JEE MAIN - Mathematics (2024 - 27th January Evening Shift - No. 20)
The position vectors of the vertices $$\mathrm{A}, \mathrm{B}$$ and $$\mathrm{C}$$ of a triangle are $$2 \hat{i}-3 \hat{j}+3 \hat{k}, 2 \hat{i}+2 \hat{j}+3 \hat{k}$$ and $$-\hat{i}+\hat{j}+3 \hat{k}$$ respectively. Let $$l$$ denotes the length of the angle bisector $$\mathrm{AD}$$ of $$\angle \mathrm{BAC}$$ where $$\mathrm{D}$$ is on the line segment $$\mathrm{BC}$$, then $$2 l^2$$ equals :
45
50
42
49
Explanation
$$\begin{aligned} & \mathrm{AB}=5 \\ & \mathrm{AC}=5 \end{aligned}$$
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$$\therefore \mathrm{D}$$ is midpoint of $$\mathrm{BC}$$
$$\begin{aligned} & \mathrm{D}\left(\frac{1}{2}, \frac{3}{2}, 3\right) \\ & \therefore l=\sqrt{\left(2-\frac{1}{2}\right)^2+\left(-3-\frac{3}{2}\right)^2+(3-3)^2} \\ & l=\sqrt{\frac{45}{2}} \\ & \therefore 2 l^2=45 \end{aligned}$$
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