Question

The vector $ P\to =a\widehat{i}+a\widehat{j}+3\widehat{k} $ and $ Q\to =a\widehat{i}+2\widehat{j}-\widehat{k} $ are perpendicular to each other, the positive value of a is:

• A. 3
• B. 2
• C. 1
• D. zero

Answer 1

Answer: (A)

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 $