Problem 61 - 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\(_{\mathbf{v}} \mathbf{u}\). \(\mathbf{u} = \langle 0, 3 \rangle\) \(\mathbf{v} = \langle 2, 15 \rangle\)

Step-by-Step Solution

Answer

The projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is \( \langle \frac{90}{229}, \frac{675}{229} \rangle \). \( \mathbf{u} \) can be written as the sum of \( \langle \frac{90}{229}, \frac{675}{229} \rangle \) and \( \langle -\frac{90}{229}, \frac{8}{229} \rangle \), which are orthogonal vectors.

Step 1: Compute Projection of u onto v

The formula to calculate the projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is given by proj\(_{\mathbf{v}} \mathbf{u}\) = \( \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{v} \|^{2}} \mathbf{v} \). First, compute the dot product \( \mathbf{u} \cdot \mathbf{v} \), which is \(0 \times 2 + 3 \times 15 = 45\). Then, calculate the magnitude of vector \( \mathbf{v} \), \(\|\mathbf{v}\|\), which equals \( \sqrt{(2)^2 + (15)^2} = \sqrt{229} \). The square of the magnitude, \( \| \mathbf{v} \|^{2} \), is then \( 229 \). So, proj\(_{\mathbf{v}} \mathbf{u}\) = \( \frac{45}{229} \times \langle 2, 15 \rangle \) = \( \langle \frac{90}{229}, \frac{675}{229} \rangle \).

Step 2: Write \( \mathbf{u} \) as the sum of two orthogonal vectors

Once we have proj\(_{\mathbf{v}} \mathbf{u}\), we can write \( \mathbf{u} \) as the sum of proj\(_{\mathbf{v}} \mathbf{u}\) and a second vector that is orthogonal to \( \mathbf{v} \). This second vector can be calculated as \( \mathbf{u} \) - proj\(_{\mathbf{v}} \mathbf{u}\). So \( \mathbf{u} = \langle 0,3 \rangle = \langle \frac{90}{229}, \frac{675}{229} \rangle + \langle 0- \frac{90}{229}, 3-\frac{675}{229} \rangle = \langle \frac{90}{229}, \frac{675}{229} \rangle + \langle -\frac{90}{229}, \frac{8}{229} \rangle \).

Key Concepts

Orthogonal Vectors

Orthogonal vectors play a fundamental role in vector mathematics. They are vectors that are at right angles to each other. If two vectors are orthogonal, their dot product is zero, which signifies no overlap in direction. This right angle relationship makes them independent in terms of vector space, meaning one cannot be expressed as a multiple of the other.
In the context of vector projection, where a vector \( \mathbf{u} \) is expressed as the sum of two orthogonal vectors, one component is the projection itself, and the other is the orthogonal component.

• Finding orthogonal vectors is crucial for operations involving projection and resolving force or direction components.
• For our task, we found \( \mathbf{u} - \text{proj}_{\mathbf{v}} \mathbf{u} \) to determine the orthogonal vector.

Understanding orthogonal vectors helps with applications in physics, computer graphics, and statistics, wherever vectors need to be decomposed into meaningful components.

Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors, resulting in a scalar. It measures the extent to which two vectors point in the same direction. The formula for the dot product of two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \) is given by \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \).
In our exercise, calculating the dot product was the first step to finding the projection of vector \( \mathbf{u} \) onto vector \( \mathbf{v} \). We computed it as:

• The result after multiplying and adding was 45, indicating the degree of alignment between \( \mathbf{u} \) and \( \mathbf{v} \).
• An essential property of the dot product is that it is commutative—that is, \( \mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a} \).

The dot product's role is significant not just for projections, but also in physics for finding work done by a force and in computer graphics for lighting calculations.

Magnitude of a Vector

The magnitude of a vector, often referred to as its length or norm, is the distance from the origin of the vector space to the point represented by the vector. For a vector \( \mathbf{v} = \langle v_1, v_2 \rangle \), its magnitude is calculated using the formula \( \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2} \).
In our example, the magnitude was used to scale the projection of vector \( \mathbf{u} \) onto \( \mathbf{v} \). The steps involved finding:

• The magnitude \( \|\mathbf{v}\| = \sqrt{(2)^2 + (15)^2} = \sqrt{229} \).
• This magnitude was then squared to find \( \|\mathbf{v}\|^2 = 229 \).

Understanding the magnitude of a vector is critical for normalizing vectors, which involves scaling them to have a length of one unit. It is an essential tool in determining the unit direction of vectors, facilitating computations in physics, engineering, and graphics.