JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the line passing through the points \((-1,2,1)\) and parallel to the line \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}\) intersect the line \(\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}\) at the point \(P\). Then the distance of \(P\) from the point \(Q(4,-5,1)\) is
- A \(5\)
- B \(5 \sqrt{5}\)
- C \(5 \sqrt{6}\)
- D \(10\)
Answer & Solution
Correct Answer
(B) \(5 \sqrt{5}\)
Step-by-step Solution
Detailed explanation
Equation of line through point \((-1,2,1)\) is \(\rightarrow\) \(\begin{aligned} & \Rightarrow \frac{x+1}{2}=\frac{y-2}{3}=\frac{z-1}{4}=\lambda-(2) \\ & \text {So, }\left[\begin{array}{l}x=2 \lambda-1 \\ y=3 \lambda+2 \\ z=4 \lambda+1\end{array}\right.\end{aligned}\)…
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