Convert u=-4(cos60degressi+sin60degressj) to component form

To convert the vector \( \mathbf{u} = -4(\cos 60^\circ \mathbf{i} + \sin 60^\circ \mathbf{j}) \) to its component form, we'll follow these steps:

### Step 1: Calculate \( \cos 60^\circ \) and \( \sin 60^\circ \)
- \( \cos 60^\circ = \frac{1}{2} \)
- \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)

### Step 2: Substitute these values into the vector equation
\[ \mathbf{u} = -4\left(\frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j}\right) \]

### Step 3: Distribute the scalar \(-4\) to each component
\[ \mathbf{u} = -4 \times \frac{1}{2} \mathbf{i} + -4 \times \frac{\sqrt{3}}{2} \mathbf{j} \]
\[ \mathbf{u} = -2 \mathbf{i} - 2\sqrt{3} \mathbf{j} \]

### Final Answer
The component form of the vector \( \mathbf{u} \) is:
\[ \mathbf{u} = \langle -2, -2\sqrt{3} \rangle \]

Would you like any further details or have any questions?

Here are some related questions:
1. How do you find the magnitude of the vector \( \mathbf{u} \)?
2. What is the direction angle of the vector \( \mathbf{u} \)?
3. How do you convert a vector from component form back to unit vector form?
4. How does the scalar multiplication affect the direction of the vector?
5. Can you express \( \mathbf{u} \) in polar coordinates?

**Tip:** Remember, multiplying a vector by a scalar changes its magnitude but not its direction unless the scalar is negative, which reverses the direction.

Learn how to convert the vector \( \mathbf{u} = -4(\cos 60^\circ \mathbf{i} + \sin 60^\circ \mathbf{j}) \) to component form with a detailed step-by-step solution. This problem involves vector components and trigonometric functions, specifically cosine and sine functions. Suitable for high school students studying vectors and trigonometry.

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