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