A) 7
B) \[-7\]
C) \[-5\]
D) 5
Correct Answer: B
Solution :
Let \[\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}\] Given, \[\overrightarrow{r}\times \overrightarrow{b}=\overrightarrow{c}\times \overrightarrow{b}\] \[\Rightarrow \] \[(x\hat{i}+y\hat{j}+z\hat{k})\times (\hat{i}+\hat{j}+\hat{k})\] \[=(4\hat{i}-3\hat{j}+7\hat{k})\times (\hat{i}+\hat{k}+\hat{k})\] \[\Rightarrow \] \[(y-z)\hat{i}-(x-z)\hat{j}+(x-y)\hat{k}\] \[=10i+3\hat{j}+7\hat{k}\] \[\Rightarrow \] \[y-z=-10,-(x-z)=3,x-y=7\] \[\Rightarrow \] \[y-z=-10,-x+z=3,x-y=7\] ...(i) and \[\overrightarrow{r}\text{.}\overrightarrow{a}=0\] \[\Rightarrow \] \[(x\hat{i}+y\hat{j}+z\hat{k}).(2\hat{i}+\hat{k})\] \[\Rightarrow \] \[2x+z=0\] ...(ii) From Eqs. (i) and (ii), we get \[x=-1,\text{ }y=-8,\text{ }z=2\] \[\overrightarrow{r}.\overrightarrow{b}=(-\hat{i}-8\hat{j}+2\hat{k}).(\hat{i}+\hat{j}+\hat{k})\] \[=-1-8+2=-7\]
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