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Let the vectors $\overline{A B}=2 \hat{i}+2 \hat{j}+\hat{k}$ and $\overline{A C}=2 \hat{i}+4 \hat{j}+4 \hat{k}$ be two sides of a triangle $A B C$. If $G$ is the centroid of $\triangle A B C$, then $\frac{22}{7}(\overline{A G})^2+5=$
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Verified Answer
The correct answer is:
38
Given vectors are $\overrightarrow{\mathrm{AB}}=2 \hat{i}+2 \hat{j}+\hat{k}$ and $\overrightarrow{\mathrm{AC}}=2 \hat{i}+4 \hat{j}+4 \hat{k}$
Centroid $\mathrm{G}$ of triangle has position vector $\overrightarrow{\mathrm{AG}}$ can be obtained by putting A as $\overrightarrow{\mathrm{O}}$, B as $2 \hat{i}+2 \hat{j}+\hat{k}$ and $\mathrm{C}$ as $2 \hat{i}+4 \hat{j}+4 \hat{k}$.
So, $\overrightarrow{\mathrm{AG}}=\frac{\overrightarrow{\mathrm{O}}+\mathrm{B}+\mathrm{C}}{3}=\frac{0+2 \hat{i}+2 \hat{j}+\hat{k}+2 \hat{i}+4 \hat{j}+4 \hat{k}}{3}$
$\begin{aligned}
& \overrightarrow{\mathrm{AG}}=\frac{4 \hat{i}+6 \hat{j}+5 \hat{k}}{4} \\
& |\overrightarrow{\mathrm{AG}}|=\sqrt{16+36+25}=\frac{\sqrt{77}}{3} \\
& \text {Now, } \frac{27}{7} \times|\overrightarrow{\mathrm{AG}}|^2+5=\frac{27}{7} \times \frac{77}{9}+5 \\
& =33+5=38
\end{aligned}$
Therefore, option (b) is correct.
Centroid $\mathrm{G}$ of triangle has position vector $\overrightarrow{\mathrm{AG}}$ can be obtained by putting A as $\overrightarrow{\mathrm{O}}$, B as $2 \hat{i}+2 \hat{j}+\hat{k}$ and $\mathrm{C}$ as $2 \hat{i}+4 \hat{j}+4 \hat{k}$.
So, $\overrightarrow{\mathrm{AG}}=\frac{\overrightarrow{\mathrm{O}}+\mathrm{B}+\mathrm{C}}{3}=\frac{0+2 \hat{i}+2 \hat{j}+\hat{k}+2 \hat{i}+4 \hat{j}+4 \hat{k}}{3}$
$\begin{aligned}
& \overrightarrow{\mathrm{AG}}=\frac{4 \hat{i}+6 \hat{j}+5 \hat{k}}{4} \\
& |\overrightarrow{\mathrm{AG}}|=\sqrt{16+36+25}=\frac{\sqrt{77}}{3} \\
& \text {Now, } \frac{27}{7} \times|\overrightarrow{\mathrm{AG}}|^2+5=\frac{27}{7} \times \frac{77}{9}+5 \\
& =33+5=38
\end{aligned}$
Therefore, option (b) is correct.
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