- This topic has 1 reply, 1 voice, and was last updated 1 month, 1 week ago by
.
-
Posts
-
Let P be a point in the plane of the vector $\overrightarrow{\mathrm{AB}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\mathrm{k}$ and $\overrightarrow{\mathrm{AC}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \mathrm{k}$
such that P is equidistant from the lines AB and AC . If $|\overrightarrow{\mathrm{AP}}|=\frac{\sqrt{5}}{2}$,
then the area of the triangle ABP is
[JEE MAIN 28 Jan 2026 S II]
(1) 2
(2) $\frac{3}{2}$
(3) $\frac{\sqrt{30}}{4}$
(4) $\frac{\sqrt{26}}{4}$Solution
As P is equidistant from lines AB and AC . It means P lies on angle bisector $\angle \mathrm{BAC}$
\[ \cos 2 \theta=\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{AC}}}{|\overrightarrow{\mathrm{AB}} \| \overrightarrow{\mathrm{AC}}|}=\frac{(3 \hat{\imath}+\hat{\jmath}-\hat{k}) \cdot(\hat{\imath}-\hat{\jmath}+3 \hat{k})}{\sqrt{9+1+1} \sqrt{1+1+9}} \]\[ \begin{aligned} & \Rightarrow \cos 2 \theta=\frac{3-1-3}{11}=-\frac{1}{11} \\ & \Rightarrow 1-2 \sin ^{2} \theta=-\frac{1}{11} \Rightarrow \sin \theta=\sqrt{\frac{6}{11}} \\ & \text { area of } \begin{aligned} \Delta \mathrm{ABP} & =\frac{1}{2}(\mathrm{AB})(\mathrm{AP}) \sin \theta \\ & =\frac{1}{2} \sqrt{9+1+1}\left(\frac{\sqrt{5}}{2}\right) \sqrt{\frac{6}{11}} \\ & =\frac{1}{2} \times \frac{\sqrt{5}}{2} \sqrt{6}=\frac{\sqrt{30}}{4} \text { Answer }(3) \end{aligned} \end{aligned} \]
- You must be logged in to reply to this topic.