Match List-I with List -II.
| List-I | List-II | ||
| (A) | $(\hat{i}×\hat{k}).\hat{j}+(\hat{j}×\hat{k}).\hat{i}=$ | (I) | 2 |
| (B) | $\hat{i}.(\hat{j}×\hat{k})+\hat{j}.(\hat{i}×\hat{k})+\hat{k}.(\hat{i}×\hat{j})=$ | (II) | -1 |
| (C) | $(\hat{k}×\hat{j}).\hat{i}+\hat{k}.\hat{j}=$ | (III) | 0 |
| (D) | $\hat{k}.(\hat{i}×\hat{j})+(\hat{j}×\hat{k}).\hat{i}=$ | (IV) | 1 |
Choose the correct answer from he options given below :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III),(B)-(IV),(C)-(II),(D)-(I)
(A) $(\hat{i}×\hat{k}).\hat{j}+(\hat{j}×\hat{k}).\hat{i}$
$=-\hat j.\hat j+\hat i.\hat i=-1+1=0$ (III)
(B) $\hat{i}.(\hat{j}×\hat{k})+\hat{j}.(\hat{i}×\hat{k})+\hat{k}.(\hat{i}×\hat{j})$
$=\hat i.\hat i+\hat j.-\hat j+\hat k.\hat k=1-1+1=1$ (IV)
(C) $(\hat{k}×\hat{j}).\hat{i}+\hat{k}.\hat{j}=-\hat i.\hat i+1=-1$ (II)
(D) $\hat{k}.(\hat{i}×\hat{j})+(\hat{j}×\hat{k}).\hat{i}$
$=\hat k.\hat k+\hat j.\hat j=1+1=2$ (I)