Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find each of the given projections if \(\mathbf{u}=3 \mathbf{i}+2 \mathbf{j}+\mathbf{k}, \mathbf{v}=2 \mathbf{i}-\mathbf{k}\), and \(\mathbf{w}=\mathbf{i}+5 \mathbf{j}-3 \mathbf{k}\). \(\operatorname{proj}_{\mathbf{u}} \mathbf{v}\)

Step-by-Step Solution

Answer

The projection of \( \mathbf{v} \) onto \( \mathbf{u} \) is \( \frac{15}{14}\mathbf{i} + \frac{10}{14}\mathbf{j} + \frac{5}{14}\mathbf{k} \).

Step 1: Find the Dot Product of \( \mathbf{u} \) and \( \mathbf{v} \)

Calculate the dot product \( \mathbf{u} \cdot \mathbf{v} \). Given \( \mathbf{u} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} \) and \( \mathbf{v} = 2\mathbf{i} - \mathbf{k} \), the dot product is calculated as follows: \( (3)(2) + (2)(0) + (1)(-1) = 6 + 0 - 1 = 5 \).

Step 2: Find the Magnitude of \( \mathbf{u} \) Squared

Calculate the square of the magnitude of \( \mathbf{u} \). The magnitude of \( \mathbf{u} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} \) is \( \sqrt{3^2 + 2^2 + 1^2} = \sqrt{14} \). Thus, the magnitude squared is \( 14 \).

Step 3: Calculate the Projection of \( \mathbf{v} \) onto \( \mathbf{u} \)

Use the formula for projection: \( \operatorname{proj}_{\mathbf{u}} \mathbf{v} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{u} \|^2} \right) \mathbf{u} \). Substitute the calculated values: \( \operatorname{proj}_{\mathbf{u}} \mathbf{v} = \left( \frac{5}{14} \right) (3\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \).

Step 4: Find the Scaled Vector

Distribute \( \frac{5}{14} \) to each component of \( \mathbf{u} \): \( \operatorname{proj}_{\mathbf{u}} \mathbf{v} = \frac{15}{14}\mathbf{i} + \frac{10}{14}\mathbf{j} + \frac{5}{14}\mathbf{k} \). Simplify if necessary.

Key Concepts

Dot Product

The dot product is an essential operation in vector algebra. It measures how much two vectors align with each other. For two vectors, \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \) and \( \mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k} \), the dot product is calculated as:

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

To find the dot product of \( \mathbf{u} \) and \( \mathbf{v} \) from our example, we apply the components:

• \( (3)(2) + (2)(0) + (1)(-1) \)
• This results in \( 6 + 0 - 1 = 5 \)

Remember that the dot product is a **scalar**, and it can also tell you about the angle between the vectors. If the dot product is zero, the vectors are perpendicular. In our case, the result indicates some degree of alignment between the vectors, but they aren't orthogonal.

Magnitude of a Vector

The magnitude of a vector, also known as its length or norm, provides a measure of the size of the vector. It's calculated using the Pythagorean theorem in three-dimensional space. For a vector \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \), its magnitude is:

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

With our vector \( \mathbf{u} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} \), the magnitude is:

• \( \sqrt{3^2 + 2^2 + 1^2} = \sqrt{14} \)

Squaring the magnitude, as done in the projection formula, gives us the denominator in our projection equations:

• \( \| \mathbf{u} \|^2 = 14 \)

This step is crucial in normalizing the direction for computing the projection, as it scales the vector into a unit vector.

Projection Formula

The projection of one vector onto another is a way of showing how much of one vector lies in the direction of another. The formula to calculate the projection of vector \( \mathbf{v} \) onto vector \( \mathbf{u} \) is formally defined as:

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

Breaking this formula down:

• The fraction \( \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{u} \|^2} \) scales vector \( \mathbf{u} \) in the direction of \( \mathbf{v} \).
• This scalar multiplier tells us how much \( \mathbf{u} \) will be stretched or shrunk.
• The result is a vector that partially "projects" \( \mathbf{v} \) onto \( \mathbf{u} \), represented as \( \operatorname{proj}_{\mathbf{u}} \mathbf{v} = \frac{15}{14}\mathbf{i} + \frac{10}{14}\mathbf{j} + \frac{5}{14}\mathbf{k} \).

Using projection helps us understand real-world scenarios, such as breaking down forces into parallel components relative to a surface or line, which is a common application in physics and engineering.