Problem 5 - 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 \(\mathbf{u}\). $$\mathbf{v}=5 \mathbf{j}-3 \mathbf{k}, \quad \mathbf{u}=\mathbf{i}+\mathbf{j}+\mathbf{k}$$

Step-by-Step Solution

Answer

a. \( \mathbf{v} \cdot \mathbf{u} = 2 \), \( |\mathbf{v}| = \sqrt{34} \), \( |\mathbf{u}| = \sqrt{3} \); b. \( \cos \theta = \frac{2\sqrt{102}}{102} \); c. \( \frac{2}{\sqrt{34}} \); d. \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{5}{17}\mathbf{j} - \frac{3}{17}\mathbf{k} \).

Step 1: Calculating the Dot Product

To find the dot product \( \mathbf{v} \cdot \mathbf{u} \), use the formula \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \). Here, \( \mathbf{v} = 0\mathbf{i} + 5\mathbf{j} - 3\mathbf{k} \) and \( \mathbf{u} = 1\mathbf{i} + 1\mathbf{j} + 1\mathbf{k} \). Thus, \( \mathbf{v} \cdot \mathbf{u} = 0 \times 1 + 5 \times 1 + (-3) \times 1 = 0 + 5 - 3 = 2 \).

Step 2: Finding Magnitudes

The magnitude \(|\mathbf{v}|\) is given by \( \sqrt{v_1^2 + v_2^2 + v_3^2} \). For \( \mathbf{v} = 0\mathbf{i} + 5\mathbf{j} - 3\mathbf{k} \),\( \ |\mathbf{v}| = \sqrt{0^2 + 5^2 + (-3)^2} = \sqrt{0 + 25 + 9} = \sqrt{34} \).Similarly, the magnitude \(|\mathbf{u}|\) is \( \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \).

Step 3: Calculating Cosine of the Angle

Use the formula for the cosine of the angle between two vectors: \( \cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}||\mathbf{u}|} \). Substitute the known values: \( \cos \theta = \frac{2}{\sqrt{34} \times \sqrt{3}} = \frac{2}{\sqrt{102}} \). Simplify it as \( \cos \theta = \frac{2\sqrt{102}}{102} \).

Step 4: Scalar Component of \(\mathbf{u}\) in Direction of \(\mathbf{v}\)

The scalar component of \( \mathbf{u} \) along \( \mathbf{v} \) is given by: \( \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|} \). Using the values found: \( \frac{2}{\sqrt{34}} \). Simplify to find the exact scalar component.

Step 5: Calculating the Vector Projection

To find the vector projection of \( \mathbf{u} \) along \( \mathbf{v} \), use the formula: \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2} \mathbf{v} \). We have: \( |\mathbf{v}|^2 = 34 \), and thus, \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{2}{34} (5\mathbf{j} - 3\mathbf{k}) = \left(\frac{10}{34}\right)\mathbf{j} - \left(\frac{6}{34}\right)\mathbf{k} \). Simplify to \( = \frac{5}{17}\mathbf{j} - \frac{3}{17}\mathbf{k} \).

Key Concepts

Dot Product

The dot product is a fundamental operation in vector calculus that helps quantify how much of one vector extends in the direction of another. When two vectors, say \( \mathbf{a} \) and \( \mathbf{b} \), are given by their components, the dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated using the formula:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)

In the provided exercise, the vectors \( \mathbf{v} = 0\mathbf{i} + 5\mathbf{j} - 3\mathbf{k} \) and \( \mathbf{u} = 1\mathbf{i} + 1\mathbf{j} + 1\mathbf{k} \) were used to find the dot product as \( 0 \times 1 + 5 \times 1 + (-3) \times 1 = 2 \). The dot product being 2 indicates that these vectors have a specific angle relationship that affects their interaction.

Magnitude of a Vector

Understanding the concept of a vector's magnitude is essential as it represents a vector's length, which is a measure of its extent in space. The magnitude for a vector \( \mathbf{a} \) with components \( (a_1, a_2, a_3) \) is computed using:

• \( |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \)

In the problem, for \( \mathbf{v} \), the magnitude is calculated as \( |\mathbf{v}| = \sqrt{0^2 + 5^2 + (-3)^2} = \sqrt{34} \). Similarly, \( \mathbf{u} \) is \( |\mathbf{u}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \). Magnitudes are crucial in quantifying the size or impact of these vectors in computations like projections.

Cosine of Angle Between Vectors

The cosine of the angle \( \theta \) between two vectors \( \mathbf{v} \) and \( \mathbf{u} \) can be found by utilizing their dot product and magnitudes in the formula:

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

In our exercise, we compute it as \( \frac{2}{\sqrt{34}\sqrt{3}} = \frac{2\sqrt{102}}{102} \). This result provides insight into how aligned or misaligned the vectors are. A cosine of zero would mean the vectors are perpendicular, while a value close to one or negative one indicates they are pointing in nearly the same or opposite directions.

Scalar Component

The scalar component of a vector, say \( \mathbf{u} \), in the direction of another vector \( \mathbf{v} \) provides a single measure of \( \mathbf{u} \) that specifically quantifies its contribution along \( \mathbf{v} \).The formula used is:

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

For the exercise, this is calculated as \( \frac{2}{\sqrt{34}} \). This scalar component shows how much of \( \mathbf{u} \) is "pushed" in the direction of \( \mathbf{v} \), which is particularly useful in physics, for instance, in determining how much work a force does when it acts along a path.

Vector Projection

Vector projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is a vector that shows how much of \( \mathbf{u} \) lies in the direction of \( \mathbf{v} \). This is calculated using:

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

In this exercise, it simplifies to:

• \( \frac{2}{34} (5\mathbf{j} - 3\mathbf{k}) = \frac{5}{17}\mathbf{j} - \frac{3}{17}\mathbf{k} \)

This form of projection is crucial in applications like physics and engineering for resolving forces or understanding the thrust direction of a velocity vector in dynamics.