If \( \hat{\imath},\hat{\jmath},\hat{k} \) represents unit vectors along x, y, z axes respectively, then
Match List-I with List-II
| List-I | List-II | ||
| (A) | \( \hat{\imath}.\hat{\imath} \) | (I) | 0 |
| (B) | \( \hat{\imath}.\hat{\jmath} \) | (II) | \( \vec{0} \) |
| (C) | \( \hat{\imath}\times\hat{\jmath} \) | (III) | 1 |
| (D) | \( \hat{\imath}\times\hat{\imath} \) | (IV) | \( \hat{k} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
A dot product is a scalar, \( |\vec{a}||\vec{b}|\cos\theta \); a cross product is a vector of magnitude \( |\vec{a}||\vec{b}|\sin\theta \), perpendicular to both vectors.
(A) \( \hat{\imath}.\hat{\imath} \)
This matches (III) 1.
\( \displaystyle \hat{\imath}.\hat{\imath}=1\times 1\times\cos 0^\circ=1 \)
(B) \( \hat{\imath}.\hat{\jmath} \)
This matches (I) 0.
\( \displaystyle \hat{\imath}.\hat{\jmath}=1\times 1\times\cos 90^\circ=0 \)
(C) \( \hat{\imath}\times\hat{\jmath} \)
This matches (IV) \( \hat{k} \). The magnitude is \( \sin 90^\circ=1 \), and by the right-hand rule the direction is along the positive z-axis.
(D) \( \hat{\imath}\times\hat{\imath} \)
This matches (II) \( \vec{0} \). The angle is 0°, so the magnitude is \( \sin 0^\circ=0 \); a cross product is a vector, so the result is the zero vector, not the number 0.
So the matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).
NCERT (2026-27 edition):
Product of Two Vectors
