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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.

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 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 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.