Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the scalar projection of \(\mathbf{u}=-\mathbf{i}+5 \mathbf{j}+3 \mathbf{k}\) on \(\mathbf{v}=-\mathbf{i}+\mathbf{j}-\mathbf{k}\)

Step-by-Step Solution

Answer

The scalar projection is \( \sqrt{3} \).

Step 1: Understand the Scalar Projection Formula

The scalar projection of vector \( \mathbf{u} \) onto vector \( \mathbf{v} \) is given by the formula: \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{v} \|} \), where \( \mathbf{u} \cdot \mathbf{v} \) is the dot product of \( \mathbf{u} \) and \( \mathbf{v} \), and \( \| \mathbf{v} \| \) is the magnitude of vector \( \mathbf{v} \).

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

The dot product \( \mathbf{u} \cdot \mathbf{v} \) is calculated as: \( (-1)(-1) + (5)(1) + (3)(-1) = 1 + 5 - 3 = 3 \).

Step 3: Find the Magnitude of \( \mathbf{v} \)

The magnitude of \( \mathbf{v} \) is found using the formula \( \| \mathbf{v} \| = \sqrt{(-1)^2 + 1^2 + (-1)^2} \). This simplifies to \( \sqrt{1 + 1 + 1} = \sqrt{3} \).

Step 4: Compute the Scalar Projection

Substitute the dot product and magnitude into the scalar projection formula: \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{3}{\sqrt{3}} \). Simplify this to get \( \sqrt{3} \).

Key Concepts

Dot Product

The dot product is an essential operation in vector calculus and linear algebra. It helps us find the angle between two vectors, among other things. To understand the dot product, consider 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 of these vectors is given by the formula:

• \( \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \)

This formula shows that you multiply corresponding components and add the results.
In the original exercise, the dot product of vectors \( \mathbf{u} = -\mathbf{i} + 5 \mathbf{j} + 3 \mathbf{k} \) and \( \mathbf{v} = -\mathbf{i} + \mathbf{j} - \mathbf{k} \) results in:

• \( (-1)(-1) + (5)(1) + (3)(-1) = 1 + 5 - 3 = 3 \)

The result, \( 3 \), is a scalar value.
Understanding this process of calculating the dot product is critical for tasks like finding the scalar projection of one vector onto another.

Magnitude of a Vector

The magnitude of a vector represents its length or size. In physics and engineering, it's often referred to as the "norm." For a vector \( \mathbf{v} = v_1 \mathbf{i} + v_2 \mathbf{j} + v_3 \mathbf{k} \), the magnitude is calculated as follows:

• \( \| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2 + v_3^2} \)

This formula involves squaring each component of the vector, summing these squares, and then taking the square root of the result.
In the original problem, the vector \( \mathbf{v} = -\mathbf{i} + \mathbf{j} - \mathbf{k} \) has a magnitude calculated as:

• \( \| \mathbf{v} \| = \sqrt{(-1)^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \)

The magnitude gives us a sense of the vector's scale, and it plays a crucial role when determining the scalar projection.

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and differential, as well as integral operations on vector spaces. This field is fundamental in physics and engineering because it provides tools to describe physical phenomena.

• Operations like dot product and calculating magnitudes are fundamental in vector calculus.
• These operations allow us to analyze forces, velocities, and other vector quantities in multi-dimensional spaces.

In the exercise, vector calculus provides the means to solve problems such as finding the scalar projection of one vector onto another.
The scalar projection, specifically, is a key concept in vector calculus. It gives us a real-life application of these operations, helping us understand how one vector behaves relative to another.

• By understanding such projections, you can interpret angles, distances, and other relational measures between vectors.

Approaching vector calculus with these basic tools ensures a solid foundation for more complex applications and real-world problem-solving.