Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Vectors

Question:

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:

Options:

(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

(A)-(IV), (B)-(III), (C)-(I), (D)-(II)

(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

Correct Answer:

(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Explanation:

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)