VectorMedium
Question
Let P be a point in the plane of the vector $\overrightarrow{AB} = 3\widehat{i} + \widehat{j} - \widehat{k}$ and $\overrightarrow{AC} = \widehat{i} - \widehat{j} + 3\widehat{k}$ 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 :
Options
A.2
B.$\frac{3}{2}$
C.$\frac{\sqrt{30}}{4}$
D.$\frac{\sqrt{26}}{4}$
Solution
$\ cos2\theta = \frac{3 - 1 - 3}{\sqrt{11} \cdot \sqrt{11}} = - \frac{1}{11}$
$${1 - 2\sin^{2}\theta = - \frac{1}{11} \Rightarrow 2\sin^{2}\theta = \frac{12}{11} \Rightarrow sin\theta = \sqrt{\frac{6}{11}} }{\therefore Area( \bigtriangleup APB) = \frac{1}{2} \times \sqrt{11} \cdot \frac{\sqrt{5}}{2} \cdot \sqrt{\frac{6}{11}} = \frac{\sqrt{30}}{4}}$$
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