Let $\theta$ be the angle between two vectors $\vec{a}$ and $\vec{b}$. Then match List-I with List-II

List-I	List-II
(A) $\sin \theta$	(I) $\dfrac{\vec{a} \cdot \vec{b}}{\vert \vec{a}\vert \vert \vec{b}\vert }$
(B) $\cos \theta$	(II) $\vert \vec{a} \times \vec{b}\vert$
(C) Area of the parallelogram with adjacent sides represented by $\vec{a}$ and $\vec{b}$	(III) $\dfrac{\vec{a} \cdot \vec{b}}{\vert \vec{a}\vert }$
(D) Projection of $\vec{a}$ on $\vec{b}$	(IV) $\dfrac{\vert \vec{a} \times \vec{b}\vert }{\vert \vec{a}\vert \vert \vec{b}\vert }$

Choose the correct answer from the options given below:

From the cross product formula:

$|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$

$\sin\theta = \dfrac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}$

(A) matches with (IV)

From the dot product formula:

$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$

$\cos\theta = \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$

(B) matches with (I)

If two adjacent sides of a parallelogram are represented by $\vec{a}$ and $\vec{b}$:

$\text{Area} = |\vec{a} \times \vec{b}|$

(C) matches with (II)

By elimination, (D) pairs with the only remaining option (III).

Note: The standard formula for the scalar projection of $\vec{a}$ on $\vec{b}$ is $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$. The expression in (III) — $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|}$ — actually represents the projection of $\vec{b}$ on $\vec{a}$. This appears to be a typo in the question, but since the other three matches are uniquely determined, (D) pairs with (III) by elimination.

(A) $\sin\theta$ → (IV) $\dfrac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}$

(B) $\cos\theta$ → (I) $\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$

(C) Area of parallelogram → (II) $|\vec{a} \times \vec{b}|$

(D) Projection → (III) $\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$

Therefore, the correct answer is (A)-(IV), (B)-(I), (C)-(II), (D)-(III).