A) 3
B) 2
C) 1
D) zero
Correct Answer: A
Solution :
Since, vectors \[\vec{P}\]and \[\vec{Q}\] are perpendicular to each other, it means \[\vec{P}.\vec{Q}=P\,Q\,\cos \theta \]Here \[\theta ={{90}^{o}}\] So, \[\vec{P}.\vec{Q}=0\] \[\therefore \] \[(a\hat{i}+a\hat{j}+3\hat{k}).(a\hat{i}-2\hat{j}-\hat{k})=0\] \[\Rightarrow \] \[{{a}^{2}}-2a-3=0\] \[{{a}^{2}}-3a+a-3=0\] \[a(a-3)+1(a-3)=0\] \[(a+1)(a-3)=0\] Therefore, positive value of \[a=3\]You need to login to perform this action.
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