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STD 12 - 10. vector algebra — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 10. vector algebra1 Mark
MCQ
Consider a $\triangle \mathrm{ABC}$ where $\mathrm{A}(1,3,2), \mathrm{B}(-2,8,0)$ and $\mathrm{C}(3,6,7)$. If the angle bisector of $\angle \mathrm{BAC}$ 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:
  • ✓
     $\frac{37}{2 \sqrt{38}}$
  • B
     $\frac{\sqrt{38}}{2}$
  • C
     $\frac{39}{2 \sqrt{38}}$
  • D
     $\sqrt{19}$
✓

Answer

Correct option: A.
 $\frac{37}{2 \sqrt{38}}$
a
$ \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 $\overrightarrow{\mathrm{AC}}$

$=\left|\frac{\overrightarrow{\mathrm{AD}} \cdot \overrightarrow{\mathrm{AC}}}{|\overrightarrow{\mathrm{AC}}|}\right|=\frac{37}{2 \sqrt{38}}$

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