Problem 30 - 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 \(\mathbf{v}\). $$\mathbf{u}=\langle 7,-4\rangle, \quad \mathbf{v}=\langle 2,1\rangle$$

Step-by-Step Solution

Answer

\( \mathbf{u}_{1} = \langle 4, 2 \rangle \), \( \mathbf{u}_{2} = \langle 3, -6 \rangle \).

Step 1 - Calculate Dot Product

Calculate the dot product \( \mathbf{u} \cdot \mathbf{v} \) to find the projection. Use the formula: \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 \). Thus, \( \mathbf{u} = \langle 7, -4 \rangle \) and \( \mathbf{v} = \langle 2, 1 \rangle \). The dot product is: \[ 7 \times 2 + (-4) \times 1 = 14 - 4 = 10 \].

Step 2 - Calculate Magnitude of v Squared

Find \( \| \mathbf{v} \|^2 \), the magnitude of \( \mathbf{v} \) squared, which is used for the projection. This is calculated as \( v_1^2 + v_2^2 = 2^2 + 1^2 = 4 + 1 = 5 \).

Step 3 - Compute Projection

Using the formula for projection, \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{v} \|^2} \mathbf{v} \), substitute the values: \[ \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{10}{5} \langle 2, 1 \rangle = 2 \langle 2, 1 \rangle \].Simplifying gives \( \text{proj}_{\mathbf{v}} \mathbf{u} = \langle 4, 2 \rangle \).

Step 4 - Determine u1 and u2

Resolve \( \mathbf{u} \) into two components. Let \( \mathbf{u}_{1} = \text{proj}_{\mathbf{v}} \mathbf{u} = \langle 4, 2 \rangle \).Calculate \( \mathbf{u}_{2} \) by subtracting \( \mathbf{u}_{1} \) from \( \mathbf{u} \):\[ \mathbf{u}_{2} = \mathbf{u} - \mathbf{u}_{1} = \langle 7, -4 \rangle - \langle 4, 2 \rangle = \langle 3, -6 \rangle \].

Key Concepts

Dot Product

The dot product is a cornerstone of vector mathematics. It quantifies how much of one vector goes in the direction of another. To calculate the dot product of two vectors, like \(\mathbf{u} = \langle 7, -4 \rangle\) and \(\mathbf{v} = \langle 2, 1 \rangle\), you use the formula:

• Multiply each pair of corresponding components: Multiply the first component of \(\mathbf{u}\) with the first component of \(\mathbf{v}\), and do the same for the second components.
• Sum the results: Add these products together.

This gives: \[ \mathbf{u} \cdot \mathbf{v} = 7 \times 2 + (-4) \times 1 = 14 - 4 = 10 \].
The dot product here is 10, indicating the extent along which \(\mathbf{u}\) aligns with \(\mathbf{v}\), crucial for calculating projections.

Magnitude of a Vector

The magnitude of a vector, often thought of as the vector's length, is crucial in calculations such as vector projections. For a vector \(\mathbf{v} = \langle 2, 1 \rangle\), the magnitude is determined using the formula:

• Square each component of the vector: This means taking the square of the first component and adding it to the square of the second component.
• Take the square root: Add these squared values, then take the square root of the sum.

Mathematically, this is expressed as:\[ \| \mathbf{v} \| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}. \]
However, for projection calculations, we often use \(\| \mathbf{v} \|^2\), which is simply 5 for our example.
This value aids in determining how \(\mathbf{u}\) projects onto \(\mathbf{v}\).

Vector Components

Vector components are the pieces of a vector that show its influence along specific directions. When decomposing a vector \(\mathbf{u} = \langle 7, -4 \rangle\) into components \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\):

• Parallel component: \(\mathbf{u}_{1} = \text{proj}_{\mathbf{v}} \mathbf{u}\), the part of \(\mathbf{u}\) that is aligned with \(\mathbf{v}\). In the example, this results in \(\langle 4, 2 \rangle\).
• Orthogonal component: \(\mathbf{u}_{2}\) is computed by subtracting \(\mathbf{u}_{1}\) from \(\mathbf{u}\). Thus, \(\mathbf{u}_{2} = \langle 7, -4 \rangle - \langle 4, 2 \rangle = \langle 3, -6 \rangle\).

This decomposition allows us to understand how a vector \(\mathbf{u}\) behaves both along and perpendicular to another vector \(\mathbf{v}\), which is fundamental in fields like physics and engineering.