AP EAMCET · Maths · Three Dimensional Geometry
If \((2,-1,3)\) is the foot of the perpendicular drawn from the origin \((0,0,0)\) to a plane then the equation of that plane is
- A \(2 x+y-3 z+6=0\)
- B \(2 x-y+3 z-14=0\)
- C \(2 x-y+3 z-13=0\)
- D \(2 x+y+3 z-10=0\)
Answer & Solution
Correct Answer
(B) \(2 x-y+3 z-14=0\)
Step-by-step Solution
Detailed explanation
Normal vector \( \vec{n} = \langle 2,-1,3 \rangle \). Point on plane \( (x_0,y_0,z_0) = (2,-1,3) \). Equation of plane: \( A(x-x_0) + B(y-y_0) + C(z-z_0) = 0 \) \( 2(x-2) - 1(y-(-1)) + 3(z-3) = 0 \) \( 2x - 4 - y - 1 + 3z - 9 = 0 \) \( 2x - y + 3z - 14 = 0 \)
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