AP EAMCET · Maths · Three Dimensional Geometry
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
- C 13
- D 9
Answer & Solution
Correct Answer
(C) 13
Step-by-step Solution
Detailed explanation
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,…
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