Problem 28 - APEX Calculus

Question

Vectors \(\vec{u}\) and \(\vec{v}\) are given. Write \(\vec{u}\) as the sum of two vectors, one of which is parallel to \(\vec{v}\) and one of which is perpendicular to \(\vec{v}\). Note: these are the same pairs of vectors as found in Exercises 21-26. \(\vec{u}=\langle 5,5\rangle, \vec{v}=\langle 1,3\rangle\)

Step-by-Step Solution

Answer

\( \vec{u} = \langle 2, 6 \rangle + \langle 3, -1 \rangle \).

Step 1: Determine the parallel component

The vector parallel to \( \vec{v} \) has the form \( k\vec{v} \), where \( k \) is a scalar. To find this, use the projection formula: \( \text{proj}_{\vec{v}} \vec{u} = \left( \frac{\vec{u} \cdot \vec{v}}{\vec{v} \cdot \vec{v}} \right) \vec{v} \).Calculate \( \vec{u} \cdot \vec{v} = 5 \times 1 + 5 \times 3 = 20 \).Calculate \( \vec{v} \cdot \vec{v} = 1^2 + 3^2 = 10 \).Thus, \( \text{proj}_{\vec{v}} \vec{u} = \left( \frac{20}{10} \right) \langle 1, 3 \rangle = 2\langle 1, 3 \rangle = \langle 2, 6 \rangle \).

Step 2: Find the perpendicular component

The vector perpendicular to \( \vec{v} \) is \( \vec{u} - \text{proj}_{\vec{v}} \vec{u} \).Calculate \( \vec{u} - \text{proj}_{\vec{v}} \vec{u} = \langle 5, 5 \rangle - \langle 2, 6 \rangle \).This gives \( \langle 5 - 2, 5 - 6 \rangle = \langle 3, -1 \rangle \).

Step 3: Express \( \vec{u} \) as a sum of two components

Now express \( \vec{u} = \langle 5, 5 \rangle \) as the sum of the parallel and perpendicular components:\( \vec{u} = \langle 2, 6 \rangle + \langle 3, -1 \rangle \).

Key Concepts

Parallel Component

When we talk about the parallel component of a vector, we're referring to the part of one vector that runs right alongside another vector. Imagine two arrows lying on a flat surface, one pointing in the exact same direction as the other. That's what we want to find.

To find the parallel component of vector \( \vec{u} \) relative to vector \( \vec{v} \), we use the concept of vector projection. The formula for this looks a bit fancy: \[\text{proj}_{\vec{v}} \vec{u} = \left( \frac{\vec{u} \cdot \vec{v}}{\vec{v} \cdot \vec{v}} \right) \vec{v}\]Here's what each part means:

• \( \vec{u} \cdot \vec{v} \) is the dot product, a special way to multiply vectors.
• \( \vec{v} \cdot \vec{v} \) is the dot product of \( \vec{v} \) with itself, which gives us a number representing "how long" the vector is.
• \( \vec{v} \) at the end scales this whole expression to stretch it in the direction of \( \vec{v} \).

After calculating, we find that the parallel component of \( \vec{u} = \langle 5,5 \rangle \) with respect to \( \vec{v} = \langle 1,3 \rangle \) is \( \langle 2, 6 \rangle \). This means if you look at \( \vec{u} \) in the direction of \( \vec{v} \), it follows the exact path of a vector stretching from the origin to \( \langle 2, 6 \rangle \).

Perpendicular Component

The perpendicular component is like the twin brother of the parallel component, but instead of going in the same direction, it stretches out at a 90-degree angle to the related vector. Finding this part helps to completely understand the behavior of the original vector in relation to another.

Once we find the parallel part (\( \text{proj}_{\vec{v}} \vec{u} \)), the perpendicular component is simply what's left over. We calculate it by subtracting the parallel component from the original vector \( \vec{u} \). In our example, this calculation looks like:\[\vec{u} - \text{proj}_{\vec{v}} \vec{u} = \langle 5, 5 \rangle - \langle 2, 6 \rangle = \langle 3, -1 \rangle\]This shows that \( \vec{u} \) leans slightly forward and down, compared to \( \vec{v} \), after accounting for the part that aligns with \( \vec{v} \). The result \( \langle 3, -1 \rangle \) reflects the unique path of \( \vec{u} \) that's at a right angle to \( \vec{v} \).

Vector Projection

Vector projection is a fundamental concept that helps us break down vectors into simpler pieces. Imagine shining a flashlight directly above one vector onto another; the shadow it casts gives us the projected vector as a slice along the surface we are considering.

Mathematically, projection simplifies to the formula we used for finding the parallel component:

• \(\text{proj}_{\vec{v}} \vec{u} = \left( \frac{\vec{u} \cdot \vec{v}}{\vec{v} \cdot \vec{v}} \right) \vec{v} \)

This projects \( \vec{u} \) along \( \vec{v} \), telling us how much of \( \vec{u} \) points in the same direction as \( \vec{v} \). It's like asking, "If I travel the path of \( \vec{v} \), how much of \( \vec{u} \) am I really tracing?"

This method is incredibly useful, not just for finding parallel vectors but also for simplifying physics problems, detecting signal directions, or even analyzing forces in engineering. The essence of the projection is to make complex vectors easier to analyze by breaking them into parts that are easier to manage or relate to one another.