Match List-I with List-II, List-I, List-II, Mathematical Statement, Value, (A) i * (j x k), (I) -k, (B) j * (i x k), (II) 1, (C) i x (j x k), (III) -1, (D) j x i, (IV) 0, Choose the correct answer from the options given below: Option 1: (A) - (II), (B) - (III), (C) - (IV), (D) - (I) Option 2: (A) - (III), (B) - (IV), (C) - (I), (D) - (II) Option 3: (A) - (IV), (B) - (III), (C) - (I), (D) - (II) Option 4: (A) - (I), (B) - (II), (C) - (III), (D) - (IV)

Answer: (A) - (II), (B) - (III), (C) - (IV), (D) - (I). Solution: Cross Product Rules: i x j = k j x k = i k x i = j j x i = -k k x j = -i i x k = -j Dot Product Rules: i * i = j * j = k * k = 1 i * j = j * k = k * i = 0 Any vector crossed with itself equals 0 (A) i * (j x k) j x k = i i * i = 1 (A) matches with (II) (B) j * (i x k) i x k = -j j * (-j) = -1 (B) matches with (III) (C) i x (j x k) j x k = i i x i = 0 (C) matches with (IV) (D) j x i j x i = -k (D) matches with (I) Final Matching: (A) → (II) (B) → (III) (C) → (IV) (D) → (I)

Algebra

Vector Algebra

Medium

core

Match List-I with List-II

List-I List-II
Mathematical Statement Value
(A) $\hat{i} \cdot (\hat{j} \times \hat{k})$ (I) $-\hat{k}$
(B) $\hat{j} \cdot (\hat{i} \times \hat{k})$ (II) 1
(C) $\hat{i} \times (\hat{j} \times \hat{k})$ (III) -1
(D) $\hat{j} \times \hat{i}$ (IV) $\vec{0}$

Choose the correct answer from the options given below:

List-I	List-II
Mathematical Statement	Value
(A) $\hat{i} \cdot (\hat{j} \times \hat{k})$	(I) $-\hat{k}$
(B) $\hat{j} \cdot (\hat{i} \times \hat{k})$	(II) 1
(C) $\hat{i} \times (\hat{j} \times \hat{k})$	(III) -1
(D) $\hat{j} \times \hat{i}$	(IV) $\vec{0}$

(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 Option: 1

Cross Product Rules:

$\hat{i} \times \hat{j} = \hat{k}$

$\hat{j} \times \hat{k} = \hat{i}$

$\hat{k} \times \hat{i} = \hat{j}$

$\hat{j} \times \hat{i} = -\hat{k}$

$\hat{k} \times \hat{j} = -\hat{i}$

$\hat{i} \times \hat{k} = -\hat{j}$

Dot Product Rules:

$\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$

$\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$

Any vector crossed with itself equals $\vec{0}$

(A) $\hat{i} \cdot (\hat{j} \times \hat{k})$

$\hat{j} \times \hat{k} = \hat{i}$

$\hat{i} \cdot \hat{i} = 1$

(A) matches with (II)

(B) $\hat{j} \cdot (\hat{i} \times \hat{k})$

$\hat{i} \times \hat{k} = -\hat{j}$

$\hat{j} \cdot (-\hat{j}) = -1$

(B) matches with (III)

$\hat{j} \times \hat{k} = \hat{i}$

$\hat{i} \times \hat{i} = \vec{0}$

(D) $\hat{j} \times \hat{i}$

$\hat{j} \times \hat{i} = -\hat{k}$

(D) matches with (I)

Final Matching:

(A) → (II)

(B) → (III)

(D) → (I)

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