To find the component form of the sum of vectors $\mathbf{u}$ and $\mathbf{v}$ with the given magnitudes and angles:

Step 1: Write the components of each vector

For a vector in component form given magnitude $|\mathbf{u}|$ and direction angle $\theta$, the components are: $\mathbf{u} = \langle |\mathbf{u}| \cos \theta_u, |\mathbf{u}| \sin \theta_u \rangle$ $\mathbf{v} = \langle |\mathbf{v}| \cos \theta_v, |\mathbf{v}| \sin \theta_v \rangle$

Substituting the given values:

• Magnitude of $\mathbf{u}$: $|\mathbf{u}| = 14$, $\theta_u = 45^\circ$
• Magnitude of $\mathbf{v}$: $|\mathbf{v}| = 30$, $\theta_v = 180^\circ$

Step 2: Compute the components

For $\mathbf{u}$: $\mathbf{u} = \langle 14 \cos 45^\circ, 14 \sin 45^\circ \rangle$ Using $\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}$: $\mathbf{u} = \langle 14 \cdot \frac{\sqrt{2}}{2}, 14 \cdot \frac{\sqrt{2}}{2} \rangle = \langle 7\sqrt{2}, 7\sqrt{2} \rangle$

For $\mathbf{v}$: $\mathbf{v} = \langle 30 \cos 180^\circ, 30 \sin 180^\circ \rangle$ Using $\cos 180^\circ = -1$ and $\sin 180^\circ = 0$: $\mathbf{v} = \langle 30 \cdot (-1), 30 \cdot 0 \rangle = \langle -30, 0 \rangle$

Step 3: Add the components of $\mathbf{u}$ and $\mathbf{v}$

$\mathbf{u} + \mathbf{v} = \langle 7\sqrt{2} + (-30), 7\sqrt{2} + 0 \rangle$ Simplify: $\mathbf{u} + \mathbf{v} = \langle 7\sqrt{2} - 30, 7\sqrt{2} \rangle$

Final Answer:

The component form of $\mathbf{u} + \mathbf{v}$ is: $\mathbf{u} + \mathbf{v} = \langle 7\sqrt{2} - 30, 7\sqrt{2} \rangle$

Let me know if you want further clarification or numerical approximations! Here are five related questions to deepen your understanding:

1. How do we determine the magnitude of the resulting vector $\mathbf{u} + \mathbf{v}$?
2. What are the steps to calculate the direction angle of $\mathbf{u} + \mathbf{v}$?
3. How would the solution change if $\theta_v$ were a different angle, such as $90^\circ$?
4. What role does trigonometric identity play in vector decomposition?
5. Can you derive the component form for three vectors instead of two?