MHT CET · Maths · Three Dimensional Geometry
The foot of the perpendicular drawn from the origin to the plane is \((4,-2,-5)\). Hence, the equation of the plane is
- A \(4 x-2 y+5 z=-5\)
- B \(4 x-2 y-5 z=45\)
- C \(4 x+2 y-5 z=37\)
- D \(4 x+2 y+5 z+13=0\)
Answer & Solution
Correct Answer
(B) \(4 x-2 y-5 z=45\)
Step-by-step Solution
Detailed explanation
D.r's of normal to the plane \(\langle 4-0,-2-0,-5-0\rangle \equiv\langle 4,-2,-5\rangle\)
Hence equation of the plane \(4 x-2 y-5 z=\lambda\)
But it passes through \((4,-2,-5)\)
\(\begin{aligned}
& \Rightarrow 4 \times 4-2 \times(-2)-5 \times(-5)=\lambda \\
& \Rightarrow \lambda=45
\end{aligned}\)
Hence equation of the plane \(4 x-2 y-5 z=\lambda\)
But it passes through \((4,-2,-5)\)
\(\begin{aligned}
& \Rightarrow 4 \times 4-2 \times(-2)-5 \times(-5)=\lambda \\
& \Rightarrow \lambda=45
\end{aligned}\)
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