Question

Find the magnitude and direction of vector 2 1

Answer 1

To find the magnitude and direction of the vector \( \mathbf{v} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} \), we can use the following formulas:

1. **Magnitude**:
The magnitude \( ||\mathbf{v}|| \) of a 2D vector \( \mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix} \) is calculated using the formula:
\[
||\mathbf{v}|| = \sqrt{x^2 + y^2}
\]
For our vector \( \mathbf{v} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} \):
\[
||\mathbf{v}|| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}
\]

2. **Direction**:
The direction (angle \( \theta \)) of the vector can be found using the tangent function:
\[
\theta = \tan^{-1}\left(\frac{y}{x}\right)
\]
For our vector:
\[
\theta = \tan^{-1}\left(\frac{1}{2}\right)
\]

The angle can be calculated using a calculator to find \( \tan^{-1}(0.5) \). This will give you the angle in radians or degrees, depending on your preference.

Let's summarize the results:
- **Magnitude**: \( ||\mathbf{v}|| = \sqrt{5} \) (approximately \( 2.236 \))
- **Direction**: \( \theta \approx \tan^{-1}(0.5) \) (approximately \( 26.57^\circ \) or \( 0.463 \) radians)

Thus, the magnitude of the vector \( \begin{pmatrix} 2 \\ 1 \end{pmatrix} \) is \( \sqrt{5} \) and the direction is approximately \( 26.57^\circ \).