Problem 32 - 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 \(\mathbf{v} .\) $$ \mathbf{u}=\langle 11,3\rangle, \quad \mathbf{v}=\langle- 3,-2\rangle $$

Step-by-Step Solution

Answer

proj on \( \mathbf{v} \) is \( \langle 9, 6 \rangle \); \( \mathbf{u}_1 = \langle 9, 6 \rangle \), \( \mathbf{u}_2 = \langle 2, -3 \rangle \).

Step 1: Calculate the Dot Product

To calculate the projection of \( \mathbf{u} \) onto \( \mathbf{v} \), we first need the dot product of the vectors \( \mathbf{u} \) and \( \mathbf{v} \). The dot product is given by: \( \mathbf{u} \cdot \mathbf{v} = 11(-3) + 3(-2) = -33 - 6 = -39 \).

Step 2: Calculate the Magnitude Squared of Vector v

Now, we need the magnitude squared of vector \( \mathbf{v} \). This is calculated as:\( \| \mathbf{v} \|^2 = (-3)^2 + (-2)^2 = 9 + 4 = 13 \).

Step 3: Calculate the Projection of u onto v

The projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is calculated using the formula:\( \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{v} \|^2} \right) \mathbf{v} \).Substituting in the values we calculated:\( \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{-39}{13} \right) \langle -3, -2 \rangle = -3 \langle -3, -2 \rangle \).This gives: \( \text{proj}_{\mathbf{v}} \mathbf{u} = \langle 9, 6 \rangle \).

Step 4: Resolve u into u1 and u2

Next, resolve \( \mathbf{u} \) into components \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \), where \( \mathbf{u}_1 \) is parallel to \( \mathbf{v} \) and \( \mathbf{u}_2 \) is orthogonal to \( \mathbf{v} \).Given that \( \mathbf{u}_1 = \text{proj}_{\mathbf{v}} \mathbf{u} \), we have \( \mathbf{u}_1 = \langle 9, 6 \rangle \).Now, find \( \mathbf{u}_2 \) by subtracting \( \mathbf{u}_1 \) from \( \mathbf{u} \):\( \mathbf{u}_2 = \mathbf{u} - \mathbf{u}_1 = \langle 11, 3 \rangle - \langle 9, 6 \rangle = \langle 2, -3 \rangle \).

Key Concepts

Dot Product

The dot product is a fundamental operation in vector mathematics. It's denoted as \( \mathbf{u} \cdot \mathbf{v} \) and is calculated by multiplying corresponding components of two vectors and then summing those products. In this example, the vectors \( \mathbf{u} = \langle 11, 3 \rangle \) and \( \mathbf{v} = \langle -3, -2 \rangle \) are given. To compute the dot product, follow these steps:

• Multiply the x-components: \( 11 \times (-3) = -33 \)
• Multiply the y-components: \( 3 \times (-2) = -6 \)
• Add the results: \(-33 + (-6) = -39 \)

The dot product \( \mathbf{u} \cdot \mathbf{v} = -39 \) is crucial for further calculations like the vector projection.

Magnitude of Vector

Understanding the magnitude of a vector is like grasping the vector's length or size. To calculate the magnitude of vector \( \mathbf{v} = \langle -3, -2 \rangle \), use the formula \( \| \mathbf{v} \| = \sqrt{x^2 + y^2} \). This example focuses on the magnitude squared, which simplifies the calculation.

• Square the x-component: \( (-3)^2 = 9 \)
• Square the y-component: \( (-2)^2 = 4 \)
• Add these squares: \( 9 + 4 = 13 \)

This means the magnitude squared is \( \| \mathbf{v} \|^2 = 13 \). This step is important for computing projections.

Orthogonal Vectors

Orthogonal vectors are vectors that meet at a right angle. In simple terms, they form a 90-degree angle with each other. Two vectors are orthogonal if their dot product is zero. In vector resolution, finding an orthogonal component helps understand how a vector can be expressed as a combination of other vectors.When resolving vector \( \mathbf{u} \) into parts, some of it runs alongside vector \( \mathbf{v} \) (parallel), while the rest diverges at a right angle (orthogonal). Calculating these parts successfully splits \( \mathbf{u} \) into \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \), with \( \mathbf{u}_2 \) being orthogonal.The orthogonal component \( \mathbf{u}_2 \) was found by subtracting \( \mathbf{u}_1 = \langle 9, 6 \rangle \) from \( \mathbf{u} = \langle 11, 3 \rangle \):

• Subtract \( x \)-components: \( 11 - 9 = 2 \)
• Subtract \( y \)-components: \( 3 - 6 = -3 \)
• Result is \( \mathbf{u}_2 = \langle 2, -3 \rangle \)

This means \( \mathbf{u}_2 \) is orthogonal to \( \mathbf{v} \).

Vector Resolution

Vector resolution is the process of splitting a vector into components. Here, the problem involves resolving \( \mathbf{u} \) into two parts: one parallel and one orthogonal to \( \mathbf{v} \).**Finding \( \mathbf{u}_1 \)**The component \( \mathbf{u}_1 \) is parallel to \( \mathbf{v} \). It's obtained via the vector projection formula. Use the projection formula:\[ \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{v} \|^2} \right) \mathbf{v} \]Plug in calculated values:

• Dot product: \( -39 \)
• Magnitude squared: \( 13 \)
• Expression becomes \( -3 \langle -3, -2 \rangle = \langle 9, 6 \rangle \)

**Finding \( \mathbf{u}_2 \)**This part is orthogonal and calculated by:\[ \mathbf{u}_2 = \mathbf{u} - \mathbf{u}_1 \]Thus:

• \( \mathbf{u}_2 = \langle 11, 3 \rangle - \langle 9, 6 \rangle = \langle 2, -3 \rangle \)

Successfully resolving vectors into components aids in visualizing their behavior in various directions.