Problem 60 - Precalculus with Limits

Question

In Exercises 59-62, find the projection of \(\mathbf{u}\) onto \(\mathbf{v}\). Then write \(\mathbf{u}\) as the sum of two orthogonal vectors, one of which is proj\(_{\mathbf{v}} \mathbf{u}\). \(\mathbf{u} = \langle 4, 2 \rangle\) \(\mathbf{v} = \langle 1, -2 \rangle\)

Step-by-Step Solution

Answer

The projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is \( \langle 0, 0 \rangle \) and \( \mathbf{u} \) can be written as the sum of two orthogonal vectors such that \( \mathbf{u} = proj_{\mathbf{v}} \mathbf{u} + \mathbf{w} = \langle 0, 0 \rangle + \langle 4, 2 \rangle = \langle 4, 2 \rangle \).

Step 1: Calculate the Projection of \( \mathbf{u} \) onto \( \mathbf{v} \)

The formula used to calculate the projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is: \[ proj_{\mathbf{v}} \mathbf{u} = \frac {\mathbf{u} \cdot \mathbf{v}} {\mathbf{v} \cdot \mathbf{v}} \mathbf{v} \] where \( \cdot \) denotes the dot product. First, find the dot products \( \mathbf{u} \cdot \mathbf{v} \) and \( \mathbf{v} \cdot \mathbf{v} \). Plugging in \( \mathbf{u} = \langle 4, 2 \rangle \) and \( \mathbf{v} = \langle 1, -2 \rangle \) gives \( \mathbf{u} \cdot \mathbf{v} = 4 \cdot 1 + 2 \cdot (-2) = 0 \) and \( \mathbf{v} \cdot \mathbf{v} = 1 \cdot 1 + (-2) \cdot (-2) = 5 \). Thus the projection is \( \frac {0} {5} \mathbf{v} = \langle 0, 0 \rangle \).

Step 2: Write \( \mathbf{u} \) as the Sum of Two Orthogonal Vectors

To write \( \mathbf{u} \) as the sum of two orthogonal vectors, one can simply subtract the projection from \( \mathbf{u} \) to get the second component (assuming the projection to be one of the two vectors). This vector will be orthogonal to \( \mathbf{v} \) because of the nature of the projection operation. The second vector \( \mathbf{w} \) can be obtained by: \[ \mathbf{w} = \mathbf{u} - proj_{\mathbf{v}} \mathbf{u} \] Here, \( \mathbf{u} = \langle 4, 2 \rangle \) and \( proj_{\mathbf{v}} \mathbf{u} = \langle 0, 0 \rangle \). Hence \( \mathbf{w} = \langle 4, 2 \rangle - \langle 0, 0 \rangle = \langle 4, 2 \rangle \). Now, \( \mathbf{u} \) is represented as the sum of two orthogonal vectors \( proj_{\mathbf{v}} \mathbf{u} \) and \( \mathbf{w} \), specifically, \( \mathbf{u} = proj_{\mathbf{v}} \mathbf{u} + \mathbf{w} = \langle 0, 0 \rangle + \langle 4, 2 \rangle = \langle 4, 2 \rangle \). As a check, \( proj_{\mathbf{v}} \mathbf{u} \cdot \mathbf{w} = \langle 0, 0 \rangle \cdot \langle 4, 2 \rangle = 0 \cdot 4 + 0 \cdot 2 = 0 \), which indicates that these two vectors are indeed orthogonal as their dot product is zero.

Key Concepts

Dot Product

The dot product is a fundamental operation in vector mathematics. It finds way into various calculations, including projections. To compute the dot product of two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), use the formula: \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \). This operation yields a single scalar value. The dot product can indicate the directional relationship of vectors:
- If it's positive, vectors often form an acute angle.
- If it's zero, vectors are orthogonal.
- If it's negative, vectors form an obtuse angle. In our exercise, the dot product \( \mathbf{u} \cdot \mathbf{v} \) was 0. This initially signals that the vectors are orthogonal, a key fact used later in creating a sum of orthogonal vectors.

Orthogonal Vectors

Orthogonal vectors might sound fancy, but it's simple at its core. Two vectors are orthogonal—or perpendicular—if their dot product is zero. This means they form a right angle with each other.
In the given solution, we found that \( proj_{\mathbf{v}} \mathbf{u} \) and \( \mathbf{w} = \mathbf{u} \) are orthogonal because their dot product is \( 0 \). This aligns with the property that

• If the dot product of two non-zero vectors is zero, the vectors are orthogonal.

Understanding orthogonality is very useful, especially in decomposing a vector into components, such as in engineering or physics applications, where one wants to separate forces or velocities into independent directions.

Vector Operations

Vector operations encompass a variety of mathematical computations done with or between vectors. Some common operations include addition, subtraction, and scaling by a scalar.
- **Addition and Subtraction**: For vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), addition implies \( \mathbf{a} + \mathbf{b} = \langle a_1 + b_1, a_2 + b_2 \rangle \) and subtraction yields \( \mathbf{a} - \mathbf{b} = \langle a_1 - b_1, a_2 - b_2 \rangle \).
- **Scalar Multiplication**: Involves multiplying each component of a vector by a scalar (constant). For instance, multiplying \( \mathbf{a} \) by \( c \) gives \( c \mathbf{a} = \langle ca_1, ca_2 \rangle \).
In the step-by-step solution, subtraction is used in decomposing vector \( \mathbf{u} \) into two orthogonal vectors: \( \mathbf{u} = proj_{\mathbf{v}} \mathbf{u} + \mathbf{w} \). Subtraction offered the orthogonal complement by \( \mathbf{w} = \mathbf{u} - proj_{\mathbf{v}} \mathbf{u} \), accurately represented by operations \( \langle 4, 2 \rangle - \langle 0, 0 \rangle = \langle 4, 2 \rangle \). Each type of operation helps solve different geometric and algebraic problems.