To find the resultant of two vectors acting at an angle to each other, we can use the law of cosines or the parallelogram law. In this case, we have two vectors of magnitude \( A = 3 \) units and \( B = 4 \) units acting at an angle \( \theta = 45° \).
Step 1: Calculate the magnitude of the resultant vector \( R \)
Using the formula:
\[ R = \sqrt{A^2 + B^2 + 2AB \cos(\theta)} \]
Substituting the values:
\[ R = \sqrt{3^2 + 4^2 + 2 \cdot 3 \cdot 4 \cdot \cos(45°)} \] \[ = \sqrt{9 + 16 + 24 \cdot \frac{\sqrt{2}}{2}} \] \[ = \sqrt{25 + 12\sqrt{2}} \]
To simplify: The approximate value of \( \sqrt{2} \) is about \( 1.414 \), so:
\[ = \sqrt{25 + 12 \cdot 1.414} \] \[ \approx \sqrt{25 + 16.968} \approx \sqrt{41.968} \approx 6.46 \text{ units} \]
Step 2: Calculate the direction of the resultant vector
To find the angle \( \phi \) of the resultant vector with respect to the first vector (3 units), we use the formula:
\[ \tan(\phi) = \frac{B \sin(\theta)}{A + B \cos(\theta)} \]
Substituting the values:
\[ \tan(\phi) = \frac{4 \sin(45°)}{3 + 4 \cos(45°)} \] \[ = \frac{4 \cdot \frac{\sqrt{2}}{2}}{3 + 4 \cdot \frac{\sqrt{2}}{2}} \] \[ = \frac{2\sqrt{2}}{3 + 2\sqrt{2}} \]
Now calculating \( 2\sqrt{2} \approx 2 \cdot 1.414 \approx 2.828 \), we substitute:
\[ = \frac{2.828}{3 + 2.828} \] \[ \approx \frac{2.828}{5.828} \approx 0.485 \]
Now, calculate \( \phi \) using the arctangent:
\[ \phi \approx \tan^{-1}(0.485) \approx 26.57° \]
Summary:
- The magnitude of the resultant vector is approximately \( 6.46 \) units.
- The direction of the resultant vector is approximately \( 26.57° \) with respect to the 3 unit vector.