JEE Mains · Maths · STD 12 - 11. three dimension geometry
The distance of the point \((-1,2,-2)\) from the line of intersection of the planes \(2 \mathrm{x}+3 \mathrm{y}+2 \mathrm{z}=0\) and \(x-2 y+z=0\) is :
- A \(\frac{1}{\sqrt{2}}\)
- B \(\frac{5}{2}\)
- C \(\frac{\sqrt{42}}{2}\)
- D \(\frac{\sqrt{34}}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\sqrt{34}}{2}\)
Step-by-step Solution
Detailed explanation
\(P_{1}: 2 x+3 y+2 z=0\) \(\Rightarrow \vec{n}_{1}=2 \hat{i}+3 \hat{j}+2 \hat{k}\) \(P_{2}: x-2 y+z=0\) \(\Rightarrow \vec{n}_{2}=\hat{i}-2 \hat{j}+k\) Direction vector of line \(\mathrm{L}\) which is line of intersection of \(P_{1} \, \&\, P_{2}\)…
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