Problem 59 - 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\(_{\mathbf{v}} \mathbf{u}\). \(\mathbf{u} = \langle 2, 2 \rangle\) \(\mathbf{v} = \langle 6, 1 \rangle\)

Step-by-Step Solution

Answer

The projection of \(\mathbf{u}\) onto \(\mathbf{v}\) is \(\langle 216/37, 14/37 \rangle\), and \(\mathbf{u}\) can be expressed as a sum of this vector and \(\langle 2/37, 60/37 \rangle\), which are orthogonal to each other.

Step 1: Compute the Dot Product of \(\mathbf{u}\) and \(\mathbf{v}\)

The dot product of \(\mathbf{u}\) and \(\mathbf{v}\) can be found by multiplying the corresponding components and sum these products. Here, it will be \(2*6 + 2*1 = 14\).

Step 2: Compute the Magnitude Squared of \(\mathbf{v}\)

This involves taking the square of each component, then sum these squares. For \(\mathbf{v}\), we get \(6^2 + 1^2 = 37\).

Step 3: Find the Projection of \(\mathbf{u}\) onto \(\mathbf{v}\)

The projection of \(\mathbf{u}\) onto \(\mathbf{v}\) is given by \(\mathbf{u} . \mathbf{v} / ||\mathbf{v}||^2 * \mathbf{v}\) = \(14/37 * \langle 6, 1 \rangle\) = \(\langle 216/37, 14/37 \rangle\)

Step 4: Find the Orthogonal Component of \(\mathbf{u}\) to \(\mathbf{v}\)

The orthogonal component of \(\mathbf{u}\) to \(\mathbf{v}\) is given by \(\mathbf{u} - \) proj\(_{\mathbf{v}} \mathbf{u}\). Using the projection calculated in step 3, this gives us \(\langle 2, 2 \rangle - \langle 216/37, 14/37 \rangle = \(\langle 2 - 216/37, 2 - 14/37 \rangle = \(\langle 2/37, 60/37 \rangle\)

Key Concepts

Dot Product

Understanding the dot product is essential for various applications in mathematics and physics. The dot product of two vectors, often represented as \( \mathbf{u} \cdot \mathbf{v} \), is a scalar quantity that is the product of the vectors' magnitudes and the cosine of the angle between them. However, when the vectors are represented in Cartesian coordinates, the dot product is simply the sum of the products of their corresponding components.

In the case of our exercise, to find the dot product of \( \mathbf{u} \), which is \( \langle 2, 2 \rangle \) and \( \mathbf{v} \), given as \( \langle 6, 1 \rangle \) we multiply each corresponding component and sum the results: \( 2 \times 6 + 2 \times 1 \), which yields 14. This operation is foundational for finding the projection of \( \mathbf{u} \) onto \( \mathbf{v} \) and for separating \( \mathbf{u} \) into components parallel and orthogonal to \( \mathbf{v} \).

The dot product is also a measure of how much one vector extends in the direction of another, and if it's zero, it indicates that the vectors are orthogonal, a concept we'll explore later.

Vector Magnitude

The magnitude of a vector is a measure of its 'length' or the distance it extends in space. For a two-dimensional vector \( \mathbf{v} = \langle v_x, v_y \rangle \), its magnitude is found using the Pythagorean theorem, resulting in \( ||\mathbf{v}|| = \sqrt{v_x^2 + v_y^2} \).

In our exercise, to compute the magnitude of vector \( \mathbf{v} = \langle 6, 1 \rangle \), we indeed square each component and then sum these squares, \( 6^2 + 1^2 = 36 + 1 = 37 \). Since we are looking for the magnitude squared, we don't need to take the square root in this case, and we’re left with 37. This value is used to normalize the vector when calculating the projection of \( \mathbf{u} \) onto \( \mathbf{v} \).

Knowing how to calculate vector magnitude is not only crucial for finding projection but also in physics for determining the strength or intensity of a quantity represented by the vector, such as velocity or force.

Orthogonal Vectors

Vectors are considered orthogonal to each other if they meet at a 90 degrees angle, in other words, if they are perpendicular. This is an important concept in mathematics because orthogonal vectors have a dot product of zero. This property is widely used as a method to check orthogonality between two vectors.

As part of the projection process in our example, once we've established the component of \( \mathbf{u} \) that lies along \( \mathbf{v} \) (the projection), the orthogonal component can then be found. To achieve this, we subtract the projection from the initial vector \( \mathbf{u} \) resulting in a vector that is perpendicular to \( \mathbf{v} \).

Following the steps provided, the orthogonal component was \( \langle 2 - 216/37, 2 - 14/37 \rangle = \langle 2/37, 60/37 \rangle \), affirming the vector is indeed orthogonal to \( \mathbf{v} \) because the dot product of this vector with \( \mathbf{v} \) would be zero. Notably, this property plays a pivotal role in various areas like computer graphics, where it is used to compute reflections and projections, and in orthogonal decompositions in vector spaces.