A) \[\hat{i}-\hat{j}+\hat{k}\]
B) \[3\hat{i}-\hat{j}+\hat{k}\]
C) \[3\hat{i}+\hat{j}-\hat{k}\]
D) \[\hat{i}-\hat{j}-\hat{k}\]
Correct Answer: C
Solution :
[c] \[\vec{r}\times \vec{a}=\vec{b}\times \vec{a}\] or \[(\vec{r}-\vec{b})\times \vec{a}=0\] \[\vec{r}\times \vec{b}=\vec{a}\times \vec{b}\] or \[(\vec{r}-\vec{b})\times \vec{b}=0\] If \[\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\], then \[\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ x-2 & y & z+1 \\ 1 & 1 & 0 \\ \end{matrix} \right|=0\] and \[\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ x-1 & y-1 & z \\ 2 & 0 & -1 \\ \end{matrix} \right|=0\] \[\Rightarrow z+1=0,x-y=2\] And \[y-1=0,x-1+2z=0\] \[\Rightarrow x=3,y=1,z=-1\]You need to login to perform this action.
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