Vector Algebra Question 2
Question 2 - 2024 (01 Feb Shift 2)
Consider a $\triangle ABC$ where $A(1,2,3), B(-2,8,0)$ and $C(3,6,7)$. If the angle bisector of $\angle BAC$ meets the line $BC$ at $D$, then the length of the projection of the vector $\overrightarrow{A D}$ on the vector $\overrightarrow{A C}$ is:
(1) $\frac{37}{2 \sqrt{38}}$
(2) $\frac{\sqrt{38}}{2}$
(3) $\frac{39}{2 \sqrt{38}}$
(4) $\sqrt{19}$
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Answer (1)
Solution
B
$D$
C $(3,6,7)$
$(-2,8,0)$
$$ \left(\frac{1}{2}, 7, \frac{7}{2}\right) $$
$A(1,3,2) ; B(-2,8,0) ; C(3,6,7)$;
$\overrightarrow{AC}=2 \hat{i}+3 \hat{j}+5 \hat{k}$
$AB=\sqrt{9+25+4}=\sqrt{38}$
$AC=\sqrt{4+9-25}=\sqrt{38}$
$\overrightarrow{AD}=\frac{1}{2} \hat{i}-4 \hat{j}-\frac{3}{2} \hat{k}=\frac{1}{2}(\hat{i}-8 \hat{j}-3 \hat{k})$
Length of projection of $\overrightarrow{AD}$ on $\overrightarrow{AC}$
$=\left|\frac{\overrightarrow{AD} \cdot \overrightarrow{AC}}{|\overrightarrow{AC}|}\right|=\frac{37}{2 \sqrt{38}}$
