Which of the following statements are correct?
(A) \(\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}\) for any two vectors \(\vec{a}\) and \(\vec{b}\)
(B) \(\hat{\imath}\cdot\hat{\jmath} = 1\)
(C) \(\vec{a}\cdot\vec{a} = |\vec{a}|^2\) for any vector \(\vec{a}\)
(D) If \(\vec{a}\cdot\vec{b} = 0\), then \(\vec{a} = \vec{0}\) or \(\vec{b} = \vec{0}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (C) only
The scalar product is \(\vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta\), where \(\theta\) is the angle between the vectors. It is commutative, and it is zero whenever the two vectors are perpendicular.
So the scalar product does not depend on the order, \(\vec{a}\cdot\vec{a} = |\vec{a}|^2\), \(\hat{\imath}\cdot\hat{\jmath} = 0\), and a zero scalar product can come from two non-zero perpendicular vectors.
Checking the statements
(A) \(\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}\) — Correct. Both equal \(|\vec{a}|\,|\vec{b}|\cos\theta\).
(B) \(\hat{\imath}\cdot\hat{\jmath} = 1\) — Incorrect. \(\hat{\imath}\) and \(\hat{\jmath}\) are perpendicular, so \(\hat{\imath}\cdot\hat{\jmath} = 0\).
(C) \(\vec{a}\cdot\vec{a} = |\vec{a}|^2\) — Correct. The angle is 0 and \(\cos 0 = 1\).
(D) If \(\vec{a}\cdot\vec{b} = 0\), then \(\vec{a} = \vec{0}\) or \(\vec{b} = \vec{0}\) — Incorrect. \(\hat{\imath}\cdot\hat{\jmath} = 0\) though neither is zero.