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If two vertices of a triangle are $\mathrm{A}(3,1,4)$ and $\mathrm{B}(-4,5,-3)$ and the centroid of the triangle is $\mathrm{G}(-1,2,1)$, then the third vertex $\mathrm{C}$ of the triangle is
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Verified Answer
The correct answer is:
$(-2,0,2)$
Let $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ and $\overline{\mathrm{g}}$ be the position vectors of $\mathrm{A}$, $\mathrm{B}, \mathrm{C}$ and $\mathrm{G}$ respectively.
$\begin{aligned}
& \overline{\mathrm{a}}=3 \hat{\mathrm{i}}+1 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}, \\
& \overline{\mathrm{b}}=-4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}, \\
& \overline{\mathrm{g}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}},
\end{aligned}$
$\mathrm{G}$ is centroid of $\triangle \mathrm{ABC}$.
$\begin{array}{rl}
\therefore \quad \bar{g} & =\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}}{3} \\
3 \bar{g} & =\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}} \\
3 & 3(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}}-4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}+\overline{\mathrm{c}} \\
\therefore \quad \overline{\mathrm{c}} & =-3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}-3 \hat{\mathrm{i}}-\hat{\mathrm{j}}-4 \hat{\mathrm{k}}+4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\
& =-2 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}
\end{array}$
$\therefore \quad$ Third vertex $\mathrm{C} \equiv(-2,0,2)$
$\begin{aligned}
& \overline{\mathrm{a}}=3 \hat{\mathrm{i}}+1 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}, \\
& \overline{\mathrm{b}}=-4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}, \\
& \overline{\mathrm{g}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}},
\end{aligned}$
$\mathrm{G}$ is centroid of $\triangle \mathrm{ABC}$.
$\begin{array}{rl}
\therefore \quad \bar{g} & =\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}}{3} \\
3 \bar{g} & =\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}} \\
3 & 3(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}}-4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}+\overline{\mathrm{c}} \\
\therefore \quad \overline{\mathrm{c}} & =-3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}-3 \hat{\mathrm{i}}-\hat{\mathrm{j}}-4 \hat{\mathrm{k}}+4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\
& =-2 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}
\end{array}$
$\therefore \quad$ Third vertex $\mathrm{C} \equiv(-2,0,2)$
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