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If $\mathrm{A}(1,2,3), \mathrm{B}(3,7,-2), \mathrm{C}(6,7,7)$ and $\mathrm{D}(-1,0,-1)$ are points in a plane, then the vector equation of the line passing through the centroids of $\triangle \mathrm{ABD}$ and $\triangle \mathrm{ACD}$ is
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Verified Answer
The correct answer is:
$\overrightarrow{\mathrm{r}}=(1+t) \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+3 \mathrm{t} \hat{\mathrm{k}}$
$\mathrm{A}(1,2,3) ; \mathrm{B}(3,7,-2) ; \mathrm{C}(6,7,7) ; \mathrm{D}(-1,0,-1)$
Centroid of $\triangle \mathrm{ABD}=(1,3,0)$
Centroid of $\triangle \mathrm{ACD}=(2,3,3)$
Equation of vector passing through $\vec{a}$ and $\vec{b}$
$\begin{aligned}
& \vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a}) \\
& \vec{r}=(\hat{i}+3 \hat{j})+\lambda(\hat{i}+3 \hat{k}) \\
& \vec{r}=(1+\lambda) \hat{i}+3 \hat{j}+3 \lambda \hat{k} \text { or } \vec{r}=(1+t) \hat{i}+3 \hat{j}+3 t \hat{k}
\end{aligned}$
Centroid of $\triangle \mathrm{ABD}=(1,3,0)$
Centroid of $\triangle \mathrm{ACD}=(2,3,3)$
Equation of vector passing through $\vec{a}$ and $\vec{b}$
$\begin{aligned}
& \vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a}) \\
& \vec{r}=(\hat{i}+3 \hat{j})+\lambda(\hat{i}+3 \hat{k}) \\
& \vec{r}=(1+\lambda) \hat{i}+3 \hat{j}+3 \lambda \hat{k} \text { or } \vec{r}=(1+t) \hat{i}+3 \hat{j}+3 t \hat{k}
\end{aligned}$
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