A) \[\pm \,\left( \frac{2\mathbf{i}+2\mathbf{j}+\mathbf{k}}{3} \right)\]
B) \[\pm \,\left( \frac{2\mathbf{i}-2\mathbf{j}+\mathbf{k}}{3} \right)\]
C) \[\pm \,\left( \frac{2\mathbf{i}-2\mathbf{j}-\mathbf{k}}{3} \right)\]
D) \[-\,\left( \frac{2\mathbf{i}+2\mathbf{j}+\mathbf{k}}{3} \right)\]
Correct Answer: A
Solution :
\[\overrightarrow{AB}=2\mathbf{i}-\mathbf{j}-2\mathbf{k},\] \[\overrightarrow{AC}=3\mathbf{i}-3\mathbf{j}+0\mathbf{k}\] \[\overrightarrow{AB}\times \overrightarrow{AC}=\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & -2 \\ 3 & -3 & 0 \\ \end{matrix} \right|=(-6\mathbf{i}-6\mathbf{j}-3\mathbf{k})\] Hence unit vector \[=\pm \left( \frac{2\mathbf{i}+2\mathbf{j}+\mathbf{k}}{3} \right)\,.\]You need to login to perform this action.
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