MHT CET · Maths · Three Dimensional Geometry
Equation of the plane passing through \((-2,2,2)\) and \((2,-2,-2)\) and perpendicular to the plane \(9 x-13 y-3 z=0\) is
- A \(5 x+3 y+2 z=0\)
- B \(5 x-3 y+2 z=0\)
- C \(5 x-3 y-2 z=0\)
- D \(5 x+3 y-2 z=0\)
Answer & Solution
Correct Answer
(A) \(5 x+3 y+2 z=0\)
Step-by-step Solution
Detailed explanation
Any plane passing through \((-2,2,2)\) is \(A(x+2)+B(y-2)+C(z-2)=0\)
\(\because\) It passes through \((2,-2,-2)\) \(\Rightarrow \quad 4 A-4 B-4 C=0\)
It is parallel to \(9 x-13 y-3 z=0\) \(\begin{array}{lll}\therefore & 9 A-13 B-3 C=0 & \ldots \text { (ii) }\end{array}\)
Solving Eqs. (i) and (ii),
\(\frac{A}{12-52}=\frac{B}{-36+12}=\frac{C}{-52+36} \)
\( \Rightarrow \frac{A}{-40}=\frac{B}{-24}=\frac{C}{-16}\)
\(\therefore\) Required equation of plane is \(-40(x+2)-24(y-2)-16(z-2)=0\)
\(\Rightarrow -40 x-80-24 y+48-16 z+32=0 \)
\( \Rightarrow 40 x+24 y+16 z=0 \)
\( \Rightarrow 5 x+3 y+2 z=0\)
\(\because\) It passes through \((2,-2,-2)\) \(\Rightarrow \quad 4 A-4 B-4 C=0\)
It is parallel to \(9 x-13 y-3 z=0\) \(\begin{array}{lll}\therefore & 9 A-13 B-3 C=0 & \ldots \text { (ii) }\end{array}\)
Solving Eqs. (i) and (ii),
\(\frac{A}{12-52}=\frac{B}{-36+12}=\frac{C}{-52+36} \)
\( \Rightarrow \frac{A}{-40}=\frac{B}{-24}=\frac{C}{-16}\)
\(\therefore\) Required equation of plane is \(-40(x+2)-24(y-2)-16(z-2)=0\)
\(\Rightarrow -40 x-80-24 y+48-16 z+32=0 \)
\( \Rightarrow 40 x+24 y+16 z=0 \)
\( \Rightarrow 5 x+3 y+2 z=0\)
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