Problem 35 - Essential Calculus Early Transcendentals

Question

If \(\mathbf{a}=\langle 3,0,-1\rangle,\) find a vector \(\mathbf{b}\) such that comp \(_{\mathbf{a}} \mathbf{b}=2\)

Step-by-Step Solution

Answer

Vector \( \mathbf{b} = \langle 1, 0, 3 - 2\sqrt{10} \rangle \) satisfies the condition.

Step 1: Understand Vector Projection

The component of vector \( \mathbf{b} \) in the direction of vector \( \mathbf{a} \) is given by \( \text{comp}_{\mathbf{a}} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \|} \). We need \( \text{comp}_{\mathbf{a}} \mathbf{b} = 2 \). This equation will help us find the dot product of \( \mathbf{a} \) and \( \mathbf{b} \).

Step 2: Calculate Magnitude of \(\mathbf{a}\)

Calculate the magnitude of vector \( \mathbf{a} = \langle 3,0,-1 \rangle \) using the formula \( \| \mathbf{a} \| = \sqrt{3^2 + 0^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \).

Step 3: Derive the Equation from the Given Component

Since \( \text{comp}_{\mathbf{a}} \mathbf{b} = 2 \), we can plug in the magnitude calculated previously: \[ \frac{\mathbf{a} \cdot \mathbf{b}}{\sqrt{10}} = 2 \]. This simplifies to \( \mathbf{a} \cdot \mathbf{b} = 2\sqrt{10} \).

Step 4: Apply Dot Product Formula

Using the definition \( \mathbf{a} \cdot \mathbf{b} = 3b_1 + 0 \cdot b_2 - 1 \cdot b_3 = 3b_1 - b_3 \), substitute into the equation derived: \( 3b_1 - b_3 = 2\sqrt{10} \).

Step 5: Choose a Variable and Solve for the Vector

To find a vector \( \mathbf{b} \), express one variable in terms of the other. Let \( b_1 = t \), then \( 3t - b_3 = 2\sqrt{10} \), so \( b_3 = 3t - 2\sqrt{10} \). Assign any value to \( t \) to get a vector \( \mathbf{b} \). For simplicity, select \( t = 1 \), which gives \( b_3 = 3 - 2\sqrt{10} \). Therefore, a vector \( \mathbf{b} \) that satisfies the condition is \( \langle 1, 0, 3 - 2\sqrt{10} \rangle \).

Step 6: Verify the Solution

Confirm the vector \( \mathbf{b} = \langle 1, 0, 3 - 2\sqrt{10} \rangle \) gives the required component. Calculate the dot product \( 3 \times 1 + 0 \times 0 - 1 \times (3 - 2\sqrt{10}) = 3 - (3 - 2\sqrt{10}) = 2\sqrt{10} \). Divide by \( \sqrt{10} \) to verify: \( \frac{2\sqrt{10}}{\sqrt{10}} = 2 \), confirming correctness.

Key Concepts

Dot Product

The dot product is a fundamental concept in vector algebra. It calculates the product of two vectors and leads to a scalar value. This operation is crucial in finding the angle between vectors and projections between vectors.
When you have two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), their dot product is given by:

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

This formula gives a scalar result, which is useful in various applications such as computing the projection of one vector onto another or determining orthogonality. For example, if the dot product equals zero, the vectors are orthogonal (perpendicular).
In our problem, the vector \( \mathbf{a} \) was \( \langle 3,0,-1 \rangle \) and the formula helped us find the expression for the dot product with vector \( \mathbf{b} \): \( 3b_1 - b_3 \). This equation became the basis for deriving further steps.

Magnitude of a Vector

The magnitude of a vector, often denoted as \( \| \mathbf{a} \| \), measures the length or size of the vector. It's like finding the distance from the origin to the point represented by the vector.
For a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \), the magnitude is calculated by:

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

This formula utilizes the Pythagorean theorem in three-dimensional space, providing an essential tool in navigation, physics, and engineering to understand how far a point is from the origin.
In our exercise, the magnitude of \( \mathbf{a} = \langle 3, 0, -1 \rangle \) was determined as \( \sqrt{10} \), a crucial step since it was used in calculating the component of \( \mathbf{b} \) in the direction of \( \mathbf{a} \). Understanding the magnitude helps bridge the transition from abstract vector concepts to real-world applications.

Vector Equation

A vector equation is used to describe a line, plane, or solution set in vector form. It's an efficient way to express complex systems or conditions between vectors.
When solving vector problems, using vector equations allows us to express relationships clearly and concisely. Each element in an equation corresponds to a component in the vector, simplifying the algebraic process. In our given problem, by setting \( \text{comp}_{\mathbf{a}} \mathbf{b} \) to 2, we organized the information using a vector equation:

• \( \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \|} = 2 \)

This equation was solved for the dot product, helping to identify an entire set of possible vectors \( \mathbf{b} \).
The flexibility of vector equations allows for assigning a variable, such as \( b_1 = t \), leading to multiple solutions satisfying the condition. This adaptability is crucial in mathematics and engineering fields, highlighting the strength and usefulness of vector-based problem-solving methods.