Problem 34 - Precalculus Mathematics for Calculus

Question

Calculate proj, u. (b) Resolve u into \(u_{1}\) and \(u_{2}\), where \(\mathbf{u}_{1}\) is parallel to \(\mathbf{v}\) and \(\mathbf{u}_{2}\) is orthogonal to \(\mathbf{v}\). $$\mathbf{u}=\langle 1,1\rangle, \quad \mathbf{v}=\langle 2,-1\rangle$$

Step-by-Step Solution

Answer

The projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is \( \langle \frac{2}{5}, -\frac{1}{5} \rangle \). \( \mathbf{u}_1 = \langle \frac{2}{5}, -\frac{1}{5} \rangle \) and \( \mathbf{u}_2 = \langle \frac{3}{5}, \frac{6}{5} \rangle \).

Step 1: Find the Dot Product

Calculate the dot product of vectors \( \mathbf{u} = \langle 1, 1 \rangle \) and \( \mathbf{v} = \langle 2, -1 \rangle \). The dot product is given by \( \mathbf{u} \cdot \mathbf{v} = 1 \cdot 2 + 1 \cdot (-1) = 2 - 1 = 1 \).

Step 2: Find the Magnitude of Vector v

Calculate the magnitude of \( \mathbf{v} = \langle 2, -1 \rangle \) using the formula \( ||\mathbf{v}|| = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \).

Step 3: Calculate proj_u (b)

Use the formula for the projection of \( \mathbf{u} \) onto \( \mathbf{v} \), \( \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \right) \mathbf{v} \). Substitute the previously calculated values: \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{1}{5} \langle 2, -1 \rangle = \langle \frac{2}{5}, -\frac{1}{5} \rangle \).

Step 4: Resolve u into u_1 and u_2

Divide \( \mathbf{u} \) into \( \mathbf{u}_1 \), which is parallel to \( \mathbf{v} \) and is the projection we just calculated, and \( \mathbf{u}_2 \), which is orthogonal to \( \mathbf{v} \).\( \mathbf{u}_1 = \langle \frac{2}{5}, -\frac{1}{5} \rangle \) and \( \mathbf{u}_2 = \mathbf{u} - \mathbf{u}_1 \).Calculate \( \mathbf{u}_2 = \langle 1, 1 \rangle - \langle \frac{2}{5}, -\frac{1}{5} \rangle = \langle 1 - \frac{2}{5}, 1 + \frac{1}{5} \rangle = \langle \frac{3}{5}, \frac{6}{5} \rangle \).

Key Concepts

Dot Product

The dot product, also known as the scalar product, is an important concept when dealing with vectors. It combines two vectors to produce a scalar, and is fundamental in vector projection and many other vector operations. To calculate the dot product of two vectors \( \mathbf{u} = \langle a, b \rangle \) and \( \mathbf{v} = \langle c, d \rangle \), we use the formula:

• \( \mathbf{u} \cdot \mathbf{v} = a \cdot c + b \cdot d \)

For example, with \( \mathbf{u} = \langle 1, 1 \rangle \) and \( \mathbf{v} = \langle 2, -1 \rangle \), the dot product is computed as:
\( \mathbf{u} \cdot \mathbf{v} = 1 \cdot 2 + 1 \cdot (-1) = 2 - 1 = 1 \).
This operation not only helps us understand the angle between the two vectors but also plays a crucial role in determining how much of one vector aligns with another. The result being 1 suggests a certain relation between \( \mathbf{u} \) and \( \mathbf{v} \), hinting they are not entirely perpendicular.

Magnitude of a Vector

The magnitude of a vector, often referred to as its length or norm, gives us a measure of how long a vector is in a geometrical sense. For any vector \( \mathbf{v} = \langle a, b \rangle \), the magnitude is calculated using the formula:

• \( ||\mathbf{v}|| = \sqrt{a^2 + b^2} \)

In the given problem, with the vector \( \mathbf{v} = \langle 2, -1 \rangle \), the magnitude is:
\( ||\mathbf{v}|| = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \).
This result, \( \sqrt{5} \), represents the length of vector \( \mathbf{v} \) in the two-dimensional space. Understanding the magnitude is essential for calculations involving the normalization of vectors and the projection of one vector onto another, as it determines the scale of projection.

Orthogonal Vectors

Orthogonal vectors are vectors that meet at right angles, meaning they are perpendicular to each other. Two vectors \( \mathbf{u} \) and \( \mathbf{v} \) are orthogonal if their dot product is zero:

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

In this exercise, we learned to resolve vector \( \mathbf{u} \) into two components: \( \mathbf{u}_1 \), which is parallel to \( \mathbf{v} \), and \( \mathbf{u}_2 \), which is orthogonal to \( \mathbf{v} \).
Calculating \( \mathbf{u}_2 \) involved finding the difference between \( \mathbf{u} \) and \( \mathbf{u}_1 \), which we found using vector projection:
\( \mathbf{u}_2 = \langle 1, 1 \rangle - \langle \frac{2}{5}, -\frac{1}{5} \rangle = \langle \frac{3}{5}, \frac{6}{5} \rangle \).
This ensures that \( \mathbf{u}_2 \) is orthogonal to \( \mathbf{v} \). Recognizing and working with orthogonal vectors is vital in many areas of geometry and physics, particularly in simplifying complex vector tasks.