Problem 3 - University Calculus: Early Transcendentals

Question

Find a. \(\quad \mathbf{v} \cdot \mathbf{u},|\mathbf{v}|,|\mathbf{u}|\) b. the cosine of the angle between \(\mathbf{v}\) and \(\mathbf{u}\) c. the scalar component of \(\mathbf{u}\) in the direction of \(\mathbf{v}\) d. the vector proje \(\mathbf{u}\). $$\mathbf{v}=10 \mathbf{i}+11 \mathbf{j}-2 \mathbf{k}, \quad \mathbf{u}=3 \mathbf{j}+4 \mathbf{k}$$

Step-by-Step Solution

Answer

a. 25, 15, 5; b. \(\cos \theta = \frac{1}{3}\); c. \(\frac{5}{3}\); d. \(\frac{10}{9} \mathbf{i} + \frac{11}{9} \mathbf{j} - \frac{2}{9} \mathbf{k}\).

Step 1: Calculate the dot product of vectors \( \mathbf{v} \) and \( \mathbf{u} \)

The dot product of two vectors \( \mathbf{v} = a_1 \mathbf{i} + b_1 \mathbf{j} + c_1 \mathbf{k} \) and \( \mathbf{u} = a_2 \mathbf{i} + b_2 \mathbf{j} + c_2 \mathbf{k} \) is given by: \( \mathbf{v} \cdot \mathbf{u} = a_1a_2 + b_1b_2 + c_1c_2 \). Substituting the values from \( \mathbf{v} = 10 \mathbf{i} + 11 \mathbf{j} - 2 \mathbf{k} \) and \( \mathbf{u} = 0 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k} \), we get: \( \mathbf{v} \cdot \mathbf{u} = (10 \times 0) + (11 \times 3) + (-2 \times 4) = 0 + 33 - 8 = 25 \). Thus, \( \mathbf{v} \cdot \mathbf{u} = 25 \).

Step 2: Compute the magnitudes \( |\mathbf{v}| \) and \( |\mathbf{u}| \)

The magnitude of a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \) is calculated as \( |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \). For \( \mathbf{v} = 10 \mathbf{i} + 11 \mathbf{j} - 2 \mathbf{k} \), \(|\mathbf{v}| = \sqrt{10^2 + 11^2 + (-2)^2} = \sqrt{100 + 121 + 4} = \sqrt{225} = 15 \).For \( \mathbf{u} = 0 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k} \), \(|\mathbf{u}| = \sqrt{0^2 + 3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \).

Step 3: Calculate the cosine of the angle between \( \mathbf{v} \) and \( \mathbf{u} \)

The cosine of the angle \( \theta \) between two vectors \( \mathbf{v} \) and \( \mathbf{u} \) is given by \( \cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}||\mathbf{u}|} \). From Steps 1 and 2, we have \( \mathbf{v} \cdot \mathbf{u} = 25 \), \(|\mathbf{v}| = 15\), and \(|\mathbf{u}| = 5\).Thus, \( \cos \theta = \frac{25}{15 \times 5} = \frac{25}{75} = \frac{1}{3} \).

Step 4: Find the scalar component of \( \mathbf{u} \) in the direction of \( \mathbf{v} \)

The scalar component of \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is given by \( \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|} \). From previous steps, \( \mathbf{v} \cdot \mathbf{u} = 25 \) and \( |\mathbf{v}| = 15 \). Therefore, the scalar component is \( \frac{25}{15} = \frac{5}{3} \).

Step 5: Calculate the vector projection of \( \mathbf{u} \) onto \( \mathbf{v} \)

The vector projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is given by \( \text{proj}_{\mathbf{v}} \mathbf{u} = \left(\frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2}\right) \mathbf{v} \). We already found \( \mathbf{v} \cdot \mathbf{u} = 25 \) and \( |\mathbf{v}| = 15 \), so \( |\mathbf{v}|^2 = 15^2 = 225 \).Thus, \( \text{proj}_{\mathbf{v}} \mathbf{u} = \left(\frac{25}{225}\right) (10 \mathbf{i} + 11 \mathbf{j} - 2 \mathbf{k}) = \frac{1}{9} (10 \mathbf{i} + 11 \mathbf{j} - 2 \mathbf{k}) = \frac{10}{9} \mathbf{i} + \frac{11}{9} \mathbf{j} - \frac{2}{9} \mathbf{k} \).

Key Concepts

Dot Product

The dot product is a mathematical operation that takes two vectors and returns a scalar. It is calculated by multiplying corresponding components of the vectors and adding the results. For two vectors \( \mathbf{v} = a_1 \mathbf{i} + b_1 \mathbf{j} + c_1 \mathbf{k} \) and \( \mathbf{u} = a_2 \mathbf{i} + b_2 \mathbf{j} + c_2 \mathbf{k} \), the dot product is expressed as:

• \( \mathbf{v} \cdot \mathbf{u} = a_1a_2 + b_1b_2 + c_1c_2 \)

This formula is useful for determining the degree to which two vectors are aligned. A dot product of zero indicates orthogonal vectors (perpendicular), while a positive or negative value shows the vectors are in a similar or opposite direction. In the exercise, the calculation of \( \mathbf{v} \cdot \mathbf{u} = 25 \) demonstrates that the vectors have some direction in common.

Magnitude of a Vector

The magnitude of a vector, also known as its length, measures how long the vector is in space. It can be understood as the distance of the vector from the origin in three-dimensional space. For a vector given by \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), its magnitude is:

• \( |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \)

The exercise specifies vectors:

• \( \mathbf{v} = 10 \mathbf{i} + 11 \mathbf{j} - 2 \mathbf{k} \)
• \( \mathbf{u} = 0 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k} \)

Calculating their magnitudes gives us \( |\mathbf{v}| = 15 \) and \( |\mathbf{u}| = 5 \). The magnitudes are important for determining how "strong" a vector is compared to others in terms of size.

Vector Projection

Vector projection refers to the process of projecting one vector onto another, which creates a new vector that indicates the part of the original vector that aligns with the second vector. The formula for projecting vector \( \mathbf{u} \) onto vector \( \mathbf{v} \) is:

• \( \text{proj}_{\mathbf{v}} \mathbf{u} = \left(\frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2}\right) \mathbf{v} \)

This is useful in applications where you want to find how much of one vector goes in the direction of another. In the exercise, after substituting values from the previous calculations, the vector projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is found to be \( \frac{10}{9} \mathbf{i} + \frac{11}{9} \mathbf{j} - \frac{2}{9} \mathbf{k} \), which represents how \( \mathbf{u} \) positions itself along \( \mathbf{v} \).

Cosine of Angle between Vectors

The cosine of the angle between two vectors provides insight into their direction relative to each other. This measurement is important in physics and engineering to understand how aligned or opposed the directions of the vectors are.
The formula to calculate the cosine of the angle \( \theta \) between the vectors \( \mathbf{v} \) and \( \mathbf{u} \) is:

• \( \cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}||\mathbf{u}|} \)

A cosine value close to 1 signifies that the vectors are nearly parallel, while values near 0 indicate perpendicular vectors. In our exercise solution, we calculated \( \cos \theta = \frac{1}{3} \), suggesting that \( \mathbf{v} \) and \( \mathbf{u} \) form an angle where they are neither completely perpendicular nor fully aligned, providing a balanced measure of similarity in direction.