Question

The vector $\overrightarrow{AB}=3\hat{i}+4\hat{k}$ and $\overrightarrow{AC}=5\hat{i}-2\hat{j}+4\hat{k}$ are the sides of a triangle ABC. The length of the median through A is

• A. $\sqrt{18}$
• B. $\sqrt{72}$
• C. $\sqrt{33}$
• D. $\sqrt{288}$

Answer 1

Answer: (C)

$\sqrt{33}$

$\overrightarrow{AB}=3\hat{i}+4\hat{k}$ and $\overrightarrow{AC}=5\hat{i}-2\hat{j}+4\hat{k}$

$Let\overline{A}=origin$

$\therefore \overline{AB}=Positionvectorof\overline{B}$

$\overline{AC}=Positionvectorof\overline{C}$

$\therefore Positionvectorofmid-pointof\overline{B}\&\overline{C}$

$=\overline{P}=\frac{\overline{AB}+\overline{AC}}{2}$

$=\left(\frac{3+5}{2}\right)\stackrel{\wedge }{i}+\left(\frac{0-2}{2}\right)\stackrel{\wedge }{j}+\left(\frac{4+4}{2}\right)\stackrel{\wedge }{k}$

$=4\stackrel{\wedge }{i}-\stackrel{\wedge }{j}+4\stackrel{\wedge }{k}$

$Median=\overline{AP}=4\stackrel{\wedge }{i}-\stackrel{\wedge }{j}+4\stackrel{\wedge }{k}$

$Lengthofmedian=\left|\overline{AP}\right|=\sqrt{16+1+16}=\sqrt{33}$