TS EAMCET · Maths · Three Dimensional Geometry
If the foot of the perpendicular drawn from the point \((2,0,-3)\) to the plane \(\pi\) is \((1,-2,0)\) and the equation of the plane \(\pi\) is \(a x+b y-3 z+d=0\) then \(a+b+d=\)
- A 0
- B 1
- C 6
- D 2
Answer & Solution
Correct Answer
(C) 6
Step-by-step Solution
Detailed explanation
Normal vector \(\vec{n} = (1-2, -2-0, 0-(-3)) = (-1, -2, 3)\). Comparing \(\vec{n}\) with \((a, b, -3)\): \((-1, -2, 3) = k(a, b, -3)\). \(3 = k(-3) \Rightarrow k = -1\). \(a = -1/k = -1/(-1) = 1\). \(b = -2/k = -2/(-1) = 2\). Foot of perpendicular \((1,-2,0)\) lies on the…
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