Obtain the scalar product of two vectors in terms of their Cartesian components.

(N/A) Let two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ be represented in Cartesian components as follows:
$\overrightarrow{A} = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}$
$\overrightarrow{B} = B_{x} \hat{i} + B_{y} \hat{j} + B_{z} \hat{k}$
Then,the scalar product is given by:
$\overrightarrow{A} \cdot \overrightarrow{B} = (A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}) \cdot (B_{x} \hat{i} + B_{y} \hat{j} + B_{z} \hat{k})$
Expanding the dot product using the distributive property:
$= A_{x} B_{x}(\hat{i} \cdot \hat{i}) + A_{x} B_{y}(\hat{i} \cdot \hat{j}) + A_{x} B_{z}(\hat{i} \cdot \hat{k}) + A_{y} B_{x}(\hat{j} \cdot \hat{i}) + A_{y} B_{y}(\hat{j} \cdot \hat{j}) + A_{y} B_{z}(\hat{j} \cdot \hat{k}) + A_{z} B_{x}(\hat{k} \cdot \hat{i}) + A_{z} B_{y}(\hat{k} \cdot \hat{j}) + A_{z} B_{z}(\hat{k} \cdot \hat{k})$
Using the properties of unit vectors: $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$ and $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{i} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{j} = \hat{k} \cdot \hat{i} = \hat{i} \cdot \hat{k} = 0$.
Substituting these values,we get:
$\overrightarrow{A} \cdot \overrightarrow{B} = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}$