Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Mathematical Statement | Value | ||
| (A) | \(\hat{\imath}\cdot(\hat{\jmath} \times \hat{k})\) | (I) | \(-\hat{k}\) |
| (B) | \(\hat{\jmath}\cdot(\hat{\imath} \times \hat{k})\) | (II) | 1 |
| (C) | \(\hat{\imath} \times (\hat{\jmath} \times \hat{k})\) | (III) | −1 |
| (D) | \(\hat{\jmath} \times \hat{\imath}\) | (IV) | \(\vec{0}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
For the unit vectors in cyclic order, \(\hat{\imath} \times \hat{\jmath} = \hat{k}\), \(\hat{\jmath} \times \hat{k} = \hat{\imath}\), \(\hat{k} \times \hat{\imath} = \hat{\jmath}\). Reversing the order changes the sign, the cross product of a vector with itself is \(\vec{0}\), and \(\hat{\imath}\cdot\hat{\imath} = \hat{\jmath}\cdot\hat{\jmath} = 1\).
(A) \(\hat{\imath}\cdot(\hat{\jmath} \times \hat{k})\)
This matches (II) 1:
\(\hat{\imath}\cdot(\hat{\jmath} \times \hat{k}) = \hat{\imath}\cdot\hat{\imath} = 1\)
(B) \(\hat{\jmath}\cdot(\hat{\imath} \times \hat{k})\)
This matches (III) −1:
\(\hat{\imath} \times \hat{k} = -\hat{\jmath}\)
\(\hat{\jmath}\cdot(-\hat{\jmath}) = -1\)
(C) \(\hat{\imath} \times (\hat{\jmath} \times \hat{k})\)
This matches (IV) \(\vec{0}\):
\(\hat{\imath} \times (\hat{\jmath} \times \hat{k}) = \hat{\imath} \times \hat{\imath} = \vec{0}\)
(D) \(\hat{\jmath} \times \hat{\imath}\)
This matches (I) \(-\hat{k}\):
\(\hat{\jmath} \times \hat{\imath} = -(\hat{\imath} \times \hat{\jmath}) = -\hat{k}\)
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).
Note: This topic was not in the CUET 2026 syllabus.