3D Geometry 3 Question 6
6. If the plane $2 x-y+2 z+3=0$ has the distances $\frac{1}{3}$ and $\frac{2}{3}$ units from the planes $4 x-2 y+4 z+\lambda=0$ and $2 x-y+2 z+\mu=0$, respectively, then the maximum value of $\lambda+\mu$ is equal to
(2019 Main, 10 April II)
(a) 13
(b) 15
(c) 5
(d) 9
Show Answer
Answer:
(a)
Solution:
- Equation of given planes are
$ 2 x-y+2 z+3=0 $
$ 4 x-2 y+4 z+\lambda=0 $
and
$ 2 x-y+2 z+\mu=0 $
$\because$ Distance between two parallel planes
$ \begin{aligned} & a x+b y+c z+d _1=0 \\ & \text { and } \quad a x+b y+c z+d _2=0 \text { is } \end{aligned} $
distance $=\frac{\left|d _1-d _2\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$
$\therefore$ Distance between planes (i) and (ii) is
$ \frac{|\lambda-2(3)|}{\sqrt{16+4+16}}=\frac{1}{3} $
[given]
$\Rightarrow|\lambda-6|=2 \Rightarrow \lambda-6= \pm 2 \Rightarrow \lambda=8$ or 4 and distance between planes (i) and (iii) is
$ \begin{array}{cc} & \frac{|\mu-3|}{\sqrt{4+1+4}}=\frac{2}{3} \\ \Rightarrow & |\mu-3|=2 \\ \Rightarrow & \mu-3= \pm 2 \Rightarrow \mu=5 \text { or } 1 \end{array} $
So, maximum value of $(\lambda+\mu)$ at $\lambda=8$ and $\mu=5$ and it is equal to 13.
