JEE Advanced

Question 15

Log In Register Lost Password
Viewing 2 posts - 1 through 2 (of 2 total)
  • Author
    Posts
  • 22 September, 2026 at 12:58 am #11618
    shahkb4
    Keymaster

    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
    (2) $\frac{3}{2}$
    (3) $\frac{\sqrt{30}}{4}$
    (4) $\frac{\sqrt{26}}{4}$

    22 September, 2026 at 12:58 am #11619
    shahkb4
    Keymaster

    Solution

    As P is equidistant from lines AB and AC . It means P lies on angle bisector $\angle \mathrm{BAC}$ $\cos 2 \theta=\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{AC}}}{|\overrightarrow{\mathrm{AB}} \| \overrightarrow{\mathrm{AC}}|}=\frac{(3 \hat{\imath}+\hat{\jmath}-\hat{k}) \cdot(\hat{\imath}-\hat{\jmath}+3 \hat{k})}{\sqrt{9+1+1} \sqrt{1+1+9}}$

    \[ \begin{aligned} & \Rightarrow \cos 2 \theta=\frac{3-1-3}{11}=-\frac{1}{11} \\ & \Rightarrow 1-2 \sin ^{2} \theta=-\frac{1}{11} \Rightarrow \sin \theta=\sqrt{\frac{6}{11}} \end{aligned} \]

    area of $\Delta \mathrm{ABP}=\frac{1}{2}(\mathrm{AB})(\mathrm{AP}) \sin \theta$

    \[ \begin{aligned} & =\frac{1}{2} \sqrt{9+1+1}\left(\frac{\sqrt{5}}{2}\right) \sqrt{\frac{6}{11}} \\ & =\frac{1}{2} \times \frac{\sqrt{5}}{2} \sqrt{6}=\frac{\sqrt{30}}{4} \text { Answer }(3) \end{aligned} \]
  • Author
    Posts
Log In Register Lost Password
Viewing 2 posts - 1 through 2 (of 2 total)
  • You must be logged in to reply to this topic.