Problem 80 - Precalculus

Question

Assume that the angle between \(\mathbf{u}\) and \(\mathbf{v}\) is \(\theta=\frac{\pi}{3} .\) Show that \(\frac{(\mathbf{u} \cdot \mathbf{v}) \mathbf{u}}{|\mathbf{v}|}-\frac{(\mathbf{v} \cdot \mathbf{u}) \mathbf{v}}{|\mathbf{u}|}=\frac{|\mathbf{u}| \mathbf{u}-|\mathbf{v}| \mathbf{v}}{2}\).

Step-by-Step Solution

Answer

The vector expression simplifies correctly to the given form.

Step 1: Understand the Problem

We need to show that the given vector expression simplifies to the right-hand side. The vectors \( \mathbf{u} \) and \( \mathbf{v} \) have an angle \( \theta = \frac{\pi}{3} \) between them.

Step 2: Use Dot Product Formula

The dot product between two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is given by \( \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos \theta \). Since \( \theta = \frac{\pi}{3} \), we have \( \cos \theta = \frac{1}{2} \). Thus, \( \mathbf{u} \cdot \mathbf{v} = \frac{1}{2} |\mathbf{u}||\mathbf{v}| \).

Step 3: Calculate Each Expression

Substitute the dot product into both expressions:1. \( \frac{(\mathbf{u} \cdot \mathbf{v}) \mathbf{u}}{|\mathbf{v}|} = \frac{\frac{1}{2} |\mathbf{u}||\mathbf{v}| \mathbf{u}}{|\mathbf{v}|} = \frac{1}{2}|\mathbf{u}|\mathbf{u} \).2. \( \frac{(\mathbf{v} \cdot \mathbf{u}) \mathbf{v}}{|\mathbf{u}|} = \frac{\frac{1}{2} |\mathbf{u}||\mathbf{v}| \mathbf{v}}{|\mathbf{u}|} = \frac{1}{2}|\mathbf{v}|\mathbf{v} \).

Step 4: Combine the Expressions

Substitute the calculated expressions back into the main equation:\[ \frac{1}{2}|\mathbf{u}|\mathbf{u} - \frac{1}{2}|\mathbf{v}|\mathbf{v} = \frac{|\mathbf{u}|\mathbf{u} - |\mathbf{v}|\mathbf{v}}{2} \].

Step 5: Verify the Simplification

Simplify the left side: \( \frac{1}{2}|\mathbf{u}|\mathbf{u} - \frac{1}{2}|\mathbf{v}|\mathbf{v} \) can be directly rewritten as \( \frac{|\mathbf{u}|\mathbf{u} - |\mathbf{v}|\mathbf{v}}{2} \), which matches with the right side of the equation we need to show.

Key Concepts

Dot Product

The dot product is a fundamental concept in vector algebra. It helps in finding relations between two vectors by linking their magnitudes with the cosine of the angle between them. For vectors \( \mathbf{u} \) and \( \mathbf{v} \), the formula is expressed as:

• \( \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos \theta \)

Here, \( \theta \) is the angle between the vectors \( \mathbf{u} \) and \( \mathbf{v} \).
The dot product results in a scalar value, which sheds light on how closely the vectors point in the same direction.
In the given problem, the angle \( \theta \) is \( \frac{\pi}{3} \), which simplifies to \( \cos \theta = \frac{1}{2} \). Therefore, the dot product ends up being \( \frac{1}{2} |\mathbf{u}||\mathbf{v}| \).
Understanding this concept helps solve the expression by simplifying the various terms in the exercise.

Angle Between Vectors

The angle between vectors is crucial when understanding how vectors interact with each other in space. This concept is harnessed using the formula for the dot product.
The relationship between two vectors \( \mathbf{u} \) and \( \mathbf{v} \) can be captured by:

• \( \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|} \)

This formula lets us find the angle \( \theta \), which reflects the orientation process of one vector relative to the other. An important aspect of learning about the angle between vectors is recognizing how different angles represent various types of vector relationships:

• \( \theta = 0 \): Vectors are parallel and point in the same direction.
• \( \theta = \frac{\pi}{2} \): Vectors are perpendicular.
• \( \theta = \pi \): Vectors are parallel and point in opposite directions.

In this specific exercise, the angle is \( \frac{\pi}{3} \), resulting in \( \cos \theta = \frac{1}{2} \), a key step in simplifying vector expressions.

Vector Expression Simplification

Simplifying vector expressions allows for an easier manipulation of vector quantities. In the given problem, the expression takes the form:

• \( \frac{(\mathbf{u} \cdot \mathbf{v}) \mathbf{u}}{|\mathbf{v}|} - \frac{(\mathbf{v} \cdot \mathbf{u}) \mathbf{v}}{|\mathbf{u}|} \)

By substituting the calculated dot product into this expression, each term simplifies:

• \( \frac{1}{2}|\mathbf{u}|\mathbf{u} \)
• \( \frac{1}{2}|\mathbf{v}|\mathbf{v} \)

The simplification proceeds by taking into account the algebraic properties they both share. The resulting form of the equation:

• \( \frac{1}{2}|\mathbf{u}|\mathbf{u} - \frac{1}{2}|\mathbf{v}|\mathbf{v} = \frac{|\mathbf{u}|\mathbf{u} - |\mathbf{v}|\mathbf{v}}{2} \)

It allows for a straightforward simplification by factoring out the common divisor. By approaching the exercise this way, students gain a deeper understanding of vector manipulation and simplification—core elements of solving vector algebra problems.