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$7 \bar{i}-4 \bar{j}+7 \bar{k}, \bar{i}-6 \bar{j}+10 \bar{k}, \overline{-i}-3 \bar{j}+4 \bar{k}, 5 \bar{i}-\bar{j}+5 \bar{k}$ respectively, then $A B C D$ is
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The correct answer is:
a quadrilateral which is not a parallelogram
$A: 7 \hat{i}-4 \hat{j}+7 \hat{k}$
$\begin{aligned} & B: \hat{i}-6 \hat{j}+10 \hat{k} \\ & C:-\hat{i}-3 \hat{j}+4 \hat{k} \\ & D: 5 \hat{i}-\hat{j}+5 \hat{k}\end{aligned}$
$\begin{aligned} & \therefore \overrightarrow{A B}=B-A=-6 \hat{i}-2 \hat{j}+3 \hat{k} \\ & \overrightarrow{B C}=C-B=-2 \hat{i}+3 \hat{j}-6 \hat{k} \\ & \overrightarrow{C D}=D-C=6 \hat{i}+2 \hat{j}+\hat{k} \\ & \overrightarrow{D A}=A-D=2 \hat{i}-3 \hat{j}+2 \hat{k} \\ & \overrightarrow{A B} \cdot \overline{C D}=-36-4+3 \neq 0\end{aligned}$
and $\overrightarrow{B C} \cdot \overrightarrow{D A}=-4-9-12 \neq 0$
as, $\mathrm{AB} \neq \mathrm{CD}$ and $\mathrm{BC} \neq \mathrm{DA}$
$\mathrm{ABCD}$ is a quadrilateral
Which is not parallelogram.
$\begin{aligned} & B: \hat{i}-6 \hat{j}+10 \hat{k} \\ & C:-\hat{i}-3 \hat{j}+4 \hat{k} \\ & D: 5 \hat{i}-\hat{j}+5 \hat{k}\end{aligned}$
$\begin{aligned} & \therefore \overrightarrow{A B}=B-A=-6 \hat{i}-2 \hat{j}+3 \hat{k} \\ & \overrightarrow{B C}=C-B=-2 \hat{i}+3 \hat{j}-6 \hat{k} \\ & \overrightarrow{C D}=D-C=6 \hat{i}+2 \hat{j}+\hat{k} \\ & \overrightarrow{D A}=A-D=2 \hat{i}-3 \hat{j}+2 \hat{k} \\ & \overrightarrow{A B} \cdot \overline{C D}=-36-4+3 \neq 0\end{aligned}$
and $\overrightarrow{B C} \cdot \overrightarrow{D A}=-4-9-12 \neq 0$
as, $\mathrm{AB} \neq \mathrm{CD}$ and $\mathrm{BC} \neq \mathrm{DA}$
$\mathrm{ABCD}$ is a quadrilateral
Which is not parallelogram.
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