Problem 84 - Precalculus with Calculus Previews

Question

In Problems \(8_{3}\) and \(84, \mathbf{u}=4 \mathbf{i}+3 \mathbf{j}\) and \(\mathbf{v}=-\mathbf{i}+\mathbf{j}\). Find the indicated vector. $$ \operatorname{proj}_{\mathbf{u}-\mathbf{v}} \mathbf{v} $$

Step-by-Step Solution

Answer

\(\operatorname{proj}_{\mathbf{u}-\mathbf{v}} \mathbf{v} = \left(\frac{-15}{29}\right)\mathbf{i} + \left(\frac{-6}{29}\right)\mathbf{j}\).

Step 1: Find the Vector Difference

Calculate \( \mathbf{u} - \mathbf{v} \) by subtracting vector \( \mathbf{v} = -\mathbf{i} + \mathbf{j} \) from \( \mathbf{u} = 4\mathbf{i} + 3\mathbf{j} \). \[ \mathbf{u} - \mathbf{v} = (4\mathbf{i} + 3\mathbf{j}) - (-\mathbf{i} + \mathbf{j}) = 4\mathbf{i} + 3\mathbf{j} + \mathbf{i} - \mathbf{j} \]Combine like terms:\[ \mathbf{u} - \mathbf{v} = (4+1)\mathbf{i} + (3-1)\mathbf{j} = 5\mathbf{i} + 2\mathbf{j} \]

Step 2: Calculate the Dot Product

Find the dot product \( \mathbf{v} \cdot (\mathbf{u} - \mathbf{v}) \).\[ \mathbf{v} = -\mathbf{i} + \mathbf{j} \] and \[ \mathbf{u} - \mathbf{v} = 5\mathbf{i} + 2\mathbf{j} \]The dot product is:\[ \mathbf{v} \cdot (\mathbf{u} - \mathbf{v}) = (-1)(5) + (1)(2) = -5 + 2 = -3 \]

Step 3: Calculate the Magnitude Squared

Compute the magnitude squared of \( \mathbf{u} - \mathbf{v} \).\[ \mathbf{u} - \mathbf{v} = 5\mathbf{i} + 2\mathbf{j} \]The magnitude squared is:\[ (5)^2 + (2)^2 = 25 + 4 = 29 \]

Step 4: Compute the Projection

Use the formula for the projection, \( \operatorname{proj}_{\mathbf{u}-\mathbf{v}} \mathbf{v} = \frac{\mathbf{v} \cdot (\mathbf{u} - \mathbf{v})}{\|\mathbf{u} - \mathbf{v}\|^2} (\mathbf{u} - \mathbf{v}) \).Substitute the results from previous steps:\[ \operatorname{proj}_{\mathbf{u}-\mathbf{v}} \mathbf{v} = \frac{-3}{29} (5\mathbf{i} + 2\mathbf{j}) \]This gives:\[ \operatorname{proj}_{\mathbf{u}-\mathbf{v}} \mathbf{v} = \left(\frac{-15}{29}\right)\mathbf{i} + \left(\frac{-6}{29}\right)\mathbf{j} \]

Key Concepts

Dot Product

The dot product is a fundamental concept in vector mathematics, especially in the context of vectors in physics. It is a way to multiply two vectors together to result in a scalar. This operation helps in understanding the directional relationship between vectors. The formula for the dot product of two vectors \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} \) and \( \mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} \) is as follows:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)

In the original problem, calculating the dot product helped determine how much of vector \(\mathbf{v}\) projects onto the vector difference \(\mathbf{u} - \mathbf{v}\). The computation \( (-1)(5) + (1)(2) = -3 \) demonstrates evaluating each component and adding the results to get the final scalar value.

Magnitude

Magnitude refers to the length or size of a vector. It's often necessary to calculate this to normalize a vector or find its projection onto another vector. For a vector \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} \), its magnitude is determined using the formula:

• \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2} \)

In the computation of the problem, the magnitude squared of \(\mathbf{u} - \mathbf{v}\) was used instead of the magnitude. The magnitude squared, \( (5)^2 + (2)^2 = 25 + 4 = 29 \), simplifies the projection computation process by removing the square root step and helping compute projections accurately and efficiently without additional calculation.

Vectors in Physics

Vectors are indispensable in physics as they describe quantities possessing both magnitude and direction. Examples include force, velocity, and acceleration. Understanding vector projection is crucial for analyzing components of these quantities in various directions.
Projection, a key vector operation, involves determining how much of one vector lies along the direction of another. This can represent real-world scenarios such as how a moving object follows a path or how forces impact each other. In reference to the exercise, the projection of \( \mathbf{v} \) onto the difference vector \( \mathbf{u} - \mathbf{v} \) indicates how much of \( \mathbf{v} \) aligns with this path. This insight allows physicists to effectively break down complex movements into manageable components.

Vector Operations

Vector operations include basic expressions such as addition, subtraction, and multiplication (both dot product and cross product). These operations allow more complex calculations and transformations of vectors in various applications:

• Addition: Combining two vectors to find resulting displacement.
• Subtraction: Finding the difference, as seen in \( \mathbf{u} - \mathbf{v} \), to assess deviation or direction.
• Dot Product: Determining alignment and energy transference between vectors.

Each of these operations plays a significant role in physics and engineering applications. The subtraction of \(\mathbf{u}\) and \(\mathbf{v}\) in the solution represents a component analysis of paths or forces, which is foundational in calculating projections and analyzing motion within fields of study.