Problem 19 - College Trigonometry

Question

Find the component form of the vector \(\vec{v}\) using the information given about its magnitude and direction. Give exact values. $$ \|\vec{v}\|=6.25 \text { ; when drawn in standard position } \vec{v} \text { lies along the negative } y \text { -axis } $$

Step-by-Step Solution

Answer

The component form of \( \vec{v} \) is \( \langle 0, -6.25 \rangle \).

Step 1: Understanding the Problem

We need to find the component form of the vector \( \vec{v} \) given its magnitude and direction. The vector is drawn in standard position along the negative \( y \)-axis.

Step 2: Identifying Direction

\( \vec{v} \) lies along the negative \( y \)-axis. This means the vector points downward and has no \( x \)-component.

Step 3: Express Magnitude in Component Form

Since \( \vec{v} \) is along the negative \( y \)-axis and has a magnitude of 6.25, the \( y \)-component is \(-6.25\) (negative because it points downward) and the \( x \)-component is 0.

Step 4: Write Vector in Component Form

The component form of \( \vec{v} \) is given by \( \vec{v} = \langle 0, -6.25 \rangle \).

Key Concepts

Magnitude

The magnitude of a vector is a measure of its length. Imagine a vector as an arrow; the magnitude is akin to the arrow's length. It is always a non-negative value and is essential in defining the vector's size.
To find the magnitude of a vector in component form, such as \[ \vec{v} = \langle x, y \rangle \] we can use the Pythagorean theorem: \[ \|\vec{v}\| = \sqrt{x^2 + y^2} \]
In our exercise, the magnitude is given as 6.25. Thus, if we made a triangle based on the vector's components, the hypotenuse (or the vector itself) would measure this length. Understanding magnitude is crucial because it gives us an idea of the distance the vector can represent. Knowing just this number allows us to describe the strength or intensity of a vector, which in this context is pointing downward along the negative y-axis.

Direction

Direction tells us where a vector is pointing. It's like navigating with a compass or describing routes with lefts and rights. For vectors in two-dimensional space, we often use angles or describe the "heading" relative to axes.
In our provided problem, the vector is stated to lie along the negative y-axis. This directional information is vital because it specifies that the vector does not have any horizontal (x-axis) movement, just a downward vertical component.
When a vector lies exactly on one of the axes, its direction is straightforward to identify. It simply parallels the axis it is on, and any deviation from this would introduce additional components. Direction sets the trajectory the vector follows when moving, which is crucial for applications such as physics and engineering.

Standard Position

Placing a vector in standard position means positioning its initial point at the origin of the coordinate system. This makes it easy to see and calculate because its direction and length (magnitude) come directly from its tip.
For example, if we draw a vector in standard position on a graph, starting from \((0, 0)\), we can easily determine its endpoint coordinates. This frame of reference simplifies calculations and comparisons between vectors because they all start from the same point.
In our exercise, having the vector in standard position at the origin helps us quickly identify that it only stretches along the y-axis, eliminating any x-component and simplifying the component identification to be \(\vec{v} = \langle 0, -6.25 \rangle\). This simplifies our understanding and computation of the vector's properties.

Y-Axis

The y-axis is one of the two main axes in the Cartesian coordinate system, running vertically. It represents the vertical component or dimension of any point or vector in the coordinate plane.
In the context of our vector, being on the negative y-axis demonstrates that the vector's movement is purely vertical, with a simple downward direction and no horizontal movement. This is why our vector's x-component is zero, confirming its exact position and movement direction straight down the y-axis.
Vectors lying along the y-axis are uniquely simple to work with in calculations because there is no need to consider any x-direction influences. When a vector has no x-component and lies wholly on the y-axis, you merely need to note any positive or negative values to determine whether it's pointing up or down.