Vectors: Calculate Operations, Angle, And Magnitude

Hey guys! Let's dive into the fascinating world of vectors! This guide will walk you through some common vector operations, like finding linear combinations, determining direction angles, and calculating magnitudes. We'll break down each step, so you can master these concepts with ease. Let's get started!

Problem Statement

We're given two vectors: u = -3i + 2j and v = i - 4j. Our mission is to calculate the following:

a) 2u + 3v b) The direction angle θ of vector v c) ||u - v||

Let's tackle each part step by step.

a) Calculating 2u + 3v

This part involves finding a linear combination of the vectors u and v. In essence, we're scaling each vector by a scalar (a constant) and then adding the results. This is a fundamental operation in vector algebra, allowing us to create new vectors from existing ones. Linear combinations are crucial in many areas, including physics (e.g., finding resultant forces) and computer graphics (e.g., transforming objects).

First, we need to perform the scalar multiplication. This involves multiplying each component of the vector by the scalar. So, 2u means multiplying both the i and j components of u by 2. Similarly, 3v means multiplying the components of v by 3.

Let's break it down:

2u = 2(-3i + 2j) = -6i + 4j

3v = 3(i - 4j) = 3i - 12j

Now that we've scaled the vectors, we can add them together. Vector addition involves adding corresponding components. We add the i components together and the j components together.

2u + 3v = (-6i + 4j) + (3i - 12j)

Combine the i components: -6i + 3i = -3i

Combine the j components: 4j - 12j = -8j

Therefore, 2u + 3v = -3i - 8j. This resulting vector represents a new vector that is a combination of the original vectors u and v, scaled and added together. It's a key concept to grasp, guys!

b) Determining the Direction Angle θ of Vector v

Now, let's figure out the direction angle of vector v. The direction angle, often denoted by θ (theta), tells us the angle that the vector makes with the positive x-axis (or the positive i-axis). This angle is measured counterclockwise from the positive x-axis. Understanding the direction angle is vital for visualizing the vector's orientation in a coordinate system.

To find the direction angle, we'll use the arctangent function (also known as the inverse tangent), denoted as tan⁻¹ or arctan. The arctangent function takes the ratio of the y-component to the x-component of the vector and returns the angle whose tangent is that ratio. However, we need to be a bit careful because the arctangent function only gives us angles in the range of -90° to +90° (or -π/2 to +π/2 radians). We might need to adjust the angle based on the quadrant in which the vector lies.

Recall that vector v = i - 4j. This means the x-component is 1, and the y-component is -4.

The formula for finding the direction angle is:

θ = tan⁻¹(y/x)

Plugging in our values:

θ = tan⁻¹(-4/1) = tan⁻¹(-4)

Using a calculator, we find that tan⁻¹(-4) ≈ -75.96°.

However, since the y-component is negative and the x-component is positive, vector v lies in the fourth quadrant. The arctangent function gave us an angle in the fourth quadrant, so we don't need to make any adjustments in this case. If the vector was in the second or third quadrant, we would need to add 180° to the result to get the correct angle.

Therefore, the direction angle θ of vector v is approximately -75.96°. We can also express this angle as 360° - 75.96° = 284.04°, which is the equivalent positive angle measured counterclockwise from the positive x-axis. Remember, visualizing the vector in the coordinate system helps to ensure the angle makes sense!

c) Calculating ||u - v||

Finally, let's calculate the magnitude of the vector u - v. The magnitude of a vector, denoted by || ||, represents its length or size. It's a scalar quantity, meaning it's just a number, not a vector with direction. In this case, we first need to find the vector u - v, and then we'll calculate its magnitude. This is a common operation when we're interested in the distance between two points represented by vectors.

First, let's find u - v. Vector subtraction is similar to vector addition – we subtract corresponding components.

u - v = (-3i + 2j) - (i - 4j)

Subtract the i components: -3i - i = -4i

Subtract the j components: 2j - (-4j) = 2j + 4j = 6j

So, u - v = -4i + 6j.

Now, to find the magnitude of this vector, we use the Pythagorean theorem. The magnitude is the square root of the sum of the squares of the components. This is based on the idea that the components of the vector form the sides of a right triangle, and the magnitude is the length of the hypotenuse.

The formula for the magnitude of a vector w = ai + bj is:

||w|| = √(a² + b²)

In our case, w = u - v = -4i + 6j, so a = -4 and b = 6.

||u - v|| = √((-4)² + 6²)

||u - v|| = √(16 + 36)

||u - v|| = √52

||u - v|| ≈ 7.21

Therefore, the magnitude of the vector u - v is approximately 7.21. This tells us the length of the vector resulting from subtracting v from u.

Conclusion

Alright, guys! We've successfully calculated 2u + 3v, the direction angle θ of vector v, and ||u - v||. We've covered some fundamental vector operations: linear combinations, finding direction angles using arctangent, and calculating magnitudes using the Pythagorean theorem. These skills are super important for anyone working with vectors in math, physics, computer graphics, and beyond. Keep practicing, and you'll become vector whizzes in no time! Keep rocking!