You are not right. Remember ##~\text{tan} =\dfrac{\text{opposite}}{\text{adjacent}}.##Ineedhelpwithphysics said:
To find the magnitude of a vector, you use the Pythagorean theorem. For a vector \( \mathbf{v} = \langle x, y \rangle \) in 2D space, the magnitude is given by \( |\mathbf{v}| = \sqrt{x^2 + y^2} \). For a vector \( \mathbf{v} = \langle x, y, z \rangle \) in 3D space, the magnitude is \( |\mathbf{v}| = \sqrt{x^2 + y^2 + z^2} \).
The direction angle \( \theta \) of a vector \( \mathbf{v} = \langle x, y \rangle \) in 2D can be found using the arctangent function: \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \). Make sure to consider the quadrant in which the vector lies to determine the correct angle.
In 3D, the direction angles \( \alpha, \beta, \) and \( \gamma \) with respect to the x, y, and z axes can be found using the dot product. For a vector \( \mathbf{v} = \langle x, y, z \rangle \), the direction cosines are \( \cos(\alpha) = \frac{x}{|\mathbf{v}|} \), \( \cos(\beta) = \frac{y}{|\mathbf{v}|} \), and \( \cos(\gamma) = \frac{z}{|\mathbf{v}|} \). The angles are then \( \alpha = \cos^{-1}\left(\frac{x}{|\mathbf{v}|}\right) \), \( \beta = \cos^{-1}\left(\frac{y}{|\mathbf{v}|}\right) \), and \( \gamma = \cos^{-1}\left(\frac{z}{|\mathbf{v}|}\right) \).
You can use a scientific calculator or software tools like MATLAB, Python (with libraries such as NumPy), or online vector calculators. These tools can perform the necessary arithmetic and trigonometric calculations to find both the magnitude and direction of a vector.
To verify your calculations, you can double-check each step. For magnitude, ensure
