Problem 31 - 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 \(\mathbf{v}\). $$\mathbf{u}=\langle 1,2\rangle, \quad \mathbf{v}=\langle 1,-3\rangle$$

Step-by-Step Solution

Answer

\( \mathbf{u}_1 = \langle -0.5, 1.5 \rangle \), \( \mathbf{u}_2 = \langle 1.5, 0.5 \rangle \).

Step 1: Calculate Dot Product of u and v

Find the dot product of the vectors \( \mathbf{u} \) and \( \mathbf{v} \). The dot product is calculated as follows: \( \mathbf{u} \cdot \mathbf{v} = 1 \cdot 1 + 2 \cdot (-3) = 1 - 6 = -5 \).

Step 2: Calculate Magnitude Squared of v

The magnitude squared of \( \mathbf{v} \) is needed to find the projection. Calculate it as follows: \( \| \mathbf{v} \|^2 = 1^2 + (-3)^2 = 1 + 9 = 10 \).

Step 3: Compute Projection of u onto v

Use the projection formula \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{v} \|^2} \mathbf{v} \). Substitute the values: \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{-5}{10} \langle 1, -3 \rangle = \langle -0.5, 1.5 \rangle \).

Step 4: Determine Orthogonal Component u2

The orthogonal component \( \mathbf{u}_2 \) is given by \( \mathbf{u}_2 = \mathbf{u} - \mathbf{u}_1 \). Since \( \mathbf{u}_1 = \langle -0.5, 1.5 \rangle \), we have \( \mathbf{u}_2 = \langle 1, 2 \rangle - \langle -0.5, 1.5 \rangle = \langle 1 + 0.5, 2 - 1.5 \rangle = \langle 1.5, 0.5 \rangle \).

Key Concepts

Dot Product

The dot product is an essential operation to understand when working with vectors. It allows us to measure how much one vector extends in the direction of another vector. To calculate the dot product between two vectors, we multiply their corresponding components and add the results together.
For example, given vectors \( \mathbf{u} = \langle 1, 2 \rangle \) and \( \mathbf{v} = \langle 1, -3 \rangle \), the dot product \( \mathbf{u} \cdot \mathbf{v} \) is computed as follows:

• Multiply the first components: \(1 \cdot 1 = 1\)
• Multiply the second components: \(2 \cdot (-3) = -6\)
• Add the results: \(1 + (-6) = -5\)

This produces a scalar value, \(-5\), which represents the dot product of the vectors \( \mathbf{u} \) and \( \mathbf{v} \). The dot product essentially captures the "closeness" of the vectors' directions.

Magnitude of a Vector

The magnitude of a vector is an important concept that helps us understand the length or size of the vector. It's similar to finding the length of a line segment in geometry. To calculate the magnitude, we use the formula that is similar to the Pythagorean theorem.
For a vector \( \mathbf{v} = \langle 1, -3 \rangle \), the magnitude is calculated as:

• Square each component: \(1^2 = 1\) and \((-3)^2 = 9\)
• Add the squares: \(1 + 9 = 10\)
• Take the square root: \( \sqrt{10} \)

However, sometimes we only need the magnitude squared, especially when calculating projections. In this case, the magnitude squared of \( \mathbf{v} \) is \(10\). Knowing the magnitude helps in determining directions, normalizing vectors, and computing projections.

Orthogonal Component

The orthogonal component of a vector is essentially the part that is perpendicular to another vector. When resolving a vector \( \mathbf{u} \) relative to another vector \( \mathbf{v} \), it is useful to decompose \( \mathbf{u} \) into two parts:

• A part that is parallel to \( \mathbf{v} \)
• A part that is orthogonal to \( \mathbf{v} \)

The orthogonal component \( \mathbf{u}_2 \) can be found by subtracting the parallel component \( \mathbf{u}_1 \) from \( \mathbf{u} \). For our example, the parallel component \( \mathbf{u}_1 \) was already calculated as \( \langle -0.5, 1.5 \rangle \). Then,

• Orthogonal component formula: \( \mathbf{u}_2 = \mathbf{u} - \mathbf{u}_1 \)
• Calculate \( \mathbf{u}_2 = \langle 1, 2 \rangle - \langle -0.5, 1.5 \rangle \)
• Result is \( \mathbf{u}_2 = \langle 1 + 0.5, 2 - 1.5 \rangle = \langle 1.5, 0.5 \rangle \)

This orthogonal component is perpendicular to \( \mathbf{v} \), ensuring that the original vector \( \mathbf{u} \) is effectively broken down in relation to \( \mathbf{v} \). Understanding orthogonal components is crucial in vector decomposition, signal processing, and solving geometrical problems.