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 ABC , 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 $\triangle A B C$.
$\begin{aligned}
\therefore \quad \overline{\mathrm{AD}} & =\frac{\overline{\mathrm{AB}}+\overline{\mathrm{AC}}}{2} \\
& =\frac{3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}}+5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}}{2} \\
& =\frac{8 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}}{2} \\
& =4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\
\therefore \quad|\overline{\mathrm{AD}}| & =\sqrt{4^2+(-1)^2+4^2}=\sqrt{33} \text { units }
\end{aligned}$