A) \[\hat{i}+5\hat{j}+2\hat{k}\]
B) \[\hat{i}-2\hat{j}+16\hat{k}\]
C) \[5\hat{i}+\hat{j}+2\hat{k}\]
D) \[5\hat{i}-\hat{j}+2\hat{k}\]
Correct Answer: C
Solution :
Let \[\vec{b}=x\,\hat{i}+y\,\hat{j}+z\,\hat{k}\] Given, \[\vec{a}=2\hat{i}+2\hat{j}+\hat{k}\] and \[\vec{a}\,.\,\vec{b}=14\] \[\Rightarrow \] \[2x+2y+z=14\] Also, given \[\vec{a}\times \vec{b}=3\hat{i}+\hat{j}-8\hat{k}\] \[\Rightarrow \] \[\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 2 & 2 & 1 \\ x & y & z \\ \end{matrix} \right|=3\hat{i}+\hat{j}-8\hat{k}\] \[\Rightarrow \] \[\hat{i}(2z-y)-\hat{j}(2z-x)+\hat{k}(2y-2x)\] \[=3\hat{i}+\hat{j}-8\hat{k}\] Equating the coefficient of \[\hat{i},\hat{j},\hat{k},\] we get \[-y+2z=3\] \[\Rightarrow \] \[2z=3+y\] ?..(ii) \[x-2z=1\] \[\Rightarrow \] \[x=1+2z\] ?..(iii) and \[-2x+2y=-8\] \[\Rightarrow \] \[-x+y=-4\] \[\Rightarrow \] \[x=4+y\] ?(iv) On putting the values from Eqs. (ii) and (iv) in Eq. (i), we get \[2(4+y)+2y+(3+y)/2=14\] \[\Rightarrow \] \[2(8+2y+2y)+3+y=28\] \[\Rightarrow \] \[9y=9\] \[\Rightarrow \] \[y=1\] \[\therefore \] \[z=2\] and \[x=5\] \[\therefore \] s\[\vec{b}=5\hat{i}+\hat{j}+2\hat{k}\]
You need to login to perform this action.
You will be redirected in
3 sec
