MHT CET · Maths · Three Dimensional Geometry
A line drawn from a point \(\mathrm{A}(-2,-2,3)\) and parallel to the line \(\frac{\mathrm{x}}{-2}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{-1}\) meets the \(\mathrm{YOZ}-\) plane in point \(\mathrm{P}\), then the co-ordinates of the point \(\mathrm{P}\) are
- A \((0,4,-4)\)
- B \((0,2,2)\)
- C \((0,-2,2)\)
- D \((0,-4,4)\)
Answer & Solution
Correct Answer
(D) \((0,-4,4)\)
Step-by-step Solution
Detailed explanation
Equation of required lines is
\(\frac{\mathrm{x}+2}{-2}=\frac{\mathrm{y}+2}{2}=\frac{\mathrm{z}-3}{-1}\) and this line meets \(\mathrm{YZ}\) plane in \(\mathrm{P}\).
Coordinates of any point on this line are \((-2 \lambda-21,2 \lambda-2,-\lambda+3)\), where \(\lambda\) is a scalar.
Since \(\mathrm{P}\) is on \(\mathrm{YZ}\) plane, we write
\(
\begin{aligned}
& -2 \lambda-2=0 \Rightarrow \lambda=-1 \\
& \therefore \mathrm{P} \equiv(0,-4,4)
\end{aligned}
\)
\(\frac{\mathrm{x}+2}{-2}=\frac{\mathrm{y}+2}{2}=\frac{\mathrm{z}-3}{-1}\) and this line meets \(\mathrm{YZ}\) plane in \(\mathrm{P}\).
Coordinates of any point on this line are \((-2 \lambda-21,2 \lambda-2,-\lambda+3)\), where \(\lambda\) is a scalar.
Since \(\mathrm{P}\) is on \(\mathrm{YZ}\) plane, we write
\(
\begin{aligned}
& -2 \lambda-2=0 \Rightarrow \lambda=-1 \\
& \therefore \mathrm{P} \equiv(0,-4,4)
\end{aligned}
\)
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