Area Of Parallelogram With Vectors: Step-by-Step Solution

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Hey guys! Today, we're diving into a fascinating problem from vector algebra: calculating the area of a parallelogram defined by two vectors. This isn't just a theoretical exercise; understanding this concept helps in various fields like physics, engineering, and computer graphics. So, let's break it down step-by-step to ensure you've got a solid grasp on the process.

Problem Statement: Vectors and Parallelogram Area

So, the core of our problem is this: We're given two vectors, a and b, expressed in terms of basis vectors m and n. Specifically, a = (1/2)m - n and b = 2m + n. We also know the magnitudes of m and n ( |m| = 4, |n| = 2 ) and the angle between them (135°). Our mission? To find the area of the parallelogram formed by vectors a and b. This involves understanding how vectors combine, what a parallelogram is in vector terms, and how to leverage the given information to arrive at the answer. Think of it like this: vectors are the building blocks, the parallelogram is the structure we're constructing, and the area is the final measurement we need to calculate.

Breaking Down the Basics of Vectors

Before we jump into the nitty-gritty calculations, let's make sure we're all on the same page about vectors. Vectors, unlike simple numbers (scalars), have both magnitude (length) and direction. They're often represented as arrows, and their operations follow specific rules. Key concepts include:

  • Vector Addition and Subtraction: Vectors can be added or subtracted component-wise. If you have vectors u = (u1, u2) and v = (v1, v2), then u + v = (u1 + v1, u2 + v2) and u - v = (u1 - v1, u2 - v2). This is crucial for understanding how the given vectors a and b are constructed from m and n.
  • Scalar Multiplication: Multiplying a vector by a scalar changes its magnitude but not its direction (unless the scalar is negative, which reverses the direction). For instance, 2v doubles the length of vector v, while -v has the same length but points in the opposite direction. This is directly applicable to the (1/2)m and 2m terms in our problem.
  • Dot Product (Scalar Product): The dot product of two vectors u and v is a scalar value given by u · v = |u| |v| cos θ, where θ is the angle between the vectors. This is a powerful tool for finding angles and projections. In our case, we know |m|, |n|, and the angle between them, which hints that the dot product will play a key role.
  • Cross Product (Vector Product): The cross product of two vectors u and v results in another vector that is perpendicular to both u and v. Its magnitude is given by |u x v| = |u| |v| sin θ, which is geometrically significant because it represents the area of the parallelogram formed by u and v. This is exactly what we need to calculate for our problem!

Parallelograms and Vectors

Now, let's connect vectors to parallelograms. A parallelogram is a four-sided figure with opposite sides parallel. If you have two vectors a and b, you can visualize them as two adjacent sides of a parallelogram. The other two sides are simply translations of these vectors. The area of this parallelogram is given by the magnitude of the cross product of the vectors: Area = |a x b|. So, our strategy boils down to finding this cross product. We need to express a x b in terms of m and n, and then use the properties of the cross product and the given information to calculate its magnitude. It's like a treasure hunt – the area is the treasure, and the vectors are the map and tools!

Step-by-Step Solution: Finding the Area

Alright, let's roll up our sleeves and get to the solution! Here’s how we'll tackle this problem:

  1. Calculate the cross product a x b: This is the heart of the problem. We need to substitute the given expressions for a and b and use the distributive property of the cross product. Remember, the cross product is not commutative (u x v = - v x u), and the cross product of a vector with itself is zero (u x u = 0).

  2. Simplify the expression: After expanding the cross product, we'll have terms involving m x m, m x n, n x m, and n x n. We'll use the properties mentioned above to simplify this expression. The key is to reduce it to terms involving only m x n (or n x m).

  3. Find the magnitude of m x n: We know |m| = 4, |n| = 2, and the angle between them is 135°. We can use the formula |m x n| = |m| |n| sin θ to calculate this magnitude. This is where the angle information comes into play.

  4. Calculate the magnitude of a x b: Once we have the simplified expression for a x b and the magnitude of m x n, we can find the magnitude of a x b, which is the area of the parallelogram.

Step 1: Calculate the Cross Product a x b

Let's start by substituting the expressions for a and b into the cross product:

a x b = ((1/2)m - n) x (2m + n)

Now, we'll use the distributive property:

a x b = (1/2)m x (2m) + (1/2)m x n - n x (2m) - n x n

Step 2: Simplify the Expression

Next, we simplify using the properties of the cross product:

  • (1/2)m x (2m) = m x m = 0
    • n x n = 0
    • n x (2m) = -2 (n x m) = 2 (m x n) (since n x m = - m x n)

So, our expression becomes:

a x b = 0 + (1/2)m x n + 2 (m x n) + 0

a x b = (1/2)m x n + 2 (m x n)

Combining the terms, we get:

a x b = (5/2) (m x n)

Step 3: Find the Magnitude of m x n

Now, we calculate the magnitude of m x n using the formula:

|m x n| = |m| |n| sin θ

We know |m| = 4, |n| = 2, and θ = 135°. So,

|m x n| = 4 * 2 * sin(135°)

Since sin(135°) = sin(180° - 45°) = sin(45°) = √2 / 2,

|m x n| = 8 * (√2 / 2) = 4√2

Step 4: Calculate the Magnitude of a x b

Finally, we find the magnitude of a x b:

|a x b| = |(5/2) (m x n)| = (5/2) |m x n|

Substituting the value of |m x n| we found earlier:

|a x b| = (5/2) * 4√2 = 10√2

Therefore, the area of the parallelogram formed by vectors a and b is 10√2 square units.

Conclusion: Mastering Vector Geometry

And there you have it! We've successfully calculated the area of the parallelogram using vector operations. The key takeaways here are:

  • Understanding the geometric interpretation of vector operations, especially the cross product.
  • Knowing the properties of the cross product (anti-commutativity, distributivity, etc.).
  • Being able to break down complex problems into smaller, manageable steps.

This problem showcases how vector algebra provides powerful tools for solving geometric problems. Practice these concepts, and you'll be able to tackle even more challenging questions with confidence. Keep exploring, keep learning, and I'll catch you in the next explanation!