JEE Mains · Maths · STD 12 - 10. vector algebra
Let P be a point in the plane of the vectors \(\overrightarrow{AB}=3\hat{i}+\hat{j}-\hat{k}\) and \(\overrightarrow{AC}=\hat{i}-\hat{j}+3\hat{k}\) such that P is equidistant from the lines AB and AC. If \({|\overrightarrow{AP}|}=\frac{\sqrt{5}}{2}\) then the area of the triangle ABP is :
- A 2
- B \(\frac{3}{2}\)
- C \(\frac{\sqrt{30}}{4}\)
- D \(\frac{\sqrt{26}}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{\sqrt{30}}{4}\)
Step-by-step Solution
Detailed explanation
\(\cos 2\theta = \frac{3-1-3}{\sqrt{11} \cdot \sqrt{11}} = -\frac{1}{11}\) \(1-2\sin^{2}\theta = -\frac{1}{11} \Rightarrow 2\sin^{2}\theta = \frac{12}{11} \Rightarrow \sin\theta = \sqrt{\frac{6}{11}}\)…
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