Question 1163824
### (a) Analysis of the Acute Case

When $\theta$ is **acute** ($0 \le \theta < 90^\circ$), the vector projection $\mathbf{w}$ points in the same direction as $\mathbf{u}$.

**1. Finding $|\mathbf{w}|$:**
By forming a right triangle where $\mathbf{v}$ is the hypotenuse and $\mathbf{w}$ is the adjacent side, we use basic trigonometry:
$$|\mathbf{w}| = |\mathbf{v}| \cos \theta$$

**2. Showing the Formula:**
To find the vector $\mathbf{w}$, we multiply its magnitude by a unit vector in the direction of $\mathbf{u}$. The unit vector for $\mathbf{u}$ is $\frac{\mathbf{u}}{|\mathbf{u}|}$.
$$\mathbf{w} = (|\mathbf{v}| \cos \theta) \frac{\mathbf{u}}{|\mathbf{u}|}$$

From the geometric definition of the dot product, $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}| \cos \theta$. We can solve for $\cos \theta$:
$$\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|}$$

Substitute this back into the equation for $\mathbf{w}$:
$$\mathbf{w} = |\mathbf{v}| \left( \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|} \right) \frac{\mathbf{u}}{|\mathbf{u}|}$$

**3. Simplification:**
The $|\mathbf{v}|$ terms cancel out:
$$\mathbf{w} = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}|^2} \mathbf{u}$$

Since $|\mathbf{u}|^2 = \mathbf{u} \cdot \mathbf{u}$, the formula simplifies to:
$$\mathbf{w} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{u} \cdot \mathbf{u}} \right) \mathbf{u}$$

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### (b) Analysis of the Obtuse Case

**1. Do $\mathbf{w}$ and $\mathbf{u}$ point in the same direction?**
**No.** If $\theta$ is **obtuse** ($90^\circ < \theta \le 180^\circ$), the vector $\mathbf{v}$ points "away" from the direction of $\mathbf{u}$. Consequently, the projection $\mathbf{w}$ will point in the **opposite direction** of $\mathbf{u}$. 

**2. Does the formula from (a) still work?**
**Yes, the formula is robust.** Here is why:
* When $\theta$ is obtuse, $\cos \theta$ is **negative**. 
* This makes the dot product $\mathbf{u} \cdot \mathbf{v}$ negative. 
* In the formula $\mathbf{w} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{u} \cdot \mathbf{u}} \right) \mathbf{u}$, the scalar multiplier becomes negative.
* A negative scalar multiplied by vector $\mathbf{u}$ automatically flips its direction, correctly resulting in a vector projection $\mathbf{w}$ that points opposite to $\mathbf{u}$.

Thus, the notation $\text{proj}_{\mathbf{u}}\mathbf{v} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u}$ is universal for any angle $\theta$ (except when $\mathbf{u}$ is the zero vector).