91Ó°ÊÓ

Coordinate geometry helps us understand the spatial connections and relationships between points in a plane. This is done by using a set of numbers, called coordinates, to define the position of these points. In the context of vectors, we specifically look at the initial point and the terminal point.

In our exercise, the initial point of vector \( \mathbf{v} \) is \( P(1,0) \), which is placed on the Cartesian plane at \( x = 1 \) and \( y = 0 \). The terminal point \( Q \) is described to be on the \( y \)-axis. Points on the \( y \)-axis have the form \( (0, y) \). This is because their \( x \)-coordinate is zero, indicating they are aligned vertically along the \( y \)-axis.

Using coordinate geometry allows us to accurately describe and analyze the properties of vectors and other geometric figures by giving them a mathematical framework on the plane.

Vectors are mathematical objects used to describe quantities that have both magnitude and direction. In our case, we are interested in the vector \( \mathbf{v} \) starting from point \( P(1,0) \) and ending at point \( Q(0, y) \).

A vector's components are derived from the differences between their starting and ending points. For vector \( \mathbf{v} \), the change in \( x \) is calculated as \( 0 - 1 = -1 \), and the change in \( y \) is \( y - 0 = y \). Hence, the components of \( \mathbf{v} \) can be expressed as \( \langle -1, y \rangle \).

These components are crucial because any vector in the Cartesian plane can be uniquely characterized by its components, representing its movement along the x and y axes.

The magnitude of a vector, sometimes called its length or norm, tells us how long the vector is. To calculate the magnitude of a vector \( \mathbf{v} = \langle a, b \rangle \), we use the formula \( \| \mathbf{v} \| = \sqrt{a^2 + b^2} \).

For our vector \( \mathbf{v} = \langle -1, y \rangle \), its magnitude is expressed as \( \sqrt{(-1)^2 + y^2} \). Simplifying this gives the expression \( \sqrt{1 + y^2} \).

In the exercise, we know the magnitude is \( \sqrt{5} \). By setting the magnitude equation as \( \sqrt{1 + y^2} = \sqrt{5} \) and solving it—through squaring both sides and simplifying—we identified that when \( y = 2 \), the vector satisfies the required magnitude. This solution process highlights how the magnitude formula helps us find specific information about vectors when certain conditions are given.