MHT CET · Maths · Three Dimensional Geometry
The point where the line \(\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}\) meets the plane \(2 x+4 y-z=1\), is
- A \((3,-1,1)\)
- B \((3,1,1)\)
- C \((1,1,3)\)
- D \((1,3,1)\)
Answer & Solution
Correct Answer
(A) \((3,-1,1)\)
Step-by-step Solution
Detailed explanation
Let point be \((a, b, c)\), then \(2 a+4 b-c=1 \quad \ldots\) (i)
and \(a=2 k+1, b=-3 k+2\) and \(c=4 k-3\)
(where \(k\) is constant)
On substituting these values in Eq. (i), we get
\(
2(2 k+1)+4(-3 k+2)-(4 k-3)=1
\)
\(\Rightarrow\) \(k=1\)
Hence, required point is \((3,-1,1)\).
and \(a=2 k+1, b=-3 k+2\) and \(c=4 k-3\)
(where \(k\) is constant)
On substituting these values in Eq. (i), we get
\(
2(2 k+1)+4(-3 k+2)-(4 k-3)=1
\)
\(\Rightarrow\) \(k=1\)
Hence, required point is \((3,-1,1)\).
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