AP EAMCET · Maths · Three Dimensional Geometry
The equation of the plane \(\pi\) through the line of intersection of the planes \(\pi_1 \equiv x+3 y-6=0\), and \(\pi_2 \equiv 3 x-y+4 z=0\) is \(\pi_1+\lambda \pi_2=0\). If the plane \(\pi\) is at unit distance from the origin, then an equation of the plane \(\pi\) is
- A \(2 x+y+2 z-3=0\)
- B \(2 x-y-2 z+3=0\)
- C \(2 x+y+2 z+3=0\)
- D \(x+2 y+2 z+3=0\)
Answer & Solution
Correct Answer
(A) \(2 x+y+2 z-3=0\)
Step-by-step Solution
Detailed explanation
Given equation of plane are \(\begin{aligned} & \pi_1=x+3 y-6=0 \\ & \pi_2=3 x-y+4 z=0 \end{aligned}\) Given, \(\pi_1+\lambda \pi_2=0\) \(\begin{aligned} (x+3 y-6)+\lambda(3 x-y+4 z) & =0 \\ (1+3 \lambda) x+(3-\lambda) y+4 \lambda z-6 & =0 \quad \ldots (i) \end{aligned}\)…
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