JEE MAIN - Mathematics (2024 - 1st February Evening Shift - No. 3)
Consider a $\triangle A B C$ where $A(1,3,2), B(-2,8,0)$ and $C(3,6,7)$. If the angle bisector of $\angle B A C$ meets
the line $B C$ at $D$, then the length of the projection of the vector $\overrightarrow{A D}$ on the vector $\overrightarrow{A C}$ is :
$\frac{37}{2 \sqrt{38}}$
$\sqrt{19}$
$\frac{39}{2 \sqrt{38}}$
$\frac{\sqrt{38}}{2}$
Explanation
_1st_February_Evening_Shift_en_3_1.png)
$D$ divides $B C$ in ratio $1: 1$
$$ D:\left(\frac{1}{2}, 7, \frac{7}{2}\right) $$
$\begin{aligned} & \overrightarrow{A D}=\left(\frac{1}{2}-1\right) \hat{i}+(7-3) \hat{j}+\left(\frac{7}{2}-2\right) \hat{k} \\\\ & =-\frac{1}{2} \hat{i}+4 \hat{j}+\frac{3}{2} \hat{k} \\\\ & \overrightarrow{A C}=2 \hat{i}+3 \hat{j}+5 \hat{k}\end{aligned}$
Projection of $\overrightarrow{A D}$ on $\overrightarrow{A C}$
$$ =\frac{-1+12+\frac{15}{2}}{\sqrt{4+9+25}}=\frac{37}{2 \sqrt{38}} $$
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