Problem 33 - 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 \(\mathbf{w}=\mathbf{i}+5 \mathbf{j}-3 \mathbf{k}\). \(\operatorname{proj}_{\mathbf{k}} \mathbf{u}\)

Step-by-Step Solution

Answer

The projection \( \operatorname{proj}_{\mathbf{k}} \mathbf{u} \) is \( \mathbf{k} \).

Step 1: Understand Projection Formula

The projection of a vector \( \mathbf{u} \) onto another vector \( \mathbf{a} \) is given by the formula \( \operatorname{proj}_{\mathbf{a}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{a}}{\|\mathbf{a}\|^2} \mathbf{a} \). To find \( \operatorname{proj}_{\mathbf{k}} \mathbf{u} \), you need to identify the direction vector, which in this case is \( \mathbf{k} = 0\mathbf{i} + 0\mathbf{j} + 1\mathbf{k} \).

Step 2: Find Dot Product

Calculate the dot product \( \mathbf{u} \cdot \mathbf{k} \). This is calculated as follows: \( (3\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \cdot (0\mathbf{i} + 0\mathbf{j} + 1\mathbf{k}) = 3 \cdot 0 + 2 \cdot 0 + 1 \cdot 1 = 1 \).

Step 3: Calculate Magnitude Squared

Find the magnitude squared \( \|\mathbf{k}\|^2 \). Since \( \mathbf{k} \) is the unit vector in the \( k \) direction, \( \|\mathbf{k}\| = 1 \). Therefore, \( \|\mathbf{k}\|^2 = 1^2 = 1 \).

Step 4: Compute Projection

Now, use the projection formula: \( \operatorname{proj}_{\mathbf{k}} \mathbf{u} = \frac{1}{1} (0\mathbf{i} + 0\mathbf{j} + 1\mathbf{k}) = \mathbf{k} = 0\mathbf{i} + 0\mathbf{j} + 1\mathbf{k} \). Therefore, the projection of \( \mathbf{u} \) onto \( \mathbf{k} \) is simply \( \mathbf{k} \).

Key Concepts

Dot Product

The dot product is a fundamental operation in vector algebra that allows you to measure the interaction between two vectors. It is represented by the symbol \( \cdot \) and involves multiplying corresponding components of two vectors and summing those products.
For 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 \( \mathbf{a} \cdot \mathbf{b} \) is given by:

• \( a_1b_1 + a_2b_2 + a_3b_3 \)

This operation is particularly useful because the dot product is zero if the vectors are perpendicular, indicating no component of one vector acts along the other.
In vector projection, we use the dot product to find how much of one vector goes in the direction of another, which is crucial for calculating projections correctly.

Magnitude of a Vector

The magnitude of a vector, often represented by \( \| \mathbf{a} \| \), is its length or size. It is calculated using the Pythagorean theorem for the components of the vector.
For a vector \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \), the magnitude is calculated as:

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

Understanding the magnitude is vital in the projection formula, where it appears in the denominator and gives the projection its scaling factor.
This tells us how large or small the resulting projection should be based on the direction vector's size.
Moreover, in this particular exercise, \( \mathbf{k} \) is a unit vector, so its magnitude is 1, simplifying the projection calculations.

Unit Vector

A unit vector is a vector with a magnitude of 1. It provides a direction without any influence of scale. To convert any non-zero vector into a unit vector, you divide it by its magnitude.
Mathematically, if \( \mathbf{a} \) is a vector, its unit vector \( \hat{\mathbf{a}} \) is given by:

• \( \hat{\mathbf{a}} = \frac{\mathbf{a}}{\| \mathbf{a} \|} \)

Unit vectors are essential for directional calculations because they only affect the direction and not the magnitude when used in formulas like vector projection.
In our exercise, the vector \( \mathbf{k} \) is already a unit vector, which means that when we project onto it, the direction is preserved without needing additional calculations for normalization.

Projection Formula

The projection formula is used to project a vector \( \mathbf{u} \) onto another vector \( \mathbf{a} \). This is visualized as the shadow that is cast by one vector onto the line defined by another.
The formula for the projection of \( \mathbf{u} \) onto \( \mathbf{a} \) is:

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

This calculation is vital in many fields such as physics and computer graphics as it isolates the portion of \( \mathbf{u} \) acting in the direction of \( \mathbf{a} \). Importantly, the formula relies heavily on the dot product to determine how much vector \( \mathbf{u} \) aligns with vector \( \mathbf{a} \) and the square of the magnitude to scale it correctly.
By understanding these components, the concept of projection becomes clear and approachable.