If $\overline{A B}=3 \hat{\imath}+5 \hat{\jmath}+4 \hat{k}, \overline{A C}=5 \hat{\imath}-5 \hat{\jmath}+2…
If $\overline{A B}=3 \hat{\imath}+5 \hat{\jmath}+4 \hat{k}, \overline{A C}=5 \hat{\imath}-5 \hat{\jmath}+2 \hat{k}$ represent the sides of triangle $\mathrm{ABC}$, then
the length of median through $\mathrm{A}$ is
$\sqrt{6}$ units
5 units
$\sqrt{5}$ units
6 units
Solution
Given
$\begin{aligned}
\frac{\overline{\mathrm{AB}}}{\mathrm{AC}} &=3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\
&=5 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}
\end{aligned}$
Let $\overline{\mathrm{AD}}$ is median
position vector of
$\begin{aligned}
\overline{\mathrm{AD}} &=\frac{(3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})+(5 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})}{2} \\
\overline{\mathrm{AD}} &=4 \hat{\mathrm{i}}+3 \hat{\mathrm{k}} \\
\therefore|\overline{\mathrm{AD}}| &=\sqrt{16+9}=\sqrt{25}=5
\end{aligned}$
