A) 1, 5, 9
B) 9, 5, 1
C) 5, 1, 9
D) none of these
Correct Answer: D
Solution :
Given\[\overrightarrow{a}=(1,p,1)\overrightarrow{b}=(q,2,2)\] \[\overrightarrow{a}.\overrightarrow{b}=r\]and\[\overrightarrow{a}\times \overrightarrow{b}=(0,-3,3)\] \[\therefore \] \[\overrightarrow{a}.\overrightarrow{b}=(\hat{i}+p\hat{j}+\hat{k}).(q\hat{i}+2\hat{j}+2\hat{k})=r\] (given) \[=q+2p+2=r\] \[\Rightarrow \] \[q+2p+2=r\] ...(i) Also, \[\overrightarrow{a}\times \overrightarrow{b}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & p & 1 \\ q & 2 & 2 \\ \end{matrix} \right|\] \[=(2p-2)\hat{i}+\hat{j}(q-2)+\hat{k}(2-pq)\] \[\{(0\hat{i}+(-3)\hat{j}+(3)\hat{k}\}\] (given) \[\Rightarrow \]\[2p-2=0;q-2=-3;\text{ }2-pq=3\] \[\Rightarrow \]\[p=1,\text{ }q=-1\] From Eq. (i) \[-1+2+2=r\] \[\Rightarrow \] \[r=3\] \[\therefore \] \[p=1,q=-1,r=3\]
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