Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find each of the given projections if \(\mathbf{u}=\mathbf{i}+2 \mathbf{j}, \mathbf{v}=2 \mathbf{i}-\mathbf{j}\), and \(\mathbf{w}=\mathbf{i}+5 \mathbf{j}\). \(\operatorname{proj}_{\mathbf{v}} \mathbf{u}\)

Step-by-Step Solution

Answer

The projection \( \operatorname{proj}_{\mathbf{v}} \mathbf{u} = \mathbf{0} \).

Step 1: Formula for Projection

The formula to find the projection of vector \( \mathbf{u} \) onto vector \( \mathbf{v} \) is given by: \[ \operatorname{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} \] This means we will need to calculate the dot products \( \mathbf{u} \cdot \mathbf{v} \) and \( \mathbf{v} \cdot \mathbf{v} \).

Step 2: Calculate \( \mathbf{u} \cdot \mathbf{v} \)

The dot product of \( \mathbf{u} = \mathbf{i} + 2 \mathbf{j} \) and \( \mathbf{v} = 2\mathbf{i} - \mathbf{j} \) is calculated as follows: \[ \mathbf{u} \cdot \mathbf{v} = (1)(2) + (2)(-1) = 2 - 2 = 0 \] This shows that \( \mathbf{u} \) and \( \mathbf{v} \) are orthogonal.

Step 3: Calculate \( \mathbf{v} \cdot \mathbf{v} \)

The dot product of \( \mathbf{v} \) with itself is as follows: \[ \mathbf{v} \cdot \mathbf{v} = (2)(2) + (-1)(-1) = 4 + 1 = 5 \] Thus, \( \mathbf{v} \cdot \mathbf{v} = 5 \).

Step 4: Compute the Projection

Now that we have the dot products, we can find \( \operatorname{proj}_{\mathbf{v}} \mathbf{u} \) using the formula from Step 1: \[ \operatorname{proj}_{\mathbf{v}} \mathbf{u} = \frac{0}{5} \mathbf{v} = 0 \times \mathbf{v} = \mathbf{0} \] Since the dot product in Step 2 was zero, the projection is the zero vector.

Key Concepts

Understanding the Dot Product

Before diving into vector projections, it's important to understand the dot product. The dot product, also known as the scalar product, is a way to multiply two vectors that results in a scalar. Essentially, it tells us to what extent two vectors are pointing in the same direction.
To compute the dot product of two vectors, we multiply corresponding components of each vector and then sum these products. For instance, if you have vectors \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} \) and \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} \), then the dot product \( \mathbf{a} \cdot \mathbf{b} \) is given by:

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

The result will be a single number (a scalar). This scalar can tell us information about the angle between the two vectors.

What are Orthogonal Vectors?

Orthogonal vectors are vectors that are perpendicular to each other. When two vectors are orthogonal, their dot product is zero. This property is not just limited to geometric interpretation but is a definition in spaces with more than two dimensions as well.
In the context of our example with vectors \( \mathbf{u} = \mathbf{i} + 2 \mathbf{j} \) and \( \mathbf{v} = 2 \mathbf{i} - \mathbf{j} \), the dot product calculation yields zero:

• \( \mathbf{u} \cdot \mathbf{v} = 2 - 2 = 0 \)

This verifies that these vectors are orthogonal, meaning they form a right angle with each other.

Projection Formula Explained

The projection of one vector onto another helps illustrate how one vector can be expressed in terms of the other. The formula for projecting vector \( \mathbf{u} \) onto vector \( \mathbf{v} \) is:

• \( \operatorname{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} \)

In this formula, \( \mathbf{u} \cdot \mathbf{v} \) determines how much of \( \mathbf{u} \) is in the direction of \( \mathbf{v} \), and \( \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \) scales \( \mathbf{v} \) to give the exact component of \( \mathbf{u} \) along \( \mathbf{v} \).
In our problem, since \( \mathbf{u} \cdot \mathbf{v} = 0 \), the projection is zero, indicating no component of \( \mathbf{u} \) aligns with \( \mathbf{v} \).

Understanding the Zero Vector

The zero vector is a vector with a magnitude of zero and no specific direction. In any dimension, it could be represented by all coordinates being zero, such as \( \mathbf{0} = 0\mathbf{i} + 0\mathbf{j} \) in two dimensions.
During vector projections, if you end up with a zero vector, it implies that the projected vector essentially "falls flat" onto the baseline it is being projected onto. In our example, since \( \operatorname{proj}_{\mathbf{v}} \mathbf{u} = \mathbf{0} \), it means \( \mathbf{u} \) does not extend in the direction of \( \mathbf{v} \) at all. Therefore, its projection is neither toward nor against \( \mathbf{v} \), just null.
The zero vector often represents vector outcomes when vectors are orthogonal or when projections yield no net component in a given direction.