If the vectors $\overline{\mathrm{AB}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}}$ and $\overline{\mathrm{AC}}=5…
If the vectors $\overline{\mathrm{AB}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}}$ and $\overline{\mathrm{AC}}=5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ are the sides of the triangle $A B C$, then the length of the median through A is
$\sqrt{45}$ units
$\sqrt{18}$ units
$\sqrt{72}$ units
$\sqrt{33}$ units
Solution
Let $A D$ be the median of $\triangle A B C$.
$\begin{aligned}
\overline{\mathrm{AD}} & =\frac{\overline{\mathrm{AB}}+\overline{\mathrm{AC}}}{2} \\
& =\frac{8 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}}{2} \\
& =4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}
\end{aligned}$
$\therefore|\overline{\mathrm{AD}}|=\sqrt{4^2+1^2+4^2}=\sqrt{33}$ units