For any vector \(\vec{a}\), the vector \((\vec{a}\cdot\hat{\imath})\,\hat{\imath}+(\vec{a}\cdot\hat{\jmath})\,\hat{\jmath}+(\vec{a}\cdot\hat{k})\,\hat{k}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\vec{a}\)
Let \(\vec{a} = a_1\hat{\imath}+a_2\hat{\jmath}+a_3\hat{k}\). Since \(\hat{\imath}, \hat{\jmath}, \hat{k}\) are mutually perpendicular unit vectors, the scalar product with each picks out one scalar component:
\(\vec{a}\cdot\hat{\imath} = a_1, \quad \vec{a}\cdot\hat{\jmath} = a_2, \quad \vec{a}\cdot\hat{k} = a_3\)
\((\vec{a}\cdot\hat{\imath})\,\hat{\imath}+(\vec{a}\cdot\hat{\jmath})\,\hat{\jmath}+(\vec{a}\cdot\hat{k})\,\hat{k} = a_1\hat{\imath}+a_2\hat{\jmath}+a_3\hat{k} = \vec{a}\)