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Lesson Explainer: Position Vectors

In this explainer, we will learn how to identify what position vectors are and write general vectors in terms of position vectors.

We begin by recalling that vectors are mathematical objects that are entirely defined by their magnitudes and directions. We can represent a vector by a directed line segment, in a suitable space, where the length of the line segment tells us the magnitude of the vector and the initial and terminal points of the directed line segment indicate the direction.

In this space, we can think of a vector as representing a displacement from its initial point 𝐴 to its terminal point 𝐡; we write this vector as 𝐴𝐡.

This allows us to represent vectors by considering their horizontal and vertical displacements separately. For example, we can measure the horizontal and vertical displacements using a diagram.

In the example above, when moving from 𝐴 to 𝐡, we move 2 units right and 4 units up. These are positive directions for the displacements, so we use positive numbers for the components of the vector. We have 𝐴𝐡=(2,4).

This is not the only way of finding the components of 𝐴𝐡 from the diagram. Let’s also consider 𝑂𝐴 and οƒŸπ‘‚π΅, where 𝑂 is the origin. These are called the position vectors of 𝐴 and 𝐡 respectively.

We can note some useful properties of position vectors. First, we see that horizontal and vertical displacements when traveling from 𝑂 to 𝐴 (or any point 𝑃) are given by the coordinates of 𝐴. Therefore, the components of οƒŸπ‘‚π‘ƒ are equal to the coordinates of 𝑃, and we can write this in terms of the unit directional vectors. If point 𝑃 is (π‘Ž,𝑏), then οƒŸπ‘‚π‘ƒ=π‘Žβƒ‘π‘–+𝑏⃑𝑗. Second, the magnitude of the position vector οƒŸπ‘‚π‘ƒ will be equal to the distance between 𝑃 and the origin.

This gives us two ways of traveling from 𝐴 to 𝐡.

We can travel directly along vector 𝐴𝐡; however, we can also travel backward along 𝑂𝐴 and then forward along οƒŸπ‘‚π΅. This is equivalent to traveling from 𝐴 to the origin and then to 𝐡. Of course, this is a longer distance to travel, but the key thing to note is that the displacements of the journeys will be the same since the start and end points are identical. We can see this in the example above.

We then recall that we can reverse the direction of a vector by multiplying it by βˆ’1. Therefore, this journey of traveling from 𝐴 to 𝐡 via the origin is the vector βˆ’οƒ π‘‚π΄+οƒŸπ‘‚π΅=οƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄.

This is an application of the triangle rule for vector addition where one of the points is the origin. We have shown the following result.

Property: Representing a Vector between Two Points in terms of Position Vectors

For any points 𝐴 and 𝐡, we have 𝐴𝐡=οƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄.

In our first example, we will find the position vector of a point given in the coordinate plane.

Example 1: Finding the Position Vector of a Point in the Coordinate Plane

Find the position vector of the point 𝐴 on the diagram.

Answer

We first recall that the position vector of point 𝐴 is the vector from the origin to the point; we write this as 𝑂𝐴. We can read the coordinates of point 𝐴 off the given diagram.

We can see that 𝐴 has coordinates (2,βˆ’3). In the same way, we can note that to travel from the origin to point 𝐴 we travel 2 units right and 3 units down.

We write this in terms of the unit directional vectors ⃑𝑖 and ⃑𝑗. This tells us that the vector from the origin must be the sum of these vectors: 2βƒ‘π‘–βˆ’3⃑𝑗.

It is worth noting that in general, if a point 𝑃 has coordinates (π‘₯,𝑦), then οƒŸπ‘‚π‘ƒ=π‘₯⃑𝑖+𝑦⃑𝑗. We can also write this as a vector οƒŸπ‘‚π‘ƒ=(π‘₯,𝑦).

Hence, 𝑂𝐴=(2,βˆ’3).

In our next example, we will find the vector between two points using their position vectors.

Example 2: Finding the Vector between Two Points Using Their Position Vectors

Find the vector from 𝐴 to 𝐡, given that 𝑂𝐴=(4,6) and οƒŸπ‘‚π΅=(3,βˆ’1).

Answer

We could answer this question by noting that the components of the position vector tell us the coordinates of the point. So, 𝐴 is the point (4,6) and 𝐡 is the point (3,βˆ’1). This allows us to determine 𝐴𝐡 graphically or otherwise.

However, it is easier to use the formula 𝐴𝐡=οƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄. We can substitute the given vectors into this formula to obtain 𝐴𝐡=(3,βˆ’1)βˆ’(4,6).

We then recall that we evaluate the subtraction of vectors componentwise. Subtracting the components of the vectors yields 𝐴𝐡=(3,βˆ’1)βˆ’(4,6)=(3βˆ’4,βˆ’1βˆ’6)=(βˆ’1,βˆ’7).

Hence, 𝐴𝐡=(βˆ’1,βˆ’7).

In our next example, we will find the vector between two points given their coordinates.

Example 3: Finding the Vector between Two Points

Find the vector from 𝐴(βˆ’2,βˆ’5) to 𝐡(βˆ’5,5).

Answer

We can find the vector between two points using the formula 𝐴𝐡=οƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄. We recall that the components of the position vector of a point are given by the coordinates of the point. In other words, if 𝑃 is the point (π‘₯,𝑦), then οƒŸπ‘‚π‘ƒ=(π‘₯,𝑦).

This allows us to find the position vectors of 𝐴 and 𝐡. We use the coordinates of each as the components of the vectors to obtain 𝑂𝐴=(βˆ’2,βˆ’5)οƒŸπ‘‚π΅=(βˆ’5,5).

Substituting these vectors into the formula yields 𝐴𝐡=(βˆ’5,5)βˆ’(βˆ’2,βˆ’5).

We then recall that we evaluate the subtraction of vectors componentwise. Subtracting the components of the vectors yields 𝐴𝐡=(βˆ’5,5)βˆ’(βˆ’2,βˆ’5)=(βˆ’5βˆ’(βˆ’2),5βˆ’(βˆ’5))=(βˆ’3,10).

Hence, 𝐴𝐡=(βˆ’3,10).

In our next example, we will find an expression for the position vector of a vertex of a regular pentagon using the position vectors of two of its other vertices.

Example 4: Position Vectors of Vertices of Regular Polygons

A regular hexagon has vertices 𝑂𝐴𝐡𝐢𝐷𝐸. Find an expression for the position vector of 𝐷 in terms of 𝑂𝐴 and οƒŸπ‘‚π΅.

Answer

To help us find an expression for the position vector of 𝐷 in terms of 𝑂𝐴 and οƒŸπ‘‚π΅, let’s start by sketching the information given.

It is worth noting that we do not know the orientation of the pentagon, and we also do not know whether to label the vertices clockwise or counterclockwise. However, this will not affect the method used.

We can add the vectors 𝑂𝐴, οƒŸπ‘‚π΅, and 𝑂𝐷 to the diagram by recalling that οƒŸπ‘‚π‘ƒ is the vector from the origin to point 𝑃.

We cannot directly find an expression for 𝑂𝐷 in terms of 𝑂𝐴 and οƒŸπ‘‚π΅. Instead, we need to consider the diagram and the geometric properties of a regular hexagon to find this expression.

We can start by using the triangle rule for vector addition to note that 𝑂𝐷=οƒŸπ‘‚πΈ+𝐸𝐷.

We can find an expression for 𝐸𝐷 by noting that opposite sides in a hexagon are parallel and have the same length. This means that 𝐸𝐷=𝐴𝐡.

We do need to be careful to choose vertices that make the vectors have the same direction.

We can find an expression for 𝐴𝐡 using the triangle rule for vector addition: 𝑂𝐴+𝐴𝐡=οƒŸπ‘‚π΅οƒ π΄π΅=οƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄.

We can add these onto the diagram.

To find an expression for οƒŸπ‘‚πΈ, we recall that we can split a regular hexagon into 6 equilateral triangles using its center, which we will call 𝑀.

We can note that each of the triangles is an equilateral triangle, so the triangles’ sides are the same length. This means that 𝑂𝑀 and 𝐴𝐡 have the same magnitude, and we know they have the same direction. Thus, 𝑂𝑀=οƒ π΄π΅βˆ’οƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄.

In the same way, we can note that 𝐸𝑀=𝑂𝐴 by noting the magnitudes and directions of the vectors. We can add these onto the diagram.

We can now find an expression for οƒŸπ‘‚πΈ using the triangle rule for vector addition: οƒŸπ‘‚πΈ+𝐸𝑀=𝑂𝑀.

Substituting 𝐸𝑀=𝑂𝐴 and 𝑂𝑀=οƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄ into the equation yields οƒŸπ‘‚πΈ+𝑂𝐴=οƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄οƒŸπ‘‚πΈ=οƒŸπ‘‚π΅βˆ’2𝑂𝐴.

Finally, we know that 𝑂𝐷=οƒŸπ‘‚πΈ+𝐸𝐷.

Substituting οƒŸπ‘‚πΈ=οƒŸπ‘‚π΅βˆ’2𝑂𝐴 and 𝐸𝐷=οƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄ into this equation gives us 𝑂𝐷=οƒŸπ‘‚π΅βˆ’2𝑂𝐴+οƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄=2οƒŸπ‘‚π΅βˆ’3𝑂𝐴.

In our next example, we will determine the magnitude of the vector between two given points.

Example 5: Finding the Magnitude of the Vector between Two Points

Find the magnitude of οƒŸπ‘ƒπ‘„, where 𝑃(2,5) and 𝑄(βˆ’2,βˆ’1). Give your answer in exact form.

Answer

We first recall that we can find the vector between two points using the formula 𝐴𝐡=οƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄. We also recall that the components of the position vector of a point are equal to the coordinates of the point, so οƒŸπ‘‚π‘ƒ=(2,5)𝑂𝑄=(βˆ’2,βˆ’1).

Substituting these vectors into the formula yields οƒŸπ‘ƒπ‘„=(βˆ’2,βˆ’1)βˆ’(2,5).

We then recall that we evaluate the subtraction of vectors componentwise. Subtracting the components of the vectors gives us οƒŸπ‘ƒπ‘„=(βˆ’2βˆ’2,βˆ’1βˆ’5)=(βˆ’4,βˆ’6).

Finally, we recall that the magnitude of a vector is given by the square root of the sum of the squares of its components. In general, β€–(π‘₯,𝑦)β€–=√π‘₯+π‘¦οŠ¨οŠ¨. We have π‘₯=βˆ’4 and 𝑦=βˆ’6, so β€–β€–οƒŸπ‘ƒπ‘„β€–β€–=(βˆ’4)+(βˆ’6)=√52=2√13.

In our next example, we will use expressions for the position vectors of two points and the magnitude of the vector between them to determine the possible values of an unknown component of one of the position vectors.

Example 6: Finding an Unknown Component

If 𝑂𝐴=(2,7), οƒŸπ‘‚π΅=βˆ’βƒ‘π‘–+π‘˜βƒ‘π‘—, and ‖‖𝐡𝐴‖‖=√34, find all possible values of π‘˜.

Answer

We want to use the fact that ‖‖𝐡𝐴‖‖=√34 to determine the possible values of π‘˜. To do this, we first need to find an expression for 𝐡𝐴, and we can do this by recalling that 𝐡𝐴=οƒ π‘‚π΄βˆ’οƒŸπ‘‚π΅. We note that 𝑂𝐴 and οƒŸπ‘‚π΅ are given in different forms, so we can start by writing both as vectors. We do this by recalling that the coefficients of the unit directional vectors will be the components of the vector, so οƒŸπ‘‚π΅=βˆ’βƒ‘π‘–+π‘˜βƒ‘π‘—=(βˆ’1,π‘˜).

We can then find 𝐡𝐴 by substituting the vectors into the formula and then evaluating the subtraction componentwise: 𝐡𝐴=(2,7)βˆ’(βˆ’1,π‘˜)=(2βˆ’(βˆ’1),7βˆ’π‘˜)=(3,7βˆ’π‘˜).

We now recall that the magnitude of a vector is given by the square root of the sum of the squares of its components. In general, β€–(π‘₯,𝑦)β€–=√π‘₯+π‘¦οŠ¨οŠ¨. We have π‘₯=3 and 𝑦=7βˆ’π‘˜, so ‖‖𝐡𝐴‖‖=β€–(3,7βˆ’π‘˜)β€–=3+(7βˆ’π‘˜).

We are told that the magnitude of this vector is equal to √34. This can only happen if the radicand is equal to 34. Thus, 34=3+(7βˆ’π‘˜).

We can expand the brackets to get 34=9+49βˆ’14π‘˜+π‘˜0=π‘˜βˆ’14π‘˜+24.

We can then factor the quadratic by finding two numbers whose product is 24 and that add up to βˆ’14. By checking the factors of 24, we can find that these numbers are βˆ’12 and βˆ’2. Therefore, we have 0=(π‘˜βˆ’2)(π‘˜βˆ’12).

Hence, either π‘˜=2 or π‘˜=12.

In our final example, we will determine the possible values of an unknown component of a vector.

Example 7: Finding an Unknown Component

Given that 𝐴 is a point that lies on the circle with equation π‘₯+𝑦=16 and that 𝑂𝐴=(π‘˜,4π‘˜), find all possible values of π‘˜. Give your answers in exact form.

Answer

There are many ways of answering this question. We will use the fact that the circle with equation π‘₯+𝑦=16 is a circle of radius 4 centered at the origin. We recall that the position vector of a point has components equal to its coordinates and that the magnitude of this position vector is the same as the point’s distance from the origin.

Since 𝐴 is on this circle, its distance from the origin is 4 and ‖‖𝑂𝐴‖‖=4. We can find the magnitude of a vector by taking the square root of the sum of the squares of the components: ‖‖𝑂𝐴‖‖=βˆšπ‘˜+(4π‘˜)4=√17π‘˜.

We can take the square root of each factor separately to get 4=√17Γ—βˆšπ‘˜.

We need to be careful since βˆšπ‘˜οŠ¨ is not necessarily equal to π‘˜. For example, consider π‘˜=βˆ’1. We have (βˆ’1)=√1=1. In general, βˆšπ‘˜=|π‘˜| since the square root function is defined as the nonnegative root of the radicand. Therefore, we can rewrite the equation as 4=√17Γ—|π‘˜|.

We can now rearrange to solve for π‘˜: |π‘˜|=4√17=4√17Γ—βˆš17√17=4√1717.

Hence, π‘˜=4√1717 or π‘˜=βˆ’4√1717.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • We call the vector 𝑂𝐴 the position vector of 𝐴. The components of 𝑂𝐴 are equal to the coordinates of 𝐴. We can also write the position vector of point 𝐴(π‘₯,𝑦) in terms of the fundamental unit vectors as 𝑂𝐴=π‘₯⃑𝑖+𝑦⃑𝑗.
  • The magnitude of the position vector of a point is the distance the point is from the origin.
  • For any points 𝐴 and 𝐡, we have 𝐴𝐡=οƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄. This means that we can represent any vector with initial point 𝐡 and terminal point 𝐴 using position vectors.

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