This section, explores the fundamental concepts of vector magnitudes and direction, crucial for understanding the spatial dynamics in mathematics and physics. We delve into the calculation of vector lengths in both two and three dimensions, the role of unit vectors, and the practical applications of these concepts in various fields.

## Vector Magnitude in 2D

### Definition

The magnitude of a 2D vector, represented as $\mathbf{v} = x\mathbf{i} + y\mathbf{j}$, is essentially the length of the vector. It is calculated using the formula:

$|\mathbf{v}| = \sqrt{x^2 + y^2}$This formula is derived from the Pythagorean theorem.

Image courtesy of Third Space Learning

## Vector Magnitude in 3D

#### Definition

The magnitude of a 3D vector, represented as $\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$, is calculated using the formula:

$|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}$This extends the Pythagorean theorem to three dimensions.

Image courtesy of Online Math Learning

## Examples

#### Example 1: Calculate the Magnitude of $6\mathbf{i} - 8\mathbf{j}$

#### Solution:

1. Identify the Vector Components: Here, the vector components are $x = 6$and $y = -8$ .

2. Apply the Magnitude Formula: The magnitude of a 2D vector is calculated using the formula $|\mathbf{v}| = \sqrt{x^2 + y^2}$.

3. Substitute the Values and Calculate: $|\mathbf{v}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

So, the magnitude of the vector $6\mathbf{i} - 8\mathbf{j}$ is 10. This represents the straight-line distance in a plane from the origin to the point (6, -8).

#### Example 2: Find the Unit Vector in the Direction of $4\mathbf{i} + 3\mathbf{j}$

#### Solution:

1. Calculate the Magnitude of the Given Vector:

$\text{Magnitude} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$2. Divide Each Component by the Magnitude: To find the unit vector, divide each component of the vector by its magnitude.

3. Compute the Unit Vector:

$\mathbf{u} = \frac{4\mathbf{i} + 3\mathbf{j}}{5} = \frac{4}{5}\mathbf{i} + \frac{3}{5}\mathbf{j}$The unit vector in the direction of is $\frac{4}{5}\mathbf{i} + \frac{3}{5}\mathbf{j}$. This vector points in the same direction as the original vector but has a standard length of 1.