Question

If the vectors $\overline{a}=3\overline{i}+\overline{j}-2\overline{k}$,$\overline{b}=-\overline{i}+3\overline{j}+4\overline{k},\overline{c}=4\overline{i}-2\overline{j}-6\overline{k}$ form the sides of the triangle then length of the median bisecting the vector $c$ is

• A. $\sqrt{12}$ units
• B. $\sqrt{6}$ units
• C. $2\sqrt{6}$ units
• D. $2\sqrt{3}$ units

Answer 1

Answer: (B)

$\sqrt{6}$ units

$\overrightarrow{a}=3\hat{i}+\hat{j}-2\hat{k}=\overrightarrow{BC}$

$\overrightarrow{b}=-\hat{i}+3\hat{j}+4\hat{k}=\overrightarrow{AC}$

$\overrightarrow{c}=4\hat{i}-2\hat{j}-6\hat{k}=\overrightarrow{AB}$

$|\overrightarrow{a}{|}^2=(3^2+1^2+2^2)=14$

$|\overrightarrow{b}{|}^2=1^2+3^2+4^2=26$

$|\overrightarrow{c}{|}^2=4^2+2^2+6^2=56$

Length of median through vertex $C$ passing through $AC$:

$L=\frac{1}{2}\sqrt{2(a^2+b^2)-c^2}$

$L=\frac{1}{2}\sqrt{2(14+26)-56}=\frac{1}{2}\sqrt{24}=\sqrt{6}$