MHT CET · Maths · Three Dimensional Geometry
Equation of the plane containing the straight line \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\) and perpendicular to the plane containing the straight lines \(\frac{x}{3}=\frac{y}{4}=\frac{z}{2}\) and \(\frac{x}{4}=\frac{y}{2}=\frac{z}{3}\) is
- A \(x+2 y-2 z=0\)
- B \(3 x+2 y-2 z=0\)
- C \(\quad x-2 y+z=0\)
- D \(5 x+2 y-4 z=0\)
Answer & Solution
Correct Answer
(C) \(\quad x-2 y+z=0\)
Step-by-step Solution
Detailed explanation
Equation of the plane containing \(\frac{x}{3}=\frac{y}{4}=\frac{z}{2}\) and \(\frac{x}{4}=\frac{y}{2}=\frac{z}{3}\) is
\(\begin{aligned}
& \left|\begin{array}{lll}
x & y & z \\
3 & 4 & 2 \\
4 & 2 & 3
\end{array}\right|=0 \\
& \Rightarrow 8 x-y-10 z=0
\end{aligned}\)
Now required plane is perpendicular to this plane.
Consider option (C)
\((8)(1)+(-1)(-2)+(-10)(1)=0\)
\(\therefore \quad\) Option (C) is correct.
\(\begin{aligned}
& \left|\begin{array}{lll}
x & y & z \\
3 & 4 & 2 \\
4 & 2 & 3
\end{array}\right|=0 \\
& \Rightarrow 8 x-y-10 z=0
\end{aligned}\)
Now required plane is perpendicular to this plane.
Consider option (C)
\((8)(1)+(-1)(-2)+(-10)(1)=0\)
\(\therefore \quad\) Option (C) is correct.
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