If \(\hat{\imath}, \hat{\jmath}\) and \(\hat{k}\) are unit vectors along co-ordinates axes OX, OY and OZ respectively, then which of the following is/are true?
(A) \(\hat{\imath} \times \hat{\imath} = \vec{0}\)
(B) \(\hat{\imath} \times \hat{k} = \hat{\jmath}\)
(C) \(\hat{\imath} \cdot \hat{\imath} = 1\)
(D) \(\hat{\imath} \cdot \hat{\jmath} = 0\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (C) and (D) only
\(\hat{\imath}, \hat{\jmath}, \hat{k}\) are mutually perpendicular unit vectors. For the cross product they follow the cyclic order \(\hat{\imath} \to \hat{\jmath} \to \hat{k} \to \hat{\imath}\):
\(\hat{\imath} \times \hat{\jmath} = \hat{k}\)
\(\hat{\jmath} \times \hat{k} = \hat{\imath}\)
\(\hat{k} \times \hat{\imath} = \hat{\jmath}\)
Reversing the order changes the sign.
Checking the statements
(A) \(\hat{\imath} \times \hat{\imath} = \vec{0}\) — Correct. The angle between \(\hat{\imath}\) and itself is 0 and \(\sin 0 = 0\), so the cross product is the zero vector.
(B) \(\hat{\imath} \times \hat{k} = \hat{\jmath}\) — Incorrect. \(\hat{k} \times \hat{\imath} = \hat{\jmath}\), so \(\hat{\imath} \times \hat{k} = -\hat{\jmath}\).
(C) \(\hat{\imath} \cdot \hat{\imath} = 1\) — Correct. \(\hat{\imath} \cdot \hat{\imath} = |\hat{\imath}|^2 = 1\).
(D) \(\hat{\imath} \cdot \hat{\jmath} = 0\) — Correct. The vectors are perpendicular and \(\cos 90^\circ = 0\).