MHT CET · Maths · Three Dimensional Geometry
The equation of plane through the point \((2,-1,-3):\) and parallel to lines \(\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}\) and \(\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}\) is
- A \(8 x+14 y+13 z-37=0\)
- B \(8 x-14 y-13 z-34=0\)
- C \(8 x-14 y-13 z+37=0\)
- D \(8 x+14 y+13 z+37=0\)
Answer & Solution
Correct Answer
(D) \(8 x+14 y+13 z+37=0\)
Step-by-step Solution
Detailed explanation
The equation of plane passing through \((2,-1,-3)\) is
\(\mathrm{a}(x-2)+\mathrm{b}(y+1)+\mathrm{c}(\mathrm{z}+3)=0\)
Also, as the plane is parallel to the given two lines,
\(\begin{aligned}
\therefore \quad & 3 a+2 b-4 c=0 \text { and } 2 a-3 b+2 c=0 \\
& \Rightarrow a=-8, b=-14, c=-13
\end{aligned}\)
\(\therefore \quad\) The equation of the required plane is
\(\begin{aligned}
& -8(x-2)-14(y+1)-13(z+3)=0 \\
& \Rightarrow 8 x+14 y+13 z+37=0
\end{aligned}\)
\(\mathrm{a}(x-2)+\mathrm{b}(y+1)+\mathrm{c}(\mathrm{z}+3)=0\)
Also, as the plane is parallel to the given two lines,
\(\begin{aligned}
\therefore \quad & 3 a+2 b-4 c=0 \text { and } 2 a-3 b+2 c=0 \\
& \Rightarrow a=-8, b=-14, c=-13
\end{aligned}\)
\(\therefore \quad\) The equation of the required plane is
\(\begin{aligned}
& -8(x-2)-14(y+1)-13(z+3)=0 \\
& \Rightarrow 8 x+14 y+13 z+37=0
\end{aligned}\)
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