Problem 58 - Algebra and Trigonometry

Question

Find a vector \(\mathbf{v}\) whose magnitude is 3 and whose component in the \(\mathbf{i}\) direction is equal to the component in the \(\mathbf{j}\) direction.

Step-by-Step Solution

Answer

The vector is \(\frac{3\sqrt{2}}{2} \textbf{i} + \frac{3\sqrt{2}}{2} \textbf{j}\) or \(-\frac{3\sqrt{2}}{2} \textbf{i} - \frac{3\sqrt{2}}{2} \textbf{j}\).

Step 1: Understand the Given Conditions

We need to find a vector \(\textbf{v}\) such that its magnitude is 3 and the components in the \(i\) and \(j\) directions are equal.

Step 2: Represent the Vector

Let \(\textbf{v} = a\textbf{i} + a\textbf{j} + b\textbf{k}\). We are given that the components in the \(i\) and \(j\) directions are equal, so both components are \(a\).

Step 3: Use Magnitude Formula

The magnitude of a vector \(\textbf{v} = a\textbf{i} + a\textbf{j} + b\textbf{k}\) is given by \[\| \textbf{v} \| = \sqrt{a^2 + a^2 + b^2}\].

Step 4: Set Magnitude Equal to 3

Since the magnitude of \(\textbf{v}\) is 3, we have \[\sqrt{a^2 + a^2 + b^2} = 3\]. This simplifies to \[\sqrt{2a^2 + b^2} = 3\].

Step 5: Solve for Components

Square both sides of the equation to get \[2a^2 + b^2 = 9\].

Step 6: Assume Simplest Case

Assume \(\textbf{v}\) lies entirely in the \(i\) and \(j\) plane, so \(\textbf{k}\) component \(b = 0\). We then get \(2a^2 = 9\).

Step 7: Solve for \(a\)

Solving for \(a\), we find \(a^2 = \frac{9}{2}\) and thus \(a = \pm \sqrt{\frac{9}{2}} = \pm \frac{3}{\sqrt{2}} = \pm \frac{3\sqrt{2}}{2}\).

Step 8: Write the Vector

Therefore, the vector can be \( \textbf{v} = \frac{3\sqrt{2}}{2} \textbf{i} + \frac{3\sqrt{2}}{2} \textbf{j} \) or \( \textbf{v} = -\frac{3\sqrt{2}}{2} \textbf{i} - \frac{3\sqrt{2}}{2} \textbf{j} \).

Key Concepts

vector components

In vector mathematics, each vector can be broken down into its individual components. These components represent the vector's effects along the coordinate axes (i.e., \textbf{i}, \textbf{j}, \textbf{k}) in a Cartesian coordinate system.
Understanding vector components allows us to simplify and manipulate vectors in equations. For example, a vector \textbf{v} can be expressed as \textbf{v} = a\textbf{i} + b\textbf{j} + c\textbf{k}, where:

• \(a\) is the component in the \textbf{i} direction (x-axis)
• \(b\) is the component in the \textbf{j} direction (y-axis)
• \(c\) is the component in the \textbf{k} direction (z-axis)

In our specific exercise, we determined that the components in the \textbf{i} and \textbf{j} directions are equal. This gives us \textbf{v} = a\textbf{i} + a\textbf{j} + b\textbf{k}, making it clear that both the x and y components have the same magnitude, \(a\).

magnitude formula

The magnitude of a vector represents its length and is crucial for understanding the vector's overall size and direction. For a given vector\(\textbf{v} = a\textbf{i} + b\textbf{j} + c\textbf{k}\), the magnitude is calculated using the formula:
\[\| \textbf{v} \| = \sqrt{a^2 + b^2 + c^2}\
\]
This formula helps us measure how long the vector is from its starting point to its end point.

In our exercise, we know the magnitude of the vector must be 3. Thus, we set the formula for magnitude equal to 3 and solve for the unknown components. After substituting the given values, our equation looks like this:\[\sqrt{a^2 + a^2 + b^2} = \sqrt{2a^2 + b^2} = 3\]
This helps us form an algebraic equation to solve for the components.

solving equations in algebra

Solving equations is a fundamental skill in algebra and essential for dealing with problems involving vectors. Whenever we're dealing with a vector or any equation, the goal is to isolate the variable and determine its value.
In our vector problem, we started with the equation from the magnitude formula:
\[\sqrt{2a^2 + b^2} = 3\
\]
We then square both sides to eliminate the square root and get:
\[2a^2 + b^2 = 9\
\]
Assuming the simplest case where the vector lies in the \(i\) and \(j\) plane, we let the \(b\) component be zero. This simplifies our equation to:
\[2a^2 = 9\
\]
From here, we just solve for \(a\) and find that:
\[a = \pm \sqrt{\frac{9}{2}} = \pm \frac{3}{\sqrt{2}} = \pm \frac{3\sqrt{2}}{2}\
\]
With these values, we can write our vector as either \(\textbf{v} = \frac{3\sqrt{2}}{2} \textbf{i} + \frac{3\sqrt{2}}{2} \textbf{j}\) or \(\textbf{v} = -\frac{3\sqrt{2}}{2} \textbf{i} - \frac{3\sqrt{2}}{2} \textbf{j}\). Solving such equations step-by-step makes it easier to understand each component's role and ensures the problem is correctly solved.