Question ID: #963
Let P be a point in the plane of the vector $\vec{AB}=3\hat{i}+\hat{j}-\hat{k}$ and $\vec{AC}=\hat{i}-\hat{j}+3\hat{k}$ such that P is equidistant from the lines AB and AC. If $|\vec{AP}|=\frac{\sqrt{5}}{2}$ then the area of the triangle ABP is :
- (1) 2
- (2) $\frac{3}{2}$
- (3) $\frac{\sqrt{30}}{4}$
- (4) $\frac{\sqrt{26}}{4}$
Solution:
Since P is equidistant from lines AB and AC, P lies on the angle bisector of $\angle BAC$. Let $\angle BAC = 2\theta$.
Then $\angle PAB = \theta$.
First, find $\cos(2\theta)$ using dot product:
$$\cos(2\theta) = \frac{\vec{AB} \cdot \vec{AC}}{|\vec{AB}| |\vec{AC}|}$$
$$|\vec{AB}| = \sqrt{3^2 + 1^2 + (-1)^2} = \sqrt{11}$$
$$|\vec{AC}| = \sqrt{1^2 + (-1)^2 + 3^2} = \sqrt{11}$$
$$\vec{AB} \cdot \vec{AC} = 3(1) + 1(-1) + (-1)(3) = 3 – 1 – 3 = -1$$
$$\cos(2\theta) = \frac{-1}{11}$$
We need $\sin\theta$ for the area of $\Delta ABP$.
Using $1 – 2\sin^2\theta = \cos(2\theta)$:
$$1 – 2\sin^2\theta = -\frac{1}{11}$$
$$2\sin^2\theta = 1 + \frac{1}{11} = \frac{12}{11}$$
$$\sin^2\theta = \frac{6}{11} \Rightarrow \sin\theta = \sqrt{\frac{6}{11}}$$
Area of $\Delta ABP = \frac{1}{2} |\vec{AB}| |\vec{AP}| \sin\theta$
$$Area = \frac{1}{2} (\sqrt{11}) \left(\frac{\sqrt{5}}{2}\right) \left(\sqrt{\frac{6}{11}}\right)$$
$$Area = \frac{1}{4} \sqrt{5} \sqrt{6} = \frac{\sqrt{30}}{4}$$
Ans. (3)
Was this solution helpful?
YesNo