SOLUTION: Let vectors u and v form an angle θ when placed tail-to-tail, and let w be the vector projection of v onto u. (a)Assume that θ is acute. Notice that w points in the same directi

Question 1163824: Let vectors u and v form an angle θ when placed tail-to-tail, and let w be the vector projection of v onto u.
(a)Assume that θ is acute. Notice that w points in the same direction as u. Find |w|, then
show that w = |v| u•v 1 u, which simplifies to just w = u•v u. |u||v| |u| u•u
(b)If θ is obtuse, do w and u point in the same direction? Does formula (a) still work? By the way, the notation projuv is sometimes used for the vector projection of v onto u.
Answer by CPhill(2264) (Show Source): You can put this solution on YOUR website!
### (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:
$$\mathbf{w} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{u} \cdot \mathbf{u}} \right) \mathbf{u}$$
---
### (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).
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