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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Let $ P $ be a point in the plane of the vector
$ \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 $ |AP| = \frac{\sqrt{5}}{2} $, then the area of the triangle $ ABP $ is:
Solution
$ \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}} $
$ \therefore \text{Area}(\triangle ABP) = \frac{1}{2} \cdot \sqrt{11} \cdot \frac{\sqrt{5}}{2} \cdot \sqrt{\frac{6}{11}} = \frac{\sqrt{30}}{4} $
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