Final answer:
The magnitude of the vector \(\vec{v} = \langle -9, 40 \rangle\) is 41.23. The direction (angle) is 282.56°, which is computed by adding 180° to the arctan of y/x because the vector lies in the second quadrant.
Step-by-step explanation:
To find the magnitude and direction of the vector \(\vec{v} = \langle -9, 40 \rangle\), we first calculate the magnitude (also known as the length or norm of the vector), which is given by the formula:
\[\text{Magnitude} = \sqrt{-9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41.23\]
Next, we determine the direction (also called the angle) of the vector. This angle \(\theta\) can be found with respect to the positive x-axis using the arctan function (inverse tangent, also labeled as tan-1):
\[\theta = \text{tan}^{-1}\left(\frac{40}{-9}\right)\]
However, since the vector points to the second quadrant (where the x-component is negative and the y-component is positive), we must add 180° to get the correct direction:
\[\theta_{\text{corrected}} = \text{tan}^{-1}\left(\frac{40}{-9}\right) + 180°\]
By performing this calculation, we find:
\[\theta_{\text{corrected}} = \text{tan}^{-1}\left(\frac{40}{-9}\right) + 180° = 102.56° + 180° = 282.56°\]
Therefore, the correct option with the magnitude and direction of vector \(\vec{v}\) is:
- Magnitude: 41.23
- Direction: 282.56°