 is given by $\dfrac{\left| d-e \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$
Hence, we have
$d\left( {{\pi }_{1}},{{\pi }_{2}} \right)=\dfrac{\left| 6-\lambda \right|}{\sqrt{{{4}^{2}}+{{2}^{2}}+{{4}^{2}}}}$, where $d\left( {{\pi }_{1}},{{\pi }_{2}} \right)$ is the distance between the planes ${{\pi }_{1}},{{\pi }_{2}}$
But given that the distance between the planes ${{\pi }_{1}},{{\pi }_{2}}$ is $\dfrac{1}{3}$
Hence, we have
$\begin{align}
  & \dfrac{1}{3}=\dfrac{\left| 6-\lambda \right|}{6} \\
 & \Rightarrow \left| \lambda -6 \right|=2 \\
\end{align}$
We know that if $\left| x \right|=a,a\ge 0\Rightarrow x=\pm a$
Hence, we have
$\begin{align}
  & \lambda -6=\pm 2 \\
 & \Rightarrow \lambda =8,4 \\
\end{align}$
Similarly, we have
$\begin{align}
  & {{\pi }_{1}}:2x-y+2z+3=0 \\
 & {{\pi }_{3}}:2x-y+2z+\mu =0 \\
\end{align}$
Hence, we have
$d\left( {{\pi }_{1}},{{\pi }_{3}} \right)=\dfrac{\left| \mu -3 \right|}{\sqrt{{{2}^{2}}+{{1}^{2}}+{{2}^{2}}}}=\dfrac{\left| \mu -3 \right|}{3}$
But given that $d\left( {{\pi }_{1}},{{\pi }_{3}} \right)=\dfrac{2}{3}$
Hence, we have
$\begin{align}
  & \dfrac{\left| \mu -3 \right|}{3}=\dfrac{2}{3} \\
 & \Rightarrow \left| \mu -3 \right|=2 \\
 & \Rightarrow \mu -3=\pm 2 \\
 & \Rightarrow \mu =5,1 \\
\end{align}$
Hence, we have
$\max \left( \lambda +\mu \right)=\max \left( \lambda \right)+\max \left( \mu \right)=8+5=13$
Hence option [c] is correct.

Note: In this question many students make mistake in solving the equation involving modulus. It must be noted that the solution of the equation $\left| x \right|=a,a\ge 0$ is $x=\pm a$. If we take only x = a, then the solution x = -a is lost and hence we arrive at incorrect conclusion.