Problem 57 - Fundamentals of Physics

Question

If \(\vec{B}\) is added to \(\vec{A}\), the result is \(6.0 \hat{\mathrm{i}}+1.0 \hat{\mathrm{j}} .\) If \(\vec{B}\) is subtracted from \(\vec{A}\), the result is \(-4.0 \hat{\mathrm{i}}+7.0 \hat{\mathrm{j}}\). What is the magnitude of \(\vec{A} ?\)

Step-by-Step Solution

Answer

The magnitude of \(\vec{A}\) is \(\sqrt{17}\).

Step 1: Understand the Problem

We are given two vector equations involving vectors \(\vec{A}\) and \(\vec{B}\). The first equation represents the sum \(\vec{A} + \vec{B} = 6.0 \hat{\mathrm{i}} + 1.0 \hat{\mathrm{j}}\). The second equation represents the difference \(\vec{A} - \vec{B} = -4.0 \hat{\mathrm{i}} + 7.0 \hat{\mathrm{j}}\). We need to find \(\vec{A}\).

Step 2: Set Up the System of Equations

Write the given vector equations:\[1. \quad \vec{A} + \vec{B} = 6.0 \hat{\mathrm{i}} + 1.0 \hat{\mathrm{j}} \2. \quad \vec{A} - \vec{B} = -4.0 \hat{\mathrm{i}} + 7.0 \hat{\mathrm{j}}\]

Step 3: Solve for \(\vec{A}\)

To find \(\vec{A}\), add the two equations:\[(\vec{A} + \vec{B}) + (\vec{A} - \vec{B}) = (6.0 \hat{\mathrm{i}} + 1.0 \hat{\mathrm{j}}) + (-4.0 \hat{\mathrm{i}} + 7.0 \hat{\mathrm{j}})\]Simplify:\[2\vec{A} = (6.0 - 4.0) \hat{\mathrm{i}} + (1.0 + 7.0) \hat{\mathrm{j}}\]\[2\vec{A} = 2.0 \hat{\mathrm{i}} + 8.0 \hat{\mathrm{j}}\]

Step 4: Isolate \(\vec{A}\)

Divide both sides by 2:\[\vec{A} = \frac{1}{2}(2.0 \hat{\mathrm{i}} + 8.0 \hat{\mathrm{j}})\]\[\vec{A} = 1.0 \hat{\mathrm{i}} + 4.0 \hat{\mathrm{j}}\]

Step 5: Calculate the Magnitude of \(\vec{A}\)

The magnitude of a vector \(\vec{V} = a \hat{\mathrm{i}} + b \hat{\mathrm{j}}\) is given by \(\sqrt{a^2 + b^2}\). Apply this formula to \(\vec{A}\):\[|\vec{A}| = \sqrt{(1.0)^2 + (4.0)^2} = \sqrt{1 + 16} = \sqrt{17}\]

Key Concepts

Vector Magnitude

When working with vectors, one of the most important aspects is understanding how to measure their size, or magnitude. The magnitude of a vector gives us an idea of its length. For a vector represented in component form like \( \vec{V} = a \hat{\mathrm{i}} + b \hat{\mathrm{j}} \), the magnitude is found using the formula \( |\vec{V}| = \sqrt{a^2 + b^2} \). This is essentially the Pythagorean theorem adapted for vectors in a two-dimensional space.
Calculating the magnitude involves squaring each component of the vector, adding these squares together, and then taking the square root of this sum. In the provided exercise, vector \( \vec{A} \) is defined as \( 1.0 \hat{\mathrm{i}} + 4.0 \hat{\mathrm{j}} \). Applying the magnitude formula, we calculate:

• Square each component: \((1.0)^2 = 1\) and \((4.0)^2 = 16\).
• Add the squares: \(1 + 16 = 17\).
• Take the square root: \(\sqrt{17} \). This is the magnitude of \( \vec{A} \).

Understanding vector magnitude is crucial in physics and engineering, as it helps in determining the size and influence of forces, velocities, and other vector quantities.

Vector Components

Vectors are often represented using their components along the axes of a coordinate system. This breakdown is useful because it simplifies calculations such as vector addition and subtraction. A vector in a two-dimensional space is typically expressed in terms of its components along the horizontal axis (\(\hat{\mathrm{i}}\)) and vertical axis (\(\hat{\mathrm{j}}\)).
For example, the vector \( \vec{A} = 1.0 \hat{\mathrm{i}} + 4.0 \hat{\mathrm{j}} \) is comprised of:

• An x-component, \(1.0 \hat{\mathrm{i}}\), which tells us how far \( \vec{A} \) extends along the x-axis.
• A y-component, \(4.0 \hat{\mathrm{j}}\), which tells us how far \( \vec{A} \) extends along the y-axis.

By focusing on components, you can tackle more complicated vector problems by dealing with one dimension at a time. The process involves using basic arithmetic on each component instead of working with the vector as a whole.
For the exercise, the vectors \( \vec{A} + \vec{B} \) and \( \vec{A} - \vec{B} \) are provided in terms of components. Analyzing these separately helps establish relations between the vectors, leading eventually to solving for the unknown vector \( \vec{A} \).

Equation System

Approaching vector problems often involves setting up systems of equations. This technique is extremely helpful when you have multiple relationships, like vector addition and subtraction, that are simultaneously true. In this exercise, the problem provides two vector equations:

• \( \vec{A} + \vec{B} = 6.0 \hat{\mathrm{i}} + 1.0 \hat{\mathrm{j}} \)
• \( \vec{A} - \vec{B} = -4.0 \hat{\mathrm{i}} + 7.0 \hat{\mathrm{j}} \)

These represent a system of equations involving vector components. To find \( \vec{A} \), adding these two equations together effectively eliminates \( \vec{B} \), leaving a single equation in terms of \( \vec{A} \):
\[ 2\vec{A} = (2.0 \hat{\mathrm{i}} + 8.0 \hat{\mathrm{j}}) \]
From this equation, \( \vec{A} \) can be isolated by dividing by 2, resulting in the components of \( \vec{A} \).
Using systems of equations is a powerful technique because it simplifies problems by reducing the number of unknowns. In the context of vectors, solving these systems allows you to find specific vector quantities when given composite operations.