Problem 55 - Precalculus with Limits

Question

Determining Orthogonal Vectors In Exercises \(53-58\) , determine whether \(u\) and \(v\) are orthogonal. $$\mathbf{u}=\frac{1}{4}(3 \mathbf{i}-\mathbf{j})$$ $$\mathbf{v}=5 \mathbf{i}+6 \mathbf{j}$$

Step-by-Step Solution

Answer

The vectors \(u\) and \(v\) are not orthogonal since their dot product is not equal to zero.

Step 1: Identify the components of the vectors

For vector \(u\), we have \(\mathbf{u}=\frac{1}{4}(3 \mathbf{i}-\mathbf{j})\), which means the components are \(u_i = 3/4\) and \(u_j = -1/4\). As for vector \(v\), we have \(\mathbf{v}=5 \mathbf{i}+6 \mathbf{j}\), which means the components are \(v_i = 5\) and \(v_j = 6\).

Step 2: Calculate the Dot Product

The formula for the dot product of two vectors is \(u \cdot v = u_i \cdot v_i + u_j \cdot v_j\). Substituting in our known values, we get \(u \cdot v = (\frac{3}{4} \cdot 5) + (-\frac{1}{4} \cdot 6)\).

Step 3: Solve the Dot Product

On solving the expression, we get \(u \cdot v = \frac{15}{4} - \frac{6}{4} = \frac{9}{4}\), which is not equal to zero.

Key Concepts

Understanding the Dot Product

The dot product is a fundamental concept in vector mathematics. It helps determine relationships between vectors, such as whether they are orthogonal. The dot product, also known as the scalar product, is calculated by multiplying the corresponding components of two vectors and summing the results. For example, vectors \( \mathbf{u} \) and \( \mathbf{v} \) are given as \( \mathbf{u} = \frac{1}{4}(3 \mathbf{i} - \mathbf{j}) \) and \( \mathbf{v} = 5 \mathbf{i} + 6 \mathbf{j} \).

• The component form of \( \mathbf{u} \) is \( (3/4, -1/4) \).
• The component form of \( \mathbf{v} \) is \( (5, 6) \).

To find their dot product, use the formula: \[u \cdot \mathbf{v} = u_i \cdot v_i + u_j \cdot v_j.\] Substituting the known values:
\[ \frac{3}{4} \cdot 5 + (-\frac{1}{4}) \cdot 6 = \frac{15}{4} - \frac{6}{4} = \frac{9}{4}. \] When the dot product is zero, vectors are orthogonal (perpendicular). Here, \( \frac{9}{4} eq 0 \), so \( \mathbf{u} \) and \( \mathbf{v} \) aren't orthogonal.

Breaking Down Vector Components

Vector components are the building blocks of any vector, represented in terms of the direction and magnitude along each axis. In a two-dimensional space, for example, the vector is expressed with its horizontal (i.e., \(x\)) and vertical (i.e., \(y\)) components. For vector \( \mathbf{u} = \frac{1}{4}(3\mathbf{i} - \mathbf{j}) \), the process is as follows:

• Multiply each scalar by the vector component: \( 3/4 \) times \( \mathbf{i} \) and \(-1/4\) times \( \mathbf{j} \).
• This gives \( u_i = 3/4 \) and \( u_j = -1/4 \).

Similarly, for \( \mathbf{v} = 5 \mathbf{i} + 6 \mathbf{j} \):

• Directly read the coefficients: \( v_i = 5 \) and \( v_j = 6 \).

Vector components allow for easier calculations of vector operations, like the dot product. Their understanding helps visualize the direction and movement within a coordinate system.

Connecting Vectors with Precalculus

Precalculus forms the bridge between basic algebra and calculus, with vectors playing a crucial role in this transition. Being familiar with vectors aids in the comprehension of multidimensional concepts essential in calculus. Precalculus covers foundational vector theories that help in solving real-world problems involving directions, forces, or movement.

• Vectors simplify complex problems by breaking them into manageable components.
• Knowing how to calculate the dot product is vital for learning more advanced concepts like vector projection and orthogonality.

In contexts such as physics and engineering, these vector operations are integral in analyzing forces or motions. Understanding vectors in precalculus ensures a robust foundation for tackling sophisticated calculus topics efficiently.