If the vectors $\mathbf{A B}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$, $\mathbf{A C}=5…
If the vectors $\mathbf{A B}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$, $\mathbf{A C}=5 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ are two sides of a triangle $A B C$, whose centroid is $G$, then $|\mathbf{A G}|=$
$\frac{2}{3} \sqrt{22}$
$\sqrt{22}$
$\frac{1}{3} \sqrt{22}$
$\sqrt{18}$
Solution
It the vectors $\mathbf{A B}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and
$\mathbf{A C}=5 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \mathbf{k}$, then
$\mathbf{A G}=\frac{(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})+(5 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})}{3}$, where $G$ is the centroid of $\triangle A B C$.
$
\begin{aligned}
& \mathbf{A G}=2 \hat{\mathbf{i}}+\frac{4}{3} \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\
& \begin{aligned}
\therefore|\mathbf{A G}| & =\sqrt{4+\frac{16}{9}+4}=\sqrt{8+\frac{16}{9}}=\sqrt{\frac{72+16}{9}} \\
& =\frac{1}{3} \sqrt{88}=\frac{2}{3} \sqrt{22}
\end{aligned}
\end{aligned}
$
Hence, option (a) is correct