Problem 31 - Algebra and Trigonometry

Question

(a) Calculate proj, \(\mathbf{u}\) . (b) Resolve \(\mathbf{u}\) into \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2},\) where \(\mathbf{u}_{1}\) is parallel to \(\mathbf{v}\) and \(\mathbf{u}_{2}\) is orthogonal to \(\mathbf{v} .\) $$ \mathbf{u}=\langle 1,2\rangle, \quad \mathbf{v}=\langle 1,-3\rangle $$

Step-by-Step Solution

Answer

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

Step 1: Calculate projection of u onto v

The projection of vector \( \mathbf{u} \) onto vector \( \mathbf{v} \) is given by the formula: \[ \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} \]First, calculate the dot product \( \mathbf{u} \cdot \mathbf{v} \): \[ \mathbf{u} \cdot \mathbf{v} = (1)(1) + (2)(-3) = 1 - 6 = -5 \]Then, calculate \( \mathbf{v} \cdot \mathbf{v} \): \[ \mathbf{v} \cdot \mathbf{v} = (1)(1) + (-3)(-3) = 1 + 9 = 10 \]Thus, \( \text{proj}_{\mathbf{v}} \mathbf{u} \) is:\[ \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{-5}{10} \mathbf{v} = \frac{-1}{2}\langle 1, -3 \rangle = \langle -\frac{1}{2}, \frac{3}{2} \rangle \]

Step 2: Find u1 and u2 vectors

Since \( \mathbf{u}_1 \) is parallel to \( \mathbf{v} \) and equivalent to the projection, we set \( \mathbf{u}_1 = \text{proj}_{\mathbf{v}} \mathbf{u} = \langle -\frac{1}{2}, \frac{3}{2} \rangle \).We find \( \mathbf{u}_2 \) by subtracting \( \mathbf{u}_1 \) from \( \mathbf{u} \):\[ \mathbf{u}_2 = \mathbf{u} - \mathbf{u}_1 = \langle 1, 2 \rangle - \langle -\frac{1}{2}, \frac{3}{2} \rangle = \langle 1 + \frac{1}{2}, 2 - \frac{3}{2} \rangle \]\[ \mathbf{u}_2 = \langle \frac{3}{2}, \frac{1}{2} \rangle \]

Step 3: Verify u1 and u2 are orthogonal and parallel

Check if \( \mathbf{u}_1 \) is parallel to \( \mathbf{v} \) and \( \mathbf{u}_2 \) is orthogonal to \( \mathbf{v} \). Since \( \mathbf{u}_1 = \text{proj}_{\mathbf{v}} \mathbf{u} \), \( \mathbf{u}_1 \) is parallel to \( \mathbf{v} \) by definition.For \( \mathbf{u}_2 \cdot \mathbf{v} = 0 \) (showing orthogonality):\[ \mathbf{u}_2 \cdot \mathbf{v} = \left( \frac{3}{2} \right) \cdot 1 + \left( \frac{1}{2} \right) \cdot (-3) = \frac{3}{2} - \frac{3}{2} = 0 \] Since \( \mathbf{u}_2 \cdot \mathbf{v} = 0 \), \( \mathbf{u}_2 \) is orthogonal to \( \mathbf{v} \).

Key Concepts

Vector Projection

Vector projection is a method to find a component of one vector along another vector. It's crucial when breaking down vectors into components which relate directly to another vector. If you have vectors \( \mathbf{u} \) and \( \mathbf{v} \), vector projection helps to identify the part of \( \mathbf{u} \) that aligns with \( \mathbf{v} \). The formula to project \( \mathbf{u} \) onto \( \mathbf{v} \) is:

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

To calculate this:

• First, find the dot product \( \mathbf{u} \cdot \mathbf{v} \), which measures how much \( \mathbf{u} \) moves in the direction of \( \mathbf{v} \).
• Divide by \( \mathbf{v} \cdot \mathbf{v} \), which projects \( \mathbf{u} \) onto \( \mathbf{v} \) giving a vector in the direction of \( \mathbf{v} \).

For \( \mathbf{u} = \langle 1, 2 \rangle \) and \( \mathbf{v} = \langle 1, -3 \rangle \), the projection \( \text{proj}_{\mathbf{v}} \mathbf{u} = \langle -\frac{1}{2}, \frac{3}{2} \rangle \). This result shows how \( \mathbf{u} \) aligns with \( \mathbf{v} \).

Dot Product

The dot product is an essential operation for determining the relationship between two vectors. It provides a way to calculate vector length and angle relative to other vectors. With vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), it's computed as follows:

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

The dot product tells us:

• If the result is zero, the vectors are orthogonal and intersect at 90 degrees.
• If positive, the vectors point in a similar direction.
• If negative, they point in opposite directions.

In our exercise, \( \mathbf{u} \cdot \mathbf{v} = -5 \) means \( \mathbf{u} \) and \( \mathbf{v} \) point partially in opposite directions. Understanding this product helps in projections and identifying vector parallels.

Orthogonal Vectors

Two vectors are orthogonal if the angle between them is 90 degrees. This means they have no component in the direction of the other. An important property of orthogonal vectors is:

• Their dot product is zero.

This property makes them essential in vector decomposition, where you break down a vector into parts:

• One part is parallel, and the other is perpendicular (orthogonal).

In our exercise, \( \mathbf{u}_2 = \langle \frac{3}{2}, \frac{1}{2} \rangle \) is orthogonal because \( \mathbf{u}_2 \cdot \mathbf{v} = 0 \). This confirms that \( \mathbf{u}_2 \) has no component in the direction of \( \mathbf{v} \). Recognizing when vectors are orthogonal is crucial in many fields, including graphics and physics.

Parallel Vectors

When vectors are parallel, they have the same or exactly opposite directions but differ in magnitude. For vectors to be parallel:

• One is a scalar multiple of the other.

In the context of projections, a vector \( \mathbf{u}_1 \) being parallel to \( \mathbf{v} \) means \( \mathbf{u}_1 \) can be expressed as \( k\mathbf{v} \), where \( k \) is some constant.
In the exercise, \( \mathbf{u}_1 = \langle -\frac{1}{2}, \frac{3}{2} \rangle \) is shown to be parallel to \( \mathbf{v} \). The projection formula itself defines \( \mathbf{u}_1 \) as the parallel component since it aligns with \( \mathbf{v} \).
Parallel vectors are significant in many applications, including physics, where forces and fields often act in a parallel manner.```