Problem 9
Question
Find a vector with the origin as initial point that is equivalent to the vector \(\overrightarrow{P Q}\). $$P=\left(\frac{4}{5},-2\right), Q=\left(\frac{17}{5},-\frac{12}{5}\right)$$
Step-by-Step Solution
Verified Answer
Question: Find an equivalent vector to PQ with the origin as its initial point, given that point P has coordinates (4/5, -2) and point Q has coordinates (17/5, -12/5).
Answer: The equivalent vector with the origin as its initial point is (13/5, -2/5).
1Step 1: Find the components of vector PQ
To find the components of the vector \(\overrightarrow{P Q}\), subtract the coordinates of P from the coordinates of Q:
$$ \overrightarrow{P Q} = \left(\frac{17}{5} - \frac{4}{5}, -\frac{12}{5} - (-2)\right) $$
2Step 2: Simplify the components
Now, simplify the components of the vector \(\overrightarrow{P Q}\):
$$ \overrightarrow{P Q} = \left(\frac{13}{5}, -\frac{2}{5}\right) $$
3Step 3: Create a new vector with the origin as the initial point
We now create a new vector with its initial point at the origin (0,0) and its terminal point at the coordinates found in step 2. This gives us a new vector equivalent to \(\overrightarrow{P Q}\):
$$ \overrightarrow{O R} = \left(0 + \frac{13}{5}, 0 - \frac{2}{5}\right) $$
4Step 4: Write the final vector
The new vector \(\overrightarrow{O R}\) with the origin as the initial point is:
$$ \overrightarrow{O R} = \left(\frac{13}{5}, -\frac{2}{5}\right) $$
This is the equivalent vector to \(\overrightarrow{P Q}\) with the origin as its initial point.
Key Concepts
Vector EquivalenceVector SubtractionCoordinate Geometry
Vector Equivalence
Understanding the concept of vector equivalence is essential in the study of physics and engineering. Two vectors are considered equivalent if they have the same magnitude and direction, regardless of their initial points. This means that they represent the same displacement or force when applied.
In the context of the given exercise, we are asked to find a vector that is equivalent to \(\overrightarrow{P Q}\). This exercise demonstrates that the vector's initial and terminal points can change while retaining its 'effect' by maintaining the same magnitude and direction. In simple terms, moving a vector around without rotating it or changing its length does not alter its essence.
The vector that starts at the origin and is equivalent to \(\overrightarrow{P Q}\) is found by preserving the same components, hence it has the same 'action' as the original vector, but just starts from a different point, in this case, the origin.
In the context of the given exercise, we are asked to find a vector that is equivalent to \(\overrightarrow{P Q}\). This exercise demonstrates that the vector's initial and terminal points can change while retaining its 'effect' by maintaining the same magnitude and direction. In simple terms, moving a vector around without rotating it or changing its length does not alter its essence.
The vector that starts at the origin and is equivalent to \(\overrightarrow{P Q}\) is found by preserving the same components, hence it has the same 'action' as the original vector, but just starts from a different point, in this case, the origin.
Vector Subtraction
Vector subtraction is a fundamental operation in vector algebra. It can be visualized by placing the tail of one vector at the head of the other and then finding the vector that connects the tail of the first vector to the head of the subtracted vector. Mathematically, subtracting the coordinates of two points allows us to find the components of the vector between them.
For instance, in our exercise, we subtract the coordinates of point P from point Q to determine the components of \(\overrightarrow{P Q}\). It is crucial to maintain the order of subtraction, as reversing it would change the direction of the resulting vector. This concept also highlights that subtraction of vectors is not commutative, meaning that \(\overrightarrow{P Q}\) is not the same as \(\overrightarrow{Q P}\).
The process of finding the equivalent vector at the origin involves subtracting each component of P from the corresponding component of Q. The result provides the components of a new vector that has the same magnitude and direction as our vector \(\overrightarrow{P Q}\) but originating from \(\overrightarrow{O}\), the origin.
For instance, in our exercise, we subtract the coordinates of point P from point Q to determine the components of \(\overrightarrow{P Q}\). It is crucial to maintain the order of subtraction, as reversing it would change the direction of the resulting vector. This concept also highlights that subtraction of vectors is not commutative, meaning that \(\overrightarrow{P Q}\) is not the same as \(\overrightarrow{Q P}\).
The process of finding the equivalent vector at the origin involves subtracting each component of P from the corresponding component of Q. The result provides the components of a new vector that has the same magnitude and direction as our vector \(\overrightarrow{P Q}\) but originating from \(\overrightarrow{O}\), the origin.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, is the study of geometry using a coordinate system. This branch of mathematics allows for the precise representation of geometric figures and their relationships on a plane using coordinates and algebraic equations.
In terms of vector analysis within coordinate geometry, points on a plane are described by pairs of numerical values, which are their coordinates. Vectors are then defined as entities that have both magnitude and direction and can be represented as directed line segments between two points in the plane.
The exercise provided uses the principles of coordinate geometry to transition from the vector \(\overrightarrow{P Q}\), defined by the coordinates of points P and Q, to an equivalent vector \(\overrightarrow{O R}\) with the same components but originating from the origin. By translating the vector to start from the origin, we maintain the geometric properties of the vector while changing its position, an operation made seamless and intuitive through the lens of coordinate geometry.
In terms of vector analysis within coordinate geometry, points on a plane are described by pairs of numerical values, which are their coordinates. Vectors are then defined as entities that have both magnitude and direction and can be represented as directed line segments between two points in the plane.
The exercise provided uses the principles of coordinate geometry to transition from the vector \(\overrightarrow{P Q}\), defined by the coordinates of points P and Q, to an equivalent vector \(\overrightarrow{O R}\) with the same components but originating from the origin. By translating the vector to start from the origin, we maintain the geometric properties of the vector while changing its position, an operation made seamless and intuitive through the lens of coordinate geometry.
Other exercises in this chapter
Problem 8
In Exercises \(1-8,\) plot the point in the complex plane corresponding to the number. $$\frac{4 i}{3}(-6-3 i)$$
View solution Problem 9
Find the dot product when \(u=\langle 4,3\rangle\) \(\boldsymbol{v}=\langle-5,2\rangle,\) and \(\boldsymbol{w}=\langle 4,-1\rangle\) $$(\mathbf{u}+\mathbf{v}) \
View solution Problem 9
Calculate the product by expressing the number in polar form and using DeMoivre's Theorem. Express your answer in the form \(a+b i\). $$(1-i)^{12}$$
View solution Problem 9
In Exercises \(9-14,\) find the absolute value. $$|5-12 i|$$
View solution