Problem 23 - Precalculus with Limits

Question

In Exercises 13-24, find the component form and the magnitude of the vector \(\mathbf{v}\).'' Initial Point - \((-1, 5)\) Terminal Point - \((15, 12)\)

Step-by-Step Solution

Answer

The component form of the vector \(\mathbf{v}\) is <16, 7> and its magnitude is \(\sqrt{305}\).

Step 1: Identify the Points

The initial point is \((-1, 5)\) and the terminal point is \((15, 12)\).

Step 2: Compute the Component Form of the Vector

To find the component form of the vector, subtract the coordinates of the initial point from the coordinates of the terminal point. In mathematical terms, \(\mathbf{v} = = <15-(-1), 12-5> = <16, 7>\). Therefore, the component form of the vector is \(\mathbf{v} = <16, 7>\)

Step 3: Compute the Magnitude of the Vector

To compute the magnitude of the vector, apply the formula \(\|\mathbf{v}\| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{16^2 + 7^2} = \sqrt{256 + 49} = \sqrt{305}\). Therefore, the magnitude of the vector \(\mathbf{v}\) is \(\sqrt{305}\)

Key Concepts

Component Form of a Vector

Vectors play a critical role in precalculus, representing quantities that have both magnitude and direction. When we talk about a vector in the component form, we're describing it in terms of its horizontal (x) and vertical (y) influences. This is akin to providing a set of instructions for how to get from one point to another.

The component form is written as \(\mathbf{v} = \), where 'x' represents the change in the horizontal direction, and 'y' corresponds to the change in the vertical direction. To find these components, we simply subtract the coordinates of the vector's initial point from its terminal point. In our exercise, the vector \(\mathbf{v}\)'s component form is found by calculating \(\mathbf{v} = <15 - (-1), 12 - 5> = <16, 7>\), which gives a clear and straightforward way to visualize the movement from the vector's start to its end.

Magnitude of a Vector

The magnitude of a vector, intuitively, is how long the vector is - in other words, the distance it spans from its initial point to its terminal point.

To compute this, we use the Pythagorean theorem because a vector's components create a right triangle with the vector itself being the hypotenuse. The formula is \(\|\mathbf{v}\| = \sqrt{x^2 + y^2}\), where 'x' and 'y' are the vector's horizontal and vertical components, respectively.

In the given problem, after finding the components as \(\mathbf{v} = <16, 7>\), the magnitude is calculated as follows: \(\|\mathbf{v}\| = \sqrt{16^2 + 7^2} = \sqrt{256 + 49} = \sqrt{305}\). Thus, our vector stretches across the distance equivalent to the square root of 305 in the coordinate plane.

Vectors in Precalculus

Vectors in precalculus are foundational for understanding more complex topics in calculus and physics. They're not just about quantities like speed or force, but also about the direction in which these quantities are applied.

Working with vectors involves a range of operations, including addition, subtraction, and scalar multiplication. Furthermore, understanding vectors helps in grasping concepts like dot product, cross product, and finding the equation of a line or plane in three dimensions.

Most importantly, solving problems with vectors requires a clear grasp of both component form and magnitude, as these aspects are essential in depicting a vector's overall effect within a particular coordinate system.

Coordinate Plane

The coordinate plane is a two-dimensional surface where every point is determined by a pair of numerical coordinates. These coordinates represent the position of the point in relation to two intersecting lines, typically referred to as the x-axis (horizontal) and y-axis (vertical).

When dealing with vectors, the coordinate plane allows us to visually interpret their behavior and perform calculations. For instance, when finding the magnitude and direction of a vector, the coordinate plane helps us map out the initial and terminal points clearly.

As seen in our exercise, by plotting points \( (-1, 5) \) and \( (15, 12) \) on the coordinate plane, we can visually assess the direction and length of the vector, aiding in our understanding of the vector's component form and magnitude.