MHT CET · Maths · Three Dimensional Geometry
The co-ordinates of the point, where the line \(\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}\) meets the plane \(2 x+4 y-z=3\), are
- A \((3,-1,-1)\)
- B \((3,1,-1)\)
- C \((3,-1,1)\)
- D \((-3,-1,-1)\)
Answer & Solution
Correct Answer
(A) \((3,-1,-1)\)
Step-by-step Solution
Detailed explanation
Given line is
\(\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}\)
Let \(\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}=\lambda\)
\(\begin{array}{ll}
\therefore \quad & x-1=2 \lambda, y-2=-3 \lambda, \mathrm{z}+5=4 \lambda \\
& x=2 \lambda+1, y=-3 \lambda+2, \mathrm{z}=4 \lambda-5 \\
\therefore \quad & 2 x+4 y-\mathrm{z}=3 \\
& \Rightarrow 2(2 \lambda+1)+4(-3 \lambda+2)-(4 \lambda-5)=3 \\
& \Rightarrow 4 \lambda+2-12 \lambda+8-4 \lambda+5=3 \\
& \Rightarrow-12 \lambda=3-15 \\
& \Rightarrow \lambda=1 \\
\therefore \quad & x=3, y=-1, \mathrm{z}=-1,
\end{array}\)
\(\therefore \quad\) Required co-ordinates are: \((3,-1,-1)\)
\(\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}\)
Let \(\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}=\lambda\)
\(\begin{array}{ll}
\therefore \quad & x-1=2 \lambda, y-2=-3 \lambda, \mathrm{z}+5=4 \lambda \\
& x=2 \lambda+1, y=-3 \lambda+2, \mathrm{z}=4 \lambda-5 \\
\therefore \quad & 2 x+4 y-\mathrm{z}=3 \\
& \Rightarrow 2(2 \lambda+1)+4(-3 \lambda+2)-(4 \lambda-5)=3 \\
& \Rightarrow 4 \lambda+2-12 \lambda+8-4 \lambda+5=3 \\
& \Rightarrow-12 \lambda=3-15 \\
& \Rightarrow \lambda=1 \\
\therefore \quad & x=3, y=-1, \mathrm{z}=-1,
\end{array}\)
\(\therefore \quad\) Required co-ordinates are: \((3,-1,-1)\)
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