JEE Mains · Maths · STD 12 - 11. three dimension geometry
The perpendicular distance, of the line \(\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}\) from the point \(\mathrm{P}(2,-10,1)\), is :
- A 6
- B \(5 \sqrt{2}\)
- C \(4 \sqrt{3}\)
- D \(3 \sqrt{5}\)
Answer & Solution
Correct Answer
(D) \(3 \sqrt{5}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \left.\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+2}{-1}=\frac{\mathrm{z}+3}{2}=\lambda \text { (let }\right) \\ & (2 \lambda+1,-\lambda-2,2 \lambda-3) \\ & \because \overrightarrow{\mathrm{PA}} \cdot \overrightarrow{\mathrm{n}}=0 \\ & \Rightarrow(2 \lambda-1)…
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