JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the foot of perpendicular from a point \(P(1,2,-1)\) to the straight line \(L: \frac{x}{1}=\frac{y}{0}=\frac{z}{-1}\) be \(N\). Let a line be drawn from \(P\) parallel to the plane \(x+y+2 z=0\) which meets \(L\) at point \(Q\). If \(\alpha\) is the acute angle between the lines \(\mathrm{PN}\) and \(\mathrm{PQ}\), then \(\cos \alpha\) is equal to \(.....\)
- A \(\frac{1}{2 \sqrt{3}}\)
- B \(\frac{1}{\sqrt{5}}\)
- C \(\frac{\sqrt{3}}{2}\)
- D \(\frac{1}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
\(\overline{\mathrm{PN}} \cdot(\hat{\mathrm{i}}-\hat{\mathrm{k}})=0\) \(\cdot \Rightarrow \mathrm{N}(1,0,-1)\) Now, \(\overrightarrow{\mathrm{PQ}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})=0\) \(\Rightarrow \mathrm{m}=-1\) \(\Rightarrow \mathrm{Q}(-1,0,1)\)…
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