Find \( \mathbf{u} \cdot \mathbf{v} \), given the magnitudes and the angle \( \theta \) be- tween the vectors. 3. \( |\mathbf{u}|=2,|\mathbf{v}|=3, \theta=135^{\circ} \) 4. \( |\mathbf{u}|=10,|\mathbf{v}|=5, \theta=180^{\circ} \)

To find the dot product \( \mathbf{u} \cdot \mathbf{v} \) given the magnitudes of the vectors and the angle between them, we use the formula: \[ \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| \cdot |\mathbf{v}| \cdot \cos(\theta) \] Let's apply this formula to both problems: --- **3. Given:** - \( |\mathbf{u}| = 2 \) - \( |\mathbf{v}| = 3 \) - \( \theta = 135^\circ \) **Solution:** \[ \mathbf{u} \cdot \mathbf{v} = 2 \times 3 \times \cos(135^\circ) \] We know that: \[ \cos(135^\circ) = -\frac{\sqrt{2}}{2} \] Substituting this value: \[ \mathbf{u} \cdot \mathbf{v} = 6 \times \left(-\frac{\sqrt{2}}{2}\right) = -3\sqrt{2} \] --- **4. Given:** - \( |\mathbf{u}| = 10 \) - \( |\mathbf{v}| = 5 \) - \( \theta = 180^\circ \) **Solution:** \[ \mathbf{u} \cdot \mathbf{v} = 10 \times 5 \times \cos(180^\circ) \] We know that: \[ \cos(180^\circ) = -1 \] Substituting this value: \[ \mathbf{u} \cdot \mathbf{v} = 50 \times (-1) = -50 \] --- **Final Answers:** 3. \( \mathbf{u} \cdot \mathbf{v} = -3\sqrt{2} \) 4. \( \mathbf{u} \cdot \mathbf{v} = -50 \)