Let P be a point in the plane of the vectors | vedclass

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(C) Let $	heta$ be the angle between $\overrightarrow{AB}$ and $\overrightarrow{AP}$. Since $P$ is equidistant from $AB$ and $AC$, $AP$ is the angle

Vedclass · Accepted Answer

$\frac{\sqrt{30}}{4}$

Vedclass · Answer

(C) Let $	heta$ be the angle between $\overrightarrow{AB}$ and $\overrightarrow{AP}$. Since $P$ is equidistant from $AB$ and $AC$, $AP$ is the angle bisector of $\angle BAC$. Let $\angle BAC = 2\alpha$. Then $\angle BAP = \alpha$.First, calculate $\cos(2\alpha) = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|} = \frac{(3)(1) + (1)(-1) + (-1)(3)}{\sqrt{3^2+1^2+(-1)^2} \sqrt{1^2+(-1)^2+3^2}} = \frac{3-1-3}{\sqrt{11} \cdot \sqrt{11}} = -\frac{1}{11}$.Using the identity $\cos(2\alpha) = 1 - 2\sin^2(\alpha)$, we have $1 - 2\sin^2(\alpha) = -\frac{1}{11}$, which implies $2\sin^2(\alpha) = \frac{12}{11}$, so $\sin^2(\alpha) = \frac{6}{11}$ and $\sin(\alpha) = \sqrt{\frac{6}{11}}$.The area of $	riangle ABP$ is given by $\frac{1}{2} |\overrightarrow{AB}| |\overrightarrow{AP}| \sin(\alpha)$.Substituting the values: $	ext{Area} = \frac{1}{2} \cdot \sqrt{11} \cdot \frac{\sqrt{5}}{2} \cdot \sqrt{\frac{6}{11}} = \frac{1}{2} \cdot \frac{\sqrt{5}}{2} \cdot \sqrt{6} = \frac{\sqrt{30}}{4}$.

Let $P$ be a point in the plane of the vectors $\overrightarrow{AB}=3\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{AC}=\hat{i}-\hat{j}+3\hat{k}$ such that $P$ is equidistant from the lines $AB$ and $AC$. If $|\overrightarrow{AP}|=\frac{\sqrt{5}}{2}$, then the area of the triangle $ABP$ is:

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