Problem 24 - Algebra and Trigonometry

Question

23-28 (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 \(\mathbf{v}\). $$\mathbf{u}=\langle 7,-4\rangle, \quad \mathbf{v}=\langle 2,1\rangle$$

Step-by-Step Solution

Answer

(a) \( \text{proj}_{\mathbf{v}} \mathbf{u} = \langle 4, 2 \rangle \). (b) \( \mathbf{u}_1 = \langle 4, 2 \rangle \), \( \mathbf{u}_2 = \langle 3, -6 \rangle \).

Step 1: Calculate the Dot Product

To find the projection of \( \mathbf{u} \) onto \( \mathbf{v} \), first calculate the dot product \( \mathbf{u} \cdot \mathbf{v} \) using the formula: \( \mathbf{u} \cdot \mathbf{v} = 7 \times 2 + (-4) \times 1 = 14 - 4 = 10 \).

Step 2: Calculate the Magnitude of v Squared

Calculate the magnitude squared of \( \mathbf{v} \) using the formula \( ||\mathbf{v}||^2 = 2^2 + 1^2 = 4 + 1 = 5 \).

Step 3: Determine the Projection of u onto v (proj u onto v)

Use the formula for the projection \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{v}||^2} \mathbf{v} \). Plug in the values: \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{10}{5} \langle 2, 1 \rangle = 2 \langle 2, 1 \rangle = \langle 4, 2 \rangle \).

Step 4: Resolve u into u₁ and u₂

\( \mathbf{u}_1 \) is the projection of \( \mathbf{u} \) onto \( \mathbf{v} \), which we found is \( \langle 4, 2 \rangle \). To find \( \mathbf{u}_2 \), use \( \mathbf{u}_2 = \mathbf{u} - \mathbf{u}_1 \): \( \mathbf{u}_2 = \langle 7, -4 \rangle - \langle 4, 2 \rangle = \langle 3, -6 \rangle \).

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector mathematics. It combines two vectors to produce a scalar (a single number) that provides insight into how much one vector is oriented in the direction of another. For two-dimensional vectors \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \), the dot product is calculated using the formula:
\[ \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 \]
In our exercise, calculating the dot product \( \mathbf{u} \cdot \mathbf{v} \) helps determine how much of vector \( \mathbf{u} \) aligns with vector \( \mathbf{v} \).

• If the dot product is positive, the vectors point in similar directions.
• If it's negative, they point in opposite directions.
• A dot product of zero indicates orthogonal (perpendicular) vectors.

Understanding the dot product is essential for applications like projecting one vector onto another.

Magnitude of a Vector

The magnitude of a vector, often referred to as its length or norm, describes how long the vector is. For a 2D vector \( \mathbf{v} = \langle v_1, v_2 \rangle \), the magnitude is calculated by:
\[ ||\mathbf{v}|| = \sqrt{v_1^2 + v_2^2} \]
In projections, the magnitude squared \( ||\mathbf{v}||^2 \) is typically used. This avoids the need for computing the square root, simplifying calculations. In the given problem, we are interested in this squared magnitude to efficiently calculate the projection of one vector onto another.

• The magnitude gives a sense of the vector's size.
• Calculating the magnitude squared is faster and useful when projecting vectors.

Remember, understanding the magnitude helps you compare the sizes of different vectors.

Orthogonal Vectors

Two vectors are said to be orthogonal if they meet at a 90-degree angle, or in simpler terms, they are perpendicular to each other. This relationship is significant, especially in vector resolution. Orthogonality is tested using the dot product: if \( \mathbf{u} \cdot \mathbf{v} = 0 \), then \( \mathbf{u} \) and \( \mathbf{v} \) are orthogonal.
In vector resolution, when decomposing a vector \( \mathbf{u} \) into components \( \mathbf{u}_1 \) parallel to another vector \( \mathbf{v} \), and \( \mathbf{u}_2 \) orthogonal to \( \mathbf{v} \):

• \( \mathbf{u}_1 \) is obtained using the projection.
• \( \mathbf{u}_2 \) represents the residual vector that is orthogonal, illustrating how \( \mathbf{u} \) deviates from being completely parallel.

This is key in understanding how to break vectors into useful components for various applications.

Vector Resolution

Vector resolution is a process of breaking down a vector into components that align with specific directions. This is extremely useful in physics and engineering. For a vector \( \mathbf{u} \) and a reference vector \( \mathbf{v} \), vector resolution involves finding two components, \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \):

• \( \mathbf{u}_1 \) is the parallel component, calculated using the projection formula \( \text{proj}_{\mathbf{v}} \mathbf{u} \). It shows how much of \( \mathbf{u} \) is along the direction of \( \mathbf{v} \).
• \( \mathbf{u}_2 \) is the orthogonal component, found by subtracting \( \mathbf{u}_1 \) from \( \mathbf{u} \). It represents the part of \( \mathbf{u} \) that does not share the direction of \( \mathbf{v} \).

This method simplifies complex vector problems by decomposing them into manageable parts, enhancing our ability to solve and understand vector scenarios.