Q89 / 160 · AP EAMCET 2023 (15 May Shift 1) · Mathematics

The distance between two parallel planes $a x+b y+c z+d_1=0$, $a x+b y+c z+d_2=0$ is given by $\frac{\left|d_1-d_2\right|}{\sqrt{a^2+b^2+c^2}}$. 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

  • A
    15
  • B
    5
  • Correct answer
    C
    13
  • D
    9

Solution

Given equation are $$ 2 x-y+2 z+3=0...(1) $$ and $4 x-2 y+4 z+\lambda=0$ $$ \Rightarrow 2 x-y+2 z+\frac{\lambda}{2}=0...(2) $$ and $2 x-y+2 z+\mu=0...(3)$ Since distance between equation (1) and (2) is $\frac{1}{3}$. Hence, $\Rightarrow \frac{\left|\frac{\lambda}{2}-3\right|}{\sqrt{4+1+4}}=\frac{1}{3} \Rightarrow \lambda=8$ Since distance between equation (1) and (3) is $\frac{2}{3}$. Hence, $\Rightarrow \frac{|\mu-3|}{\sqrt{4+1+4}}=\frac{2}{3} \Rightarrow \mu=5$ Therefore $\lambda+\mu=8+5=13$

Hence, the correct answer is option: 'C'.

Relevant Questions