A) 23
B) 9
C) 8
D) 3
Correct Answer: D
Solution :
Key Idea: Scalar product of two vectors\[\vec{A}\] and \[\vec{B}\] is \[AB\,\cos \,\theta ,\]where \[\theta \] is angle between \[\vec{A}\] and\[\vec{B}\]. The scalar product of two vectors is defined as a scalar quantity equal to the product of their magnitudes and the cosine of the angle between them. i.e., \[\vec{A}.\vec{B}=|\vec{A}|\vec{B}|cos\theta =AB\cos \theta \] Given, \[\theta ={{90}^{o}},\,\cos {{90}^{o}}=0\] \[\therefore \] \[\vec{A}=\vec{P}=a\hat{i}+a\hat{j}+3\hat{k}\] and \[\vec{B}=\vec{Q}=a\hat{i}-2\hat{j}-\hat{k}\] \[\vec{P}.\vec{Q}=(a\hat{i}+a\hat{j}+3\hat{k}).(a\hat{i}-2\hat{j}-\hat{k})=0\] \[\Rightarrow \] \[{{a}^{2}}-2a-3=0\] \[(a-3)(a+1)=0\] \[\Rightarrow \]\[a=3\]or \[a=-1\] -ve value should be neglected, so possible value of a is 3.
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