Problem 75 - Precalculus with Calculus Previews

Question

Verify that the vector \(\mathbf{w}=\mathbf{v}-\frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{u}|^{2}} \mathbf{u}\) is orthogonal to the vector \(\mathbf{u}\).

Step-by-Step Solution

Answer

The vector \( \mathbf{w} \) is orthogonal to \( \mathbf{u} \).

Step 1: Understand the Problem

We need to prove that the vector \( \mathbf{w} = \mathbf{v} - \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{u}|^{2}} \mathbf{u} \) is orthogonal to \( \mathbf{u} \). This means that the dot product of \( \mathbf{w} \) and \( \mathbf{u} \) should be zero.

Step 2: Set Up the Dot Product

Calculate the dot product \( \mathbf{w} \cdot \mathbf{u} = \left( \mathbf{v} - \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{u}|^{2}} \mathbf{u} \right) \cdot \mathbf{u} \). This expands to \( \mathbf{v} \cdot \mathbf{u} - \left( \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{u}|^{2}} \mathbf{u} \right) \cdot \mathbf{u} \).

Step 3: Calculate the Components

Start by evaluating the term \( \left( \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{u}|^{2}} \mathbf{u} \right) \cdot \mathbf{u} \). This is equivalent to \( \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{u}|^{2}} \left( \mathbf{u} \cdot \mathbf{u} \right) \).

Step 4: Simplify the Dot Product

Since \( \mathbf{u} \cdot \mathbf{u} = |\mathbf{u}|^2 \), substitute it back into the expression from Step 3. The term becomes \( \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{u}|^{2}} |\mathbf{u}|^{2} = \mathbf{v} \cdot \mathbf{u} \).

Step 5: Conclude the Orthogonality

Substitute the result from Step 4 back into the expression from Step 2: \( \mathbf{v} \cdot \mathbf{u} - \mathbf{v} \cdot \mathbf{u} = 0 \). Since the result of the dot product is zero, \( \mathbf{w} \) is orthogonal to \( \mathbf{u} \).

Key Concepts

Dot Product

The dot product is a key operation in vector mathematics. It helps determine the relationship between two vectors, such as their angle, length, and orthogonality. You compute the dot product of vectors \( \mathbf{a} \) and \( \mathbf{b} \) by multiplying their corresponding components and summing up the results. Mathematically, the dot product is represented as:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + \, ... \, + a_nb_n \)

The dot product reflects the cosine of the angle between vectors when they are non-zero. Specifically, two vectors are orthogonal (i.e., perpendicular) if their dot product equals zero. This happens because cosine of 90 degrees is zero, resulting in:

• \( \mathbf{a} \cdot \mathbf{b} = 0 \)

Understanding the dot product is crucial for proving vector relationships. For this exercise, we use it to show that a certain vector \( \mathbf{w} \) is orthogonal to \( \mathbf{u} \) by demonstrating that their dot product is zero.

Vector Subtraction

Vector subtraction is a simple yet important operation when working with vectors. It allows you to find the difference between two vectors. Given vectors \( \mathbf{a} \) and \( \mathbf{b} \), the subtraction \( \mathbf{a} - \mathbf{b} \) is performed by subtracting each component of \( \mathbf{b} \) from the corresponding component of \( \mathbf{a} \). Mathematically, this looks like:

• \( \mathbf{a} - \mathbf{b} = (a_1-b_1, a_2-b_2, ..., a_n-b_n) \)

Vector subtraction is useful in finding the vector that points from one vector to another. In the given exercise, vector subtraction helps define vector \( \mathbf{w} \) as the result of subtracting a scaled projection of vector \( \mathbf{u} \) from vector \( \mathbf{v} \). This operation is vital to forming the vector \( \mathbf{w} \), which allows us to then check its orthogonality to \( \mathbf{u} \) using a dot product.

Vector Magnitude

The magnitude of a vector provides information about its length. It is the measure of how long a vector is in space, irrespective of its direction. For a vector \( \mathbf{a} \) with components \( (a_1, a_2, ..., a_n) \), its magnitude \( |\mathbf{a}| \) is calculated as:

• \(||\mathbf{a}|| = \sqrt{a_1^2 + a_2^2 + \, ... \, + a_n^2} \)

The magnitude helps in normalizing vectors and is also used in calculating the dot product when determining vector orthogonality. In this exercise, vector magnitude is essential to calculate the scale factor for the projection of \( \mathbf{v} \) onto \( \mathbf{u} \). We use this magnitude to adjust the vector \( \mathbf{u} \) so its length fits perfectly into our orthogonality formula, ensuring that when subtracted from \( \mathbf{v} \), it aids in the construction of \( \mathbf{w} \). Calculating the correct magnitude is necessary to simplify expressions mathematically and achieve the desired outcomes like proving two vectors' orthogonality.