We need to show that the dot product of a vector with itself is equal to the square of the magnitude of the vector.

1. Problem Description

We need to show that the dot product of a vector with itself is equal to the square of the magnitude of the vector.

2. Solution Steps

Let $\vec{v}$ be a vector in $n$-dimensional space, represented as $\vec{v} = (v_1, v_2, ..., v_n)$.

The dot product of $\vec{v}$ with itself is given by:

$\vec{v} \cdot \vec{v} = v_1^2 + v_2^2 + ... + v_n^2$

The magnitude of the vector $\vec{v}$, denoted as $||\vec{v}||$, is defined as:

$||\vec{v}|| = \sqrt{v_1^2 + v_2^2 + ... + v_n^2}$

The square of the magnitude of $\vec{v}$ is:

$||\vec{v}||^2 = (\sqrt{v_1^2 + v_2^2 + ... + v_n^2})^2 = v_1^2 + v_2^2 + ... + v_n^2$

Comparing the expression for $\vec{v} \cdot \vec{v}$ and $||\vec{v}||^2$, we can see that:

$\vec{v} \cdot \vec{v} = ||\vec{v}||^2$

This shows that the dot product of a vector with itself gives the square of the magnitude of the vector.

3. Final Answer

The dot product of a vector with itself is equal to the square of the magnitude of the vector. This is shown by $\vec{v} \cdot \vec{v} = ||\vec{v}||^2$.