**$\overrightarrow{V_1} . \overrightarrow{V_1}$ Express in terms of $V_1$.****$\overrightarrow{V_1} . \overrightarrow{V_2}$ When they are perpendicular.****$\overrightarrow{V_1} . \overrightarrow{V_2}$ When they are parallel.**

This question aims to find the dot product of two vectors when they are parallel and also when they are perpendicular.

The question can be solved by revising the concept of vector multiplication, exclusively the dot product between two vectors. The dot product is also called the scalar product of vectors. It is the product of the magnitude of both vectors with the cosine of the angle between those vectors.

## Expert Answer:

The dot product or the scalar product of two vectors is the product of their magnitude and the cosine of the angle between them. If $\overrightarrow{A}$ and $\overrightarrow{B}$ are two vectors, their dot product is given by:

\[ \overrightarrow{A} . \overrightarrow{B} = |A| |B| \cos \theta \]

$|A|$ and $|B|$ are the magnitude of $\overrightarrow{A}$ and $\overrightarrow{B}$ respectively and $\theta$ is the angle between those vectors.

The given problem has two vectors $\overrightarrow{V_1}$ and $\overrightarrow{V_2}$ with magnitudes $V_1$ and $V_2$, respectively.

a) The dot product of $\overrightarrow{V_1}$ with itself is given by:

\[ \overrightarrow{V_1} . \overrightarrow{V_1} = |V_1| |V_1| \cos (0^{\circ}) \]

The angle of the vector with itself is zero.

\[ \cos (0^{\circ}) = 1 \]

\[ \overrightarrow{V_1} . \overrightarrow{V_1} = (V_1) (V_1) 1 \]

\[ \overrightarrow{V_1} . \overrightarrow{V_1} = V_1^{2} \]

The dot product of the vector with itself is its magnitude squared.

b) The dot product of $\overrightarrow{V_1}$ with $\overrightarrow{V_2}$ when they are perpendicular to each other. Then the angle between these vectors will be $90^{\circ}$.

\[ \overrightarrow{V_1} . \overrightarrow{V_2} = |V_1| |V_2| \cos (90^{\circ}) \]

As,

\[ \cos (90^{\circ}) = 0 \]

\[ \overrightarrow{V_1} . \overrightarrow{V_2} = 0 \]

The dot product of two perpendicular vectors is zero.

c) The dot product of $\overrightarrow{V_1}$ with $\overrightarrow{V_2}$ when they are parallel to each other. Then the angle between these two vectors will be zero.

\[ \overrightarrow{V_1} . \overrightarrow{V_2} = |V_1| |V_2| \cos (0^{\circ}) \]

\[ \overrightarrow{V_1} . \overrightarrow{V_2} = (V_1) (V_2) 1 \]

\[ \overrightarrow{V_1} . \overrightarrow{V_2} = V_1 V_2 \]

The dot product of two parallel vectors is the product of their magnitudes.

## Numerical Results:

The dot product of a vector with itself gives its magnitude squared.

\[ \overrightarrow{V_1} . \overrightarrow{V_1} = V_1^{2} \]

The dot product of two perpendicular vectors gives out zero.

\[ \overrightarrow{V_1} . \overrightarrow{V_2} = 0 \]

The dot product of two parallel vectors provides the product of the magnitudes of those vectors.

\[ \overrightarrow{V_1} . \overrightarrow{V_2} = V_1 V_2 \]

## Example:

We have $\overrightarrow{V_1}$ and $\overrightarrow{V_2}$ with magnitude $4$ and $6$, respectively. The angle between these two vectors is $45^{\circ}$.

The dot product between $\overrightarrow{V_1}$ and $\overrightarrow{V_2}$ is given by:

\[ |V_1| = 4 \]

\[ |V_2| = 6 \]

\[ \overrightarrow{V_1} . \overrightarrow{V_2} = |V_1| |V_2| \cos (\theta) \]

By substituting the values, we get:

\[ \overrightarrow{V_1} . \overrightarrow{V_2} = (4) (6) \cos 45^{\circ} \]

\[ \overrightarrow{V_1} . \overrightarrow{V_2} = 24 (0.707) \]

\[ \overrightarrow{V_1} . \overrightarrow{V_2} = 16.97 \text{units}^{2} \]

Image/Mathematical Drawings are created with Geogebra.