Finding A Vector With Magnitude 6 And Component Ratios

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In mathematics, particularly in linear algebra and vector calculus, finding vectors that satisfy certain conditions is a fundamental task. This often involves working with the magnitude and components of vectors in a coordinate system. Let's delve into the problem of finding a vector v with a given magnitude and specific relationships between its components. This article will explore the process, mathematical principles, and practical steps involved in solving this type of problem.

Problem Statement

Consider the problem where we need to find a vector v that meets the following criteria:

  1. The magnitude (or length) of v is 6.
  2. The component of v in the i direction (x-component) is twice the component in the j direction (y-component).

Our goal is to determine the vector v that satisfies both of these conditions. This involves understanding the relationship between a vector's magnitude, its components, and how to express these conditions mathematically.

Understanding Vectors and Components

Before diving into the solution, it’s crucial to understand the basic concepts of vectors and their components.

Vector Representation

A vector in a two-dimensional space (like the Cartesian plane) can be represented as an ordered pair v = (x, y), where x and y are the components of the vector along the i (x-axis) and j (y-axis) directions, respectively. These components represent the vector's projection onto the respective axes. For instance, if we have a vector v = (4, 2), the x-component is 4, and the y-component is 2. This means the vector extends 4 units along the x-axis and 2 units along the y-axis.

Magnitude of a Vector

The magnitude (or length) of a vector v = (x, y) is denoted as ||v|| and can be calculated using the Pythagorean theorem. The formula is:

||v|| = √(x² + y²)

This formula essentially calculates the Euclidean distance from the origin (0, 0) to the point (x, y) in the Cartesian plane. The magnitude gives us a scalar value representing the length of the vector, regardless of its direction. For example, if v = (3, 4), its magnitude is:

||v|| = √(3² + 4²) = √(9 + 16) = √25 = 5

So, the length of the vector v is 5 units.

Setting up the Equations

To find the vector v that satisfies the given conditions, we need to translate the conditions into mathematical equations. Let v = (x, y) be the vector we are looking for. We have two conditions:

  1. The magnitude of v is 6, which gives us the equation:

    √(x² + y²) = 6

  2. The x-component is twice the y-component, which gives us:

    x = 2y

Now we have a system of two equations with two variables (x and y), which we can solve to find the components of the vector v.

Equation 1: Magnitude Condition

The first condition states that the magnitude of v must be 6. Using the formula for the magnitude of a vector, we have:

√(x² + y²) = 6

To simplify, we can square both sides of the equation to eliminate the square root:

x² + y² = 36

This equation represents a circle centered at the origin with a radius of 6. Any vector whose endpoint lies on this circle will have a magnitude of 6.

Equation 2: Component Relationship

The second condition states that the x-component (x) is twice the y-component (y). This can be written as:

x = 2y

This equation represents a line in the Cartesian plane. It tells us that for any point on this line, the x-coordinate is twice the y-coordinate.

Solving the System of Equations

Now that we have our two equations:

  1. x² + y² = 36
  2. x = 2y

We can solve this system to find the values of x and y that satisfy both conditions. The method of substitution is a straightforward approach for this system.

Using Substitution Method

Since we already have x expressed in terms of y in the second equation, we can substitute this expression into the first equation:

(2y)² + y² = 36

Expanding and simplifying, we get:

4y² + y² = 36

5y² = 36

Now, divide both sides by 5:

y² = 36/5

Taking the square root of both sides gives us two possible values for y:

y = ±√(36/5)

y = ±(6/√5)

To rationalize the denominator, we multiply the numerator and denominator by √5:

y = ±(6√5)/5

So we have two values for y: y = (6√5)/5 and y = -(6√5)/5.

Finding the Corresponding x-values

Now that we have the values for y, we can use the equation x = 2y to find the corresponding values for x.

  1. For y = (6√5)/5:

    x = 2 * (6√5)/5

    x = (12√5)/5

  2. For y = -(6√5)/5:

    x = 2 * -(6√5)/5

    x = -(12√5)/5

Thus, we have two possible vectors that satisfy the given conditions:

  1. v₁ = ((12√5)/5, (6√5)/5)
  2. v₂ = (-(12√5)/5, -(6√5)/5)

Verifying the Solution

To ensure our solutions are correct, we need to verify that both vectors satisfy the original conditions:

  1. Magnitude of v is 6.
  2. x-component is twice the y-component.

Verifying Vector v₁ = ((12√5)/5, (6√5)/5)

  1. Magnitude: ||v₁|| = √[((12√5)/5)² + ((6√5)/5)²]

    ||v₁|| = √[(144 * 5)/25 + (36 * 5)/25]

    ||v₁|| = √[(720/25) + (180/25)]

    ||v₁|| = √(900/25)

    ||v₁|| = √36 = 6

    The magnitude condition is satisfied.

  2. Component Relationship: x = (12√5)/5 and y = (6√5)/5

    x = 2 * y

    (12√5)/5 = 2 * (6√5)/5

    (12√5)/5 = (12√5)/5

    The component relationship is also satisfied.

Verifying Vector v₂ = (-(12√5)/5, -(6√5)/5)

  1. Magnitude: ||v₂|| = √[(-(12√5)/5)² + (-(6√5)/5)²]

    ||v₂|| = √[(144 * 5)/25 + (36 * 5)/25]

    ||v₂|| = √[(720/25) + (180/25)]

    ||v₂|| = √(900/25)

    ||v₂|| = √36 = 6

    The magnitude condition is satisfied.

  2. Component Relationship: x = -(12√5)/5 and y = -(6√5)/5

    x = 2 * y

    -(12√5)/5 = 2 * -(6√5)/5

    -(12√5)/5 = -(12√5)/5

    The component relationship is also satisfied.

Both vectors v₁ and v₂ satisfy the given conditions, confirming our solutions.

Conclusion

We successfully found two vectors that meet the specified conditions: a magnitude of 6 and an x-component twice the y-component. The vectors are:

  1. v₁ = ((12√5)/5, (6√5)/5)
  2. v₂ = (-(12√5)/5, -(6√5)/5)

This problem demonstrates the application of vector concepts, magnitude calculation, and solving systems of equations. Understanding these principles is crucial for further studies in mathematics, physics, and engineering, where vectors are frequently used to model various phenomena.

Practical Applications

Finding vectors with specific magnitudes and component relationships has many practical applications in various fields:

Physics

In physics, vectors are used to represent forces, velocities, and accelerations. Determining vectors with specific properties is essential in solving problems related to motion, equilibrium, and dynamics. For instance, when analyzing projectile motion, physicists need to find the initial velocity vector that satisfies certain conditions, such as the launch angle and initial speed.

Engineering

Engineers use vectors in structural analysis, where they need to determine the forces acting on a structure. They also use vectors in control systems, robotics, and computer graphics. For example, in robotics, vectors are used to represent the position and orientation of robotic arms, and engineers need to calculate the joint angles to achieve desired movements. Similarly, in computer graphics, vectors are used to represent the vertices of 3D models, and transformations are applied to these vectors to create animations and visual effects.

Computer Science

Vectors are used extensively in computer graphics, game development, and machine learning. In computer graphics, vectors represent the position, direction, and scaling of objects. Game developers use vectors to simulate the movement of game characters and objects. In machine learning, vectors are used to represent data points in high-dimensional spaces, and algorithms are used to find patterns and relationships between these vectors.

Navigation

In navigation systems, vectors are used to represent the direction and speed of movement. Global Positioning System (GPS) devices use vectors to calculate the user's position and direction of travel. Pilots and sailors use vectors to plot courses and navigate through the air and sea.

Further Exploration

To deepen your understanding of vectors, consider exploring these related topics:

Vector Addition and Subtraction

Learn how to add and subtract vectors geometrically and algebraically. Understanding vector addition is crucial for analyzing the combined effect of multiple forces or velocities.

Dot Product and Cross Product

Explore the dot product and cross product of vectors, which are essential tools for calculating angles, projections, and areas in vector spaces. The dot product is used to find the component of one vector along another, while the cross product is used to find a vector perpendicular to two given vectors.

Linear Algebra

Study linear algebra to understand the broader context of vectors, matrices, and linear transformations. Linear algebra provides the mathematical foundation for many areas of science and engineering.

Vector Calculus

Delve into vector calculus to learn about vector-valued functions, derivatives, and integrals, which are essential for analyzing curves, surfaces, and fields in three-dimensional space.

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions of multiple variables. It provides the tools to analyze vector fields, gradients, and line integrals, which are crucial in physics and engineering.

By mastering these concepts and techniques, you can tackle more complex problems involving vectors and their applications in various fields. Vectors are a fundamental tool in mathematics, science, and engineering, and a solid understanding of their properties and operations is essential for success in these disciplines.

In conclusion, the process of finding a vector with a specific magnitude and component ratio is a valuable exercise in understanding vector properties and their applications. By setting up equations based on the given conditions and solving them, we can determine the vector that satisfies the criteria. This skill is applicable in various fields, making it an essential concept for students and professionals alike.