Let P be a point in the plane of the vector $\overrightarrow{AB} = 3\widehat{i} + \widehat{j} - \wid

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 :

Accepted Answer

$\ cos2	heta = \frac{3 - 1 - 3}{\sqrt{11} \cdot \sqrt{11}} = - \frac{1}{11}$$${1 - 2\sin^{2}	heta = - \frac{1}{11} \Rightarrow 2\sin^{2}	heta = \frac{12}{11} \Rightarrow sin	heta = \sqrt{\frac{6}{11}} }{	herefore Area( \bigtriangleup APB) = \frac{1}{2} 	imes \sqrt{11} \cdot \frac{\sqrt{5}}{2} \cdot \sqrt{\frac{6}{11}} = \frac{\sqrt{30}}{4}}$$

Question

Options

Solution

More Properties of Triangle Questions