Find the resultant of two vectors 3 units and 4 units acting at a point at an angle of 45° with each other

1 answer

To find the resultant of two vectors acting at an angle to each other, we can use the law of cosines. Given two vectors \( A \) and \( B \) with magnitudes \( |A| = 3 \) units and \( |B| = 4 \) units, and the angle \( \theta = 45^\circ \) between them, the formula for the magnitude of the resultant vector \( R \) is:

\[ R = \sqrt{A^2 + B^2 + 2AB \cos(\theta)} \]

Substituting the known values:

  • \( A = 3 \)
  • \( B = 4 \)
  • \( \theta = 45^\circ \)
  • \( \cos(45^\circ) = \frac{\sqrt{2}}{2} \)

Now, substituting into the formula:

\[ R = \sqrt{3^2 + 4^2 + 2 \cdot 3 \cdot 4 \cdot \frac{\sqrt{2}}{2}} \]

Calculating each term:

  1. \( 3^2 = 9 \)
  2. \( 4^2 = 16 \)
  3. \( 2 \cdot 3 \cdot 4 \cdot \frac{\sqrt{2}}{2} = 12\sqrt{2} \)

So, we have:

\[ R = \sqrt{9 + 16 + 12\sqrt{2}} \]

This simplifies to:

\[ R = \sqrt{25 + 12\sqrt{2}} \]

To find the numerical value, we compute \( 12\sqrt{2} \) (approximately \( 12 \times 1.414 \approx 16.97 \)):

\[ R \approx \sqrt{25 + 16.97} \approx \sqrt{41.97} \]

Now, calculating the square root:

\[ R \approx 6.46 \text{ units} \]

Thus, the resultant vector has a magnitude of approximately \( 6.46 \) units.