A) \[4\hat{i}+8\hat{j}-4\hat{k}\]
B) \[8\hat{i}+4\hat{j}+4\hat{k}\]
C) \[3\hat{i}+\hat{j}+2\hat{k}\]
D) \[\hat{i}+\hat{j}-\hat{k}\]
Correct Answer: B
Solution :
We know that a vector perpendicular to the plane containing the points \[\vec{A},\]\[\vec{B},\]\[\vec{C}\] is given by \[\vec{A}\times \vec{B}+\vec{B}\times \vec{C}+\vec{C}\times \vec{A}.\] We have, \[\vec{A}=\hat{i}-\hat{j}+2\hat{k},\]\[\vec{B}=2\hat{i}+0\hat{j}-\hat{k}\] and \[\vec{C}=0\hat{i}+2\hat{j}+\hat{k}\] Now, \[\vec{A}\times \vec{B}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & -1 & 2 \\ 2 & 0 & -1 \\ \end{matrix} \right|=\hat{i}+5\hat{j}+2\hat{k}\] \[\vec{B}\times \vec{C}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 2 & 0 & -1 \\ 0 & 2 & 1 \\ \end{matrix} \right|=2\hat{i}-2\hat{j}+4\hat{k}\] \[\vec{C}\times \vec{A}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 0 & 2 & 1 \\ 1 & -1 & 2 \\ \end{matrix} \right|=5\hat{i}+\hat{j}-2\hat{k}\] Thus, \[\vec{A}\times \vec{B}+\vec{B}\times \vec{C}+\vec{C}\times \vec{A}=(\hat{i}+5\hat{j}+2\hat{k})\] \[+(2\hat{i}-2\hat{j}+4\hat{k})+(5\hat{i}+\hat{j}-2\hat{k})\] \[=8\hat{i}+4\hat{j}+4\hat{k}\]
You need to login to perform this action.
You will be redirected in
3 sec
