Problem 57 - Algebra and Trigonometry

Question

Find a vector \(\mathbf{v}\) whose magnitude is 4 and whose component in the \(\mathbf{i}\) direction is twice the component in the \(\mathbf{j}\) direction.

Step-by-Step Solution

Answer

The vector \( \textbf{v} \) is either \( \frac{8 \sqrt{5}}{5} \textbf{i} + \frac{4 \sqrt{5}}{5} \textbf{j} \) or \( -\frac{8 \sqrt{5}}{5} \textbf{i} - \frac{4 \sqrt{5}}{5} \textbf{j} \).

Step 1: Define the Vector Components

Define the components of the vector \(\textbf{v}\) in terms of \(\textbf{i}\) and \(\textbf{j}\). Let the component in the \(\textbf{j}\) direction be \(y\). Then, the component in the \(\textbf{i}\) direction will be \((2y)\). Therefore, we can write the vector \(\textbf{v}\) as \(\textbf{v} = 2y \textbf{i} + y \textbf{j}\).

Step 2: Calculate the Magnitude of the Vector

Use the formula for the magnitude of a vector, \(|\textbf{v}| = \sqrt{(2y)^2 + y^2} = \sqrt{4y^2 + y^2} = \sqrt{5y^2} = \sqrt{5} \abs{y}\). Given that the magnitude of \(\textbf{v}\) is 4, we set \(\textbf{4} = \sqrt{5} \abs{y}\).

Step 3: Solve for y

Solve for \(y\) in the equation \(\textbf{4} = \sqrt{5} \abs{y}\). First, isolate \(\textbf{y}\) by dividing both sides by \(\textbf{ \sqrt{5} }\): \( y = \pm{ \frac{4}{ \sqrt{5} }} \.\) Simplify by rationalizing the denominator: \( y = \pm{ \frac{4 \sqrt{5}}{5} }.\)

Step 4: Form the Vector \(\textbf{v}\)

Substitute the value of \(\textbf{y}\) back into the vector components. For \(y = \frac{4 \sqrt{5}}{5}\), \( 2y = \frac{8 \sqrt{5}}{5}\). Thus, \( \textbf{v} = \frac{8 \sqrt{5}}{5} \textbf{i} + \frac{4 \sqrt{5}}{5} \textbf{j}\). Similarly, for \(-y = -\frac{4 \sqrt{5}}{5}\), \(-2y = -\frac{8 \sqrt{5}}{5}\), so \( \textbf{v} = -\frac{8 \sqrt{5}}{5} \textbf{i} - \frac{4 \sqrt{5}}{5} \textbf{j}\).

Key Concepts

Vector Components

Vectors can be broken down into their components along standard axes. In typical 2D space, the vector \(\textbf{v}\) is described by its components along the \(\textbf{i}\) (horizontal) and \(\textbf{j}\) (vertical) directions. For this exercise, let's assign the component in the \(\textbf{j}\) direction as \(y\). According to the problem, the component in the \(\textbf{i}\) direction is twice that in the \(\textbf{j}\) direction, making it \(2y\). Therefore, the vector can be written mathematically as \(\textbf{v} = 2y \textbf{i} + y \textbf{j}\). Breaking vectors into components helps simplify complex problems and calculations.

Magnitude Calculation

To compute the magnitude of a vector, we use the formula \(\text{Magnitude of } \textbf{v} = \sqrt{x^2 + y^2}\). For a vector \(\textbf{v} = 2y \textbf{i} + y \textbf{j}\), we substitute the components into this formula. Thus, we get \(|\textbf{v}| = \sqrt{(2y)^2 + y^2} = \sqrt{4y^2 + y^2} = \sqrt{5y^2} = \sqrt{5} \abs{y}\). In the given problem, the magnitude is specified as 4. We then set up the equation \(\text{4} = \sqrt{5} \abs{y}\) and prepare to solve for \(y\).

Rationalizing the Denominator

When solving the equation \(\text{4} = \sqrt{5} \abs{y}\), we isolate \(y\) by dividing both sides by \(\text{\text{√5}}\). This yields \(y = \pm \frac{4}{\text{\text{√5}}}\). To make the expression easier to work with, we rationalize the denominator. This involves multiplying both the numerator and the denominator by \(\text{√5}\). Our equation then becomes \(y = \pm \frac{4 \text{\text{√5}}}{5}\). Rationalizing helps avoid irrational numbers in the denominator, simplifying further calculations.

Solving Equations

After rationalizing, we substitute the values of \(y\) back into the vector components to form the vector \(\textbf{v}\). There are two possible vectors since \(y\) can be positive or negative. If \(y = \frac{4 \text{\text{√5}}}{5}\), then \(2y = \frac{8 \text{\text{√5}}}{5}\). Substituting these results, we get \(\textbf{v} = \frac{8 \text{\text{√5}}}{5} \textbf{i} + \frac{4 \text{\text{√5}}}{5} \textbf{j}\). Alternatively, if \(y = -\frac{4 \text{\text{√5}}}{5}\), then \(2y = -\frac{8 \text{\text{√5}}}{5}\). Thus, the vector becomes \(\textbf{v} = -\frac{8 \text{\text{√5}}}{5} \textbf{i} - \frac{4 \text{\text{√5}}}{5} \textbf{j}\). Solving these equations provides us with the two potential vectors for the given problem.