AP EAMCET · Maths · Three Dimensional Geometry
Let \(\pi\) be the plane passing through the point \((3,-3,1)\) and perpendicular to the line joining the points \((3,4,-1)\) and \((2,-1,5)\). If the equation of the plane containing the points \((3,4,-1),(-1,2,5)\) and perpendicular to the plane \(\pi\) is \(a x+y+c z-d=0\), then \(3(a+c)=\)
- A \(-d\)
- B \(2 d\)
- C \(d\)
- D \(-2 d\)
Answer & Solution
Correct Answer
(C) \(d\)
Step-by-step Solution
Detailed explanation
Plane \(\pi\) passes through \((3,-3,1)\) and perpendicular to the line joining the points \((3,4,-1)\) and \((2,-1,5)\). \(\therefore \mathrm{DR}\) 's of normal to the plane are \((3-2),(4,+1),(-1-5)\) \(\equiv(1,5,-6)\) Equation of plane \(\pi\) is given by \(x+5 y-6 z+d=0\)…
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