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Chapter 9: Problem 10

The coordinates of two points \(P\) and \(Q\) are given. In case, determine the components of the vector \(\overrightarrow{P Q}\). Write your answers in the form \(\langle a, b\rangle .\) $$P(0,-4) \text { and } Q(0,-8)$$

Short Answer

Expert verified
The vector \(\overrightarrow{P Q}\) is \(\langle 0, -4 \rangle\).

Step by step solution

01

Understand the Given Points

We are given two points: \(P(0, -4)\) and \(Q(0, -8)\). These are coordinates on a Cartesian coordinate plane, where the first number is the x-coordinate and the second number is the y-coordinate.
02

Recall Vector Components

The vector \(\overrightarrow{P Q}\) represents the direction and distance from point \(P\) to point \(Q\). The components of this vector are found by subtracting the coordinates of \(P\) from \(Q\).
03

Subtract X-Coordinates

Compute the x-component of the vector by subtracting the x-coordinate of \(P\) from the x-coordinate of \(Q\). Here, both x-coordinates are 0, so the x-component is: \[ x_Q - x_P = 0 - 0 = 0 \]
04

Subtract Y-Coordinates

Compute the y-component of the vector by subtracting the y-coordinate of \(P\) from the y-coordinate of \(Q\). Here, the y-coordinates are \(-8\) and \(-4\), respectively, so the y-component is: \[ y_Q - y_P = -8 - (-4) = -8 + 4 = -4 \]
05

Write the Vector in Component Form

Combine the x and y components found in steps 3 and 4. The vector \(\overrightarrow{P Q}\) is written in the component form as \(\langle 0, -4 \rangle\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cartesian Coordinate Plane
The Cartesian coordinate plane is an essential tool in mathematics, helping us to visually represent and solve problems involving points and vectors.
It is structured by two perpendicular lines: the x-axis (horizontal) and the y-axis (vertical), which intersect at the point known as the origin (0,0).
Each point on this plane is defined by a pair of numbers, called coordinates, written in the form \(x, y\).

This system allows us to easily map relationships and interactions between points and lines.
  • The x-coordinate indicates the position relative to the vertical y-axis - moving left or right.
  • The y-coordinate shows the position in relation to the horizontal x-axis - moving up or down.
This makes it quite simple to understand where points are located and how vectors run between them.
It is especially useful for solving geometry problems, as it gives us a clear method for visualizing and calculating distances and directions between points.
Coordinate Subtraction
Coordinate subtraction is a fundamental technique used to find the components of a vector.
This method helps us determine the direction and magnitude of a vector created by two distinct points.
To retrieve these components, we subtract the coordinates of the initial point from the coordinates of the terminal point.

Here's how it's done:
  • Subtract the x-coordinate of the initial point from the x-coordinate of the terminal point to find the x-component of the vector.
  • Subtract the y-coordinate of the initial point from the y-coordinate of the terminal point to find the y-component of the vector.
By carrying out these subtractions step by step, we construct a new vector pointing from the initial position directly to the terminal position.
In our problem with points \(P(0, -4)\) and \(Q(0, -8)\), the x-component is \(0 - 0 = 0\), and the y-component is \(-8 - (-4) = -4\).
This process gives us the simplified form of the vector.
Component Form
The component form of a vector provides a simplified way to represent the vector's properties, particularly its direction and magnitude.
It is an expression consisting of two numbers enclosed in angle brackets, like this: \(\).
Each number in the component form corresponds to one of the coordinate axes on the Cartesian plane.

The formula for finding the component form of a vector \(\overrightarrow{P Q}\) between two points \(P(x_P, y_P)\) and \(Q(x_Q, y_Q)\) is:
  • Calculate the component along the x-axis: \(x_Q - x_P\).
  • Calculate the component along the y-axis: \(y_Q - y_P\).
Hence, you get the vector in this concise form: \(\).
For example, in our case with \(P(0, -4)\) and \(Q(0, -8)\), the component form is \(<0, -4>\).
This form is crucial for calculations in physics and engineering, where vectors represent forces, velocities, or other vector quantities.
By using component form, it becomes much easier to work with vectors in a mathematical and graphical sense.

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Most popular questions from this chapter

Round each answer to one decimal place. In \(\triangle A B C, \angle A=40^{\circ}, b=6.1 \mathrm{cm},\) and \(c=3.2 \mathrm{cm}\) (a) Find \(a\) using the law of cosines. (b) Find \(\angle C\) using the law of sines. (c) Find \(\angle B\)

We study the dot product of two vectors. Given two vectors \(\mathbf{A}=\left\langle x_{1}, y_{1}\right\rangle\) and \(\mathbf{B}=\left\langle x_{2}, y_{2}\right\rangle,\) we define the dot product \(\mathbf{A} \cdot \mathbf{B}\) as follows: $$\mathbf{A} \cdot \mathbf{B}=x_{1} x_{2}+y_{1} y_{2}$$ For example, if \(\mathbf{A}=\langle 3,4\rangle\) and \(\mathbf{B}=\langle-2,5\rangle,\) then \(\mathbf{A} \cdot \mathbf{B}=(3)(-2)+(4)(5)=14 .\) Notice that the dot product of two vectors is a real number. For this reason, the dot product is also known as the scalar product. For Exercises \(61-63\) the vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are defined as follows: $$\mathbf{u}=\langle-4,5\rangle \quad \mathbf{v}=\langle 3,4\rangle \quad \mathbf{w}=\langle 2,-5\rangle$$ (a) Compute \(\mathbf{v} \cdot \mathbf{v}\) and \(|\mathbf{v}|^{2}\). (b) Compute \(\mathbf{w} \cdot \mathbf{w}\) and \(|\mathbf{w}|^{2}\).

Round each answer to one decimal place. A regular pentagon is inscribed in a circle of radius 1 unit. Find the perimeter of the pentagon. Hint: First find the length of a side using the law of cosines.

Graph the parametric equations using the given range for the parameter t. In each case, begin with the standard viewing rectangle and then make adjustments, as necessary, so that the graph utilizes as much of the viewing screen as possible. For example, in graphing the circle given by \(x=\cos t\) and \(y=\sin t,\)it would be natural to choose a viewing rectangle extending from -1 to 1 in both the \(x\) - and \(y\) -directions. \(x=\sin (0.8 t+\pi), y=\sin t, \quad 0 \leq t \leq 10 \pi\) (Bowditch curve)

Prove the following identity for \(\triangle A B C:\) $$ \frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{a^{2}+b^{2}+c^{2}}{2 a b c} $$ Suggestion: Use the law of cosines to substitute for \(a^{2},\) for \(b^{2},\) and for \(c^{2}\) in the numerator of the expression on the right-hand side.
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