AP EAMCET · Maths · Three Dimensional Geometry
Let \(\pi\) be the plane that passes through the point \((-2,1,-1)\) and parallel to the plane \(2 x-y+2 z=0\). Then the foot of perpendicular drawn from the point \((1,2,1)\) to the plane \(\pi\) is
- A \((-3,-1,1)\)
- B \((-1,1,-3)\)
- C \((-3,3,-1)\)
- D \((-1,3,-1)\)
Answer & Solution
Correct Answer
(D) \((-1,3,-1)\)
Step-by-step Solution
Detailed explanation
Equation of plane parallel to \(2 x-y+2 z=0\) is \(2 x-y+2 z+k=0\) It also passes through \((-2,1,-1)\) \(\begin{aligned} & \Rightarrow-4-1-2+k=0 \Rightarrow k=7 \\ & \therefore \pi \equiv 2 x-y+2 z+7=0 \end{aligned}\) Let \((a, b, c)\) be the foot of perpendicular from…
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