A) \[-8\hat{i}+4\hat{j}+8\hat{k}\]
B) \[8\hat{i}+4\hat{j}+8\hat{k}\]
C) \[8\hat{i}-4\hat{j}+8\hat{k}\]
D) \[8\hat{i}-4\hat{j}-8\hat{k}\]
E) \[4i-8\hat{j}-8\hat{k}\]
Correct Answer: C
Solution :
Let\[\overrightarrow{a}=4\hat{i}+6\hat{j}-\hat{k}\]and \[\overrightarrow{b}=3\hat{i}+8\hat{j}+\hat{k}\] \[\therefore \] \[\overrightarrow{c}=|\overrightarrow{a}\times \overrightarrow{b}|=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 4 & 6 & -1 \\ 3 & 8 & 1 \\ \end{matrix} \right|\] \[=14\hat{i}-7\hat{j}+14\hat{k}\] \[\Rightarrow \] \[\hat{c}=\frac{14\hat{i}-7\hat{j}+14\hat{k}}{21}\] \[\Rightarrow \] \[\overrightarrow{d}=\hat{c}\times 12\] \[=12.\frac{(14\hat{i}-7\hat{j}+14\hat{k})}{21}\] \[=8\hat{i}-4\hat{j}+8\hat{k}\]
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