Match List-I with List-II
|
List-I Mathematical Statement |
List-II Value |
|
(A) $\hat i.(\hat j×\hat k)$ |
(I) $-\hat k$ |
|
(B) $\hat j.(\hat i×\hat k)$ |
(II) 1 |
|
(C) $\hat i× (\hat j×\hat k)$ |
(III) -1 |
|
(D) $\hat j×\hat i$ |
(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)
|
List-I Mathematical Statement |
List-II Value |
|
(A) $\hat i.(\hat j×\hat k)$ |
(II) 1 |
|
(B) $\hat j.(\hat i×\hat k)$ |
(III) -1 |
|
(C) $\hat i× (\hat j×\hat k)$ |
(IV) $\vec 0$ |
|
(D) $\hat j×\hat i$ |
(I) $-\hat k$ |
(A) $\hat{i} \cdot (\hat{j} \times \hat{k})$: $\hat{j} \times \hat{k} = \hat{i}$, so $\hat{i} \cdot \hat{i} = 1$ → (II)
(B) $\hat{j} \cdot (\hat{i} \times \hat{k})$: $\hat{i} \times \hat{k} = -\hat{j}$, so $\hat{j} \cdot (-\hat{j}) = -1$ → (III)
(C) $\hat{i} \times (\hat{j} \times \hat{k})$: $\hat{j} \times \hat{k} = \hat{i}$, so $\hat{i} \times \hat{i} = \vec{0}$ → (IV)
(D) $\hat{j} \times \hat{i} = -\hat{k}$ → (I)