Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 29-34, 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

\( \operatorname{proj}_{\mathbf{u}} \mathbf{v} = \frac{15}{14}\mathbf{i} + \frac{10}{14}\mathbf{j} + \frac{5}{14}\mathbf{k} \).

Step 1: Understanding the Problem

We need to find the projection of vector \( \mathbf{v} \) onto vector \( \mathbf{u} \). The projection of a vector \( \mathbf{b} \) onto another vector \( \mathbf{a} \) is given by the formula: \( \operatorname{proj}_{\mathbf{a}} \mathbf{b} = \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{a} \cdot \mathbf{a}} \right) \mathbf{a} \).

Step 2: Calculating Dot Product \( \mathbf{u} \cdot \mathbf{v} \)

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

Step 3: Calculating Dot Product \( \mathbf{u} \cdot \mathbf{u} \)

Calculate the dot product of \( \mathbf{u} \) with itself:\( \mathbf{u} \cdot \mathbf{u} = (3)(3) + (2)(2) + (1)(1) = 9 + 4 + 1 = 14 \).

Step 4: Finding the Projection Scalar

Use the projection formula to find the scalar multiplier:\( \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{u} \cdot \mathbf{u}} = \frac{5}{14} \).

Step 5: Calculating \( \operatorname{proj}_{\mathbf{u}} \mathbf{v} \)

Multiply vector \( \mathbf{u} \) by the scalar from the previous step:\( \operatorname{proj}_{\mathbf{u}} \mathbf{v} = \frac{5}{14} \times (3\mathbf{i} + 2\mathbf{j} + \mathbf{k}) = \frac{15}{14}\mathbf{i} + \frac{10}{14}\mathbf{j} + \frac{5}{14}\mathbf{k} \).

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector algebra. It involves two vectors and results in a scalar, a single number. This operation is particularly useful when you need to determine angles between vectors, or in our case, when projecting one vector onto another.

To compute the dot product between two vectors, you multiply their corresponding components and then sum the results. For example, given vectors \(\mathbf{u} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) and \(\mathbf{v} = 2\mathbf{i} - \mathbf{k}\), the dot product \(\mathbf{u} \cdot \mathbf{v}\) is calculated as follows:

\[\mathbf{u} \cdot \mathbf{v} = 3 \times 2 + 2 \times 0 + 1 \times (-1) = 6 + 0 - 1 = 5.\]

This result is pivotal as it is used in calculating projections and other vector operations. Remember, if the dot product is zero, the vectors are perpendicular.

Vectors in 3D Space

Vectors in 3D space have three components and are expressed in terms of the unit vectors \(\mathbf{i}, \mathbf{j}, \text{ and } \mathbf{k}\), which represent the x, y, and z directions respectively. Each vector in this space can be considered as an arrow pointing from the origin to a point given by its components.

In this exercise, you encounter vectors like \(\mathbf{u} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k}\), which means it moves 3 units along the x-axis, 2 units along the y-axis, and 1 unit along the z-axis from the origin.

Understanding vectors in 3D involves:

• Visualizing their direction and magnitude.
• Performing operations such as addition, subtraction, and multiplication by scalars.
• Computation of dot products and cross products for finding angles and areas.

Remember that the three-dimensional vector space allows us to describe an object's orientation and position in a more comprehensive and spatial way.

Projection Formula

The projection of one vector onto another represents the component of the first vector that runs in the direction of the second. It's like casting a shadow of one vector onto the line created by another, measuring how much of the vector aligns with the other one.

The formula for the projection of vector \(\mathbf{b}\) onto vector \(\mathbf{a}\) is:
\[\operatorname{proj}_{\mathbf{a}} \mathbf{b} = \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{a} \cdot \mathbf{a}} \right) \mathbf{a}.\]

This formula can be broken down into simple steps:

• Calculate the dot product \(\mathbf{a} \cdot \mathbf{b}\) to determine how much \(\mathbf{b}\) is directed along \(\mathbf{a}\).
• Compute \(\mathbf{a} \cdot \mathbf{a}\) to find the magnitude squared of \(\mathbf{a}\).
• Divide and multiply the vector \(\mathbf{a}\) by this ratio to get the projection vector.

In our scenario, this approach yields \(\operatorname{proj}_{\mathbf{u}} \mathbf{v} = \frac{5}{14} \times (3\mathbf{i} + 2\mathbf{j} + \mathbf{k})\), which produces the vector \(\frac{15}{14}\mathbf{i} + \frac{10}{14}\mathbf{j} + \frac{5}{14}\mathbf{k}\). This projection vector illustrates the part of \(\mathbf{v}\) that runs in the direction of \(\mathbf{u}\).