Match List-I with List-II

Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$.

List-I	List-II
(A) $\vec{a} \cdot \vec{b}$	(I) $\dfrac{\vec{a} \cdot \vec{b}}{\vert \vec{b}\vert ^2} \vec{b}$
(B) $\vec{a} \times \vec{b}$	(II) $\vec{a} \cdot \vec{b} = 0$
(C) Projection vector of $\vec{a}$ on $\vec{b}$ ($\ne{0}$)	(III) $\vert \vec{a}\vert \vert \vec{b}\vert \sin \theta \, \hat{n}$ where $\hat{n}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$
(D) $\vec{a}$ and $\vec{b}$ are orthogonal vectors	(IV) $\vert \vec{a}\vert \vert \vec{b}\vert \cos \theta$

The dot product of two vectors is defined as:

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

This gives a scalar quantity.

$\text{(A)} \rightarrow \text{(IV)}$

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The cross product of two vectors is defined as:

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

where $\hat{n}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$.

This gives a vector quantity.

$\text{(B)} \rightarrow \text{(III)}$

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The projection vector of $\vec{a}$ onto $\vec{b}$ is:

Scalar projection $= \dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$

Multiplying by the unit vector $\dfrac{\vec{b}}{|\vec{b}|}$ to get the vector projection:

$\dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \times \dfrac{\vec{b}}{|\vec{b}|}$

$= \dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\;\vec{b}$

$\text{(C)} \rightarrow \text{(I)}$

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Orthogonal vectors are perpendicular, so $\theta = 90°$:

$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos 90°$

$= |\vec{a}||\vec{b}| \times 0$

$= 0$

Two vectors are orthogonal if and only if $\vec{a} \cdot \vec{b} = 0$.

$\text{(D)} \rightarrow \text{(II)}$

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$\text{(A)} \rightarrow \text{(IV)},\quad \text{(B)} \rightarrow \text{(III)},\quad \text{(C)} \rightarrow \text{(I)},\quad \text{(D)} \rightarrow \text{(II)}$