Lesson Explainer: Vector Operations in 2D | Nagwa Lesson Explainer: Vector Operations in 2D | Nagwa

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Lesson Explainer: Vector Operations in 2D Mathematics • First Year of Secondary School

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In this explainer, we will learn how to perform operations on vectors algebraically such as vector addition, vector subtraction, and scalar multiplication in two dimensions.

Let us recall the vector operations that we will be using, beginning with the addition and subtraction of 2D vectors.

Definition: Addition and Subtraction of 2D Vectors

To add or subtract a pair of 2D vectors, we add or subtract the corresponding components of the two vectors. If โƒ‘๐‘ฃ=(๐‘Ž,๐‘) and โƒ‘๐‘ค=(๐‘,๐‘‘), then โƒ‘๐‘ฃ+โƒ‘๐‘ค=(๐‘Ž+๐‘,๐‘+๐‘‘),โƒ‘๐‘ฃโˆ’โƒ‘๐‘ค=(๐‘Žโˆ’๐‘,๐‘โˆ’๐‘‘).

Next, recall the scalar multiplication of a 2D vector.

Definition: Scalar Multiplication of 2D Vectors

To multiply a 2D vector by a scalar, we multiply each component of the vector by the scalar. Given a vector โƒ‘๐‘ฃ=(๐‘Ž,๐‘) and a scalar ๐‘, ๐‘โƒ‘๐‘ฃ=(๐‘๐‘Ž,๐‘๐‘).

We also recall that a 2D vector can be expressed in terms of fundamental unit vectors โƒ‘๐‘– and โƒ‘๐‘—, which are defined by โƒ‘๐‘–=(1,0),โƒ‘๐‘—=(0,1).

Starting from the component form of a vector, we can apply vector addition and scalar multiplication to write (๐‘Ž,๐‘)=(๐‘Ž,0)+(0,๐‘)=๐‘Ž(1,0)+๐‘(0,1)=๐‘Žโƒ‘๐‘–+๐‘โƒ‘๐‘—.

This explains how the conversion between these two forms of a vector is achieved. We can also perform algebraic operations of vectors using fundamental vectors. In this case, we can treat the fundamental unit vectors โƒ‘๐‘– and โƒ‘๐‘— as indeterminates and collect the like terms together. For instance, the addition of two vectors can be written as ๏€บ๐‘Žโƒ‘๐‘–+๐‘โƒ‘๐‘—๏†+๏€บ๐‘โƒ‘๐‘–+๐‘‘โƒ‘๐‘—๏†=๐‘Žโƒ‘๐‘–+๐‘โƒ‘๐‘—+๐‘โƒ‘๐‘–+๐‘‘โƒ‘๐‘—=๏€บ๐‘Žโƒ‘๐‘–+๐‘โƒ‘๐‘–๏†+๏€บ๐‘โƒ‘๐‘—+๐‘‘โƒ‘๐‘—๏†=(๐‘Ž+๐‘)โƒ‘๐‘–+(๐‘+๐‘‘)โƒ‘๐‘—.

This process leads to the same formula for addition of two vectors as the one given above in component form. Similarly, the scalar multiplication can be written as ๐‘๏€บ๐‘Žโƒ‘๐‘–+๐‘โƒ‘๐‘—๏†=๐‘๐‘Žโƒ‘๐‘–+๐‘๐‘โƒ‘๐‘—, which again is the same as the formula given above for a vector in component form. As we can see here, one of the benefits of using the fundamental unit vectors is that it makes algebraic operations of vectors more closely resemble the corresponding algebraic operations of real numbers.

We can combine multiple algebraic operations involving 2D vectors. Since algebraic operations of vectors resemble the corresponding operations of real numbers, we can determine the order of operations by following the order of operations of real numbers. Remember that the order of operations of real numbers is given by the acronym PEMDAS, which stands for parentheses, exponent, multiplication, division, addition, and subtraction. We also remember that the ordering between addition and subtraction is often performed in the order that the expression is written. For instance, to compute 9โˆ’2+4, we would first compute the subtraction 9โˆ’2=7 and then compute the addition 7+4=11.

Since the vector operations do not include exponentiation and division, we only need to consider these four operations: parentheses, multiplication, addition, and subtraction.

Rule: Order of Vector Operations

When we compute an algebraic expression involving 2D vectors, which includes scalar multiplication, addition, and subtraction, we must follow the order of operations as given below:

  1. Parentheses
  2. Scalar multiplication
  3. Addition and subtraction

In our first example, we will combine scalar multiplication with subtraction of 2D vectors.

Example 1: Evaluating Expressions Involving the Subtraction and Scalar Multiplication of Given Vectors in Two Dimensions

Given that โƒ‘๐ด=(2,โˆ’4) and โƒ‘๐ต=(โˆ’7,โˆ’6), find โƒ‘๐ดโˆ’4โƒ‘๐ต.

Answer

In this example, we need to compute โƒ‘๐ดโˆ’4โƒ‘๐ต, which involves the subtraction and scalar multiplication of 2D vectors. Recall that the order of vector operations is given by

  1. parentheses,
  2. scalar multiplication,
  3. addition and subtraction.

So, we can begin by computing the scalar multiplication 4โƒ‘๐ต. Recall that given a vector โƒ‘๐‘ฃ=(๐‘Ž,๐‘) and a scalar ๐‘, the scalar multiplication is defined by ๐‘โƒ‘๐‘ฃ=(๐‘๐‘Ž,๐‘๐‘).

Hence, 4โƒ‘๐ต=4(โˆ’7,โˆ’6)=(โˆ’28,โˆ’24).

Next, we can compute the subtraction. Recall that we can subtract a pair of vectors by subtracting the corresponding components of the two vectors. This gives us โƒ‘๐ดโˆ’4โƒ‘๐ต=(2,โˆ’4)โˆ’(โˆ’28,โˆ’24)=(2โˆ’(โˆ’28),โˆ’4โˆ’(โˆ’24))=(30,20).

In the previous example, we computed an algebraic expression involving 2D vectors that used the scalar multiplication and subtraction of vectors.

Using these concepts, we can sometimes write a vector in terms of two vectors. This idea is similar to how we can write any vector in terms of the fundamental unit vectors โƒ‘๐‘– and โƒ‘๐‘—, where we write (๐‘Ž,๐‘)=๐‘Žโƒ‘๐‘–+๐‘โƒ‘๐‘—.

Given two 2D vectors โƒ‘๐ด and โƒ‘๐ต, we can sometimes write a third 2D vector โƒ‘๐ถ in terms of โƒ‘๐ด and โƒ‘๐ต if we find scalars ๐‘Ž and ๐‘ satisfying โƒ‘๐ถ=๐‘Žโƒ‘๐ด+๐‘โƒ‘๐ต.

If we compute the scalar multiplication and addition of 2D vectors on the right-hand side of this equation, we can obtain a pair of simultaneous equations involving ๐‘Ž and ๐‘. We should keep in mind that this system of equations does not always have a solution, in which case we cannot express vector โƒ‘๐ถ in terms of โƒ‘๐ด and โƒ‘๐ต. But when this system of equations leads to solutions ๐‘Ž and ๐‘, then we can express โƒ‘๐ถ in terms of โƒ‘๐ด and โƒ‘๐ต.

While this process of expressing a 2D vector in terms of two vectors is purely algebraic, it is helpful to visualize this using a graph to understand why this may be possible. Consider the three vectors โƒ‘๐ด, โƒ‘๐ต, and โƒ‘๐ถ given below.

Let us consider what these operations mean graphically to express โƒ‘๐ถ as ๐‘Žโƒ‘๐ด+๐‘โƒ‘๐ต=โƒ‘๐ถ.

We know that the scalar multiplication of a vector results in stretching or contracting a vector and also reversing the direction of a vector if the scalar is negative. This means that the vectors ๐‘Žโƒ‘๐ด and ๐‘โƒ‘๐ต are scaled (or reversed) versions of vectors โƒ‘๐ด and โƒ‘๐ต respectively. We also know that the sum of two 2D vectors forms the diagonal of the parallelogram formed by the two vectors. Hence, the equation above means that we can scale vectors โƒ‘๐ด and โƒ‘๐ต to draw a parallelogram having โƒ‘๐ถ as its diagonal, as seen below.

In the next example, we will write a vector in terms of two other vectors.

Example 2: Using Operations on Vectors to Express a Vector in Terms of Two Other Vectors

Given that โƒ‘๐ด=(โˆ’4,โˆ’1) and โƒ‘๐ต=(โˆ’2,โˆ’1), express โƒ‘๐ถ=(โˆ’8,โˆ’1) in terms of โƒ‘๐ด and โƒ‘๐ต.

Answer

In this example, we need to express vector โƒ‘๐ถ in terms of two vectors โƒ‘๐ด and โƒ‘๐ต. This means that we want to find scalars ๐‘Ž and ๐‘ such that ๐‘Žโƒ‘๐ด+๐‘โƒ‘๐ต=โƒ‘๐ถ.

The left-hand side of this equation contains scalar multiplication and addition of 2D vectors. We know that the scalar multiplication should be computed before the addition of vectors. Recall that given a vector โƒ‘๐‘ฃ=(๐‘Ž,๐‘) and a scalar ๐‘, the scalar multiplication is defined by ๐‘โƒ‘๐‘ฃ=(๐‘๐‘Ž,๐‘๐‘).

This leads to ๐‘Žโƒ‘๐ด=๐‘Ž(โˆ’4,โˆ’1)=(โˆ’4๐‘Ž,โˆ’๐‘Ž),๐‘โƒ‘๐ต=๐‘(โˆ’2,โˆ’1)=(โˆ’2๐‘,โˆ’๐‘).

Next, let us add the two vectors. Recall that we can add a pair of vectors by adding the corresponding components of the two vectors. Hence, ๐‘Žโƒ‘๐ด+๐‘โƒ‘๐ต=(โˆ’4๐‘Ž,โˆ’๐‘Ž)+(โˆ’2๐‘,โˆ’๐‘)=(โˆ’4๐‘Žโˆ’2๐‘,โˆ’๐‘Žโˆ’๐‘).

Since this needs to equal vector โƒ‘๐ถ, we obtain (โˆ’4๐‘Žโˆ’2๐‘,โˆ’๐‘Žโˆ’๐‘)=(โˆ’8,โˆ’1).

We know that two vectors are equal if the corresponding components of the vectors are equal. This leads to a pair of simultaneous equations: โˆ’4๐‘Žโˆ’2๐‘=โˆ’8,โˆ’๐‘Žโˆ’๐‘=โˆ’1.

To solve this pair of simultaneous equations, let us first divide both sides of the first equation by โˆ’2: 2๐‘Ž+๐‘=4,โˆ’๐‘Žโˆ’๐‘=โˆ’1.

Then, adding these two equations leads to ๐‘Ž=3. Substituting this value into the second equation leads to โˆ’3โˆ’๐‘=โˆ’1.

Rearranging this equation so ๐‘ is the subject, we obtain ๐‘=โˆ’2. Hence, โƒ‘๐ถ can be written as 3โƒ‘๐ดโˆ’2โƒ‘๐ต.

Next, we will consider an example involving vector operations where vectors are defined by their endpoints. If we are given points ๐ด(๐‘ฅ,๐‘ฆ)๏Šง๏Šง and ๐ต(๐‘ฅ,๐‘ฆ)๏Šจ๏Šจ, vector ๏ƒ ๐ด๐ต is defined by the difference of the position vectors ๏ƒŸ๐‘‚๐ต and ๏ƒ ๐‘‚๐ด. This leads to ๏ƒ ๐ด๐ต=๏ƒŸ๐‘‚๐ตโˆ’๏ƒ ๐‘‚๐ด=(๐‘ฅ,๐‘ฆ)โˆ’(๐‘ฅ,๐‘ฆ)=(๐‘ฅโˆ’๐‘ฅ,๐‘ฆโˆ’๐‘ฆ).๏Šจ๏Šจ๏Šง๏Šง๏Šจ๏Šง๏Šจ๏Šง

We also recall that ๏ƒ ๐ด๐ต+๏ƒŸ๐ต๐ถ=๏ƒ ๐ด๐ถ.

This equation can be visualized by the following graph.

In the next example, we will find the missing coordinates of a point when we are given an equation involving 2D vector operations.

Example 3: Finding Missing Coordinates from Vector Equations

On a lattice, where ๏ƒ ๐ด๐ถ=(3,3), ๏ƒŸ๐ต๐ถ=(13,โˆ’7), and 2๏ƒ ๐‘‚๐ถ+2๏ƒ ๐ด๐ต=(โˆ’4,โˆ’4), find the coordinates of the point ๐ถ.

Answer

In this example, we need to find the coordinates of point ๐ถ from the given vector equation. On the left-hand side of the given equation, we see two vectors ๏ƒ ๐‘‚๐ถ and ๏ƒ ๐ด๐ต. We recall ๏ƒ ๐‘‚๐ถ is the position vector of point ๐ถ, which means that the components of this vector is given by the corresponding coordinates of ๐ถ. Since we do not know the coordinates of ๐ถ, let us denote the coordinates of ๐ถ by (๐‘ฅ,๐‘ฆ). Then, ๏ƒ ๐‘‚๐ถ=(๐‘ฅ,๐‘ฆ).

Next, consider the other vector in the given equation, ๏ƒ ๐ด๐ต. We note that this is not one of the vectors that are provided. However, we can find this vector using the given vectors by recalling ๏ƒ ๐ด๐ต+๏ƒŸ๐ต๐ถ=๏ƒ ๐ด๐ถ.

Let us denote ๏ƒ ๐ด๐ต=(๐‘Ž,๐‘). Substituting ๏ƒŸ๐ต๐ถ=(13,โˆ’7) and ๏ƒ ๐ด๐ถ=(3,3) into the equation above leads to (๐‘Ž,๐‘)+(13,โˆ’7)=(3,3).

We recall that we can add a pair of vectors by adding the corresponding components of the two vectors. This leads to (๐‘Ž+13,๐‘โˆ’7)=(3,3).

Hence, we must have ๐‘Ž+13=3,๐‘โˆ’7=3.

Rearranging these two equations leads to ๐‘Ž=โˆ’10 and ๐‘=10. This gives us ๏ƒ ๐ด๐ต=(โˆ’10,10).

Then, the given equation can be written as 2(๐‘ฅ,๐‘ฆ)+2(โˆ’10,10)=(โˆ’4,โˆ’4).

The left-hand side of this equation contains the scalar multiplication and addition of 2D vectors. We know that we need to compute the scalar multiplications before the addition. Recall that we can compute the scalar multiplication of a 2D vector by multiplying each component of the vector by the scalar. This means 2(๐‘ฅ,๐‘ฆ)=(2๐‘ฅ,2๐‘ฆ),2(โˆ’10,10)=(โˆ’20,20).

Then, the equation can be written as (2๐‘ฅ,2๐‘ฆ)+(โˆ’20,20)=(โˆ’4,โˆ’4).

Adding the vectors on the left-hand side, (2๐‘ฅโˆ’20,2๐‘ฆ+20)=(โˆ’4,โˆ’4).

Setting the corresponding components of the two vectors equal leads to a pair of equations: 2๐‘ฅโˆ’20=โˆ’4,2๐‘ฆ+20=โˆ’4.

Rearranging these equations leads to ๐‘ฅ=8,๐‘ฆ=โˆ’12, which gives the ๐‘ฅ- and ๐‘ฆ-coordinates of ๐ถ. Hence, the coordinates of ๐ถ are (8,โˆ’12).

In the next example, we will consider a vector operation involving three vectors.

Example 4: Addition, Subtraction, and Scalar Multiplication of Vectors

Given that โƒ‘๐ต=(โˆ’9,โˆ’3), โƒ‘๐ถ=(โˆ’4,โˆ’2), and โƒ‘๐ท=(โˆ’2,9), determine the vector โƒ‘๐ด that satisfies the equation โƒ‘๐ด=โˆ’4โƒ‘๐ต+2โƒ‘๐ถโˆ’6โƒ‘๐ท.

Answer

In this example, we need to compute the right-hand side of the equation to find vector โƒ‘๐ด. The expression we need to compute involves scalar multiplication, addition, and subtraction of 2D vectors. Recall that the order of vector operations is given by

  1. parentheses,
  2. scalar multiplication,
  3. addition and subtraction.

Since there are no parentheses in this expression, let us begin by computing the scalar multiplications. Recall that we can multiply a vector by a scalar by multiplying each component of the vector by the scalar. Hence, โˆ’4โƒ‘๐ต=โˆ’4(โˆ’9,โˆ’3)=(36,12),2โƒ‘๐ถ=2(โˆ’4,โˆ’2)=(โˆ’8,โˆ’4),6โƒ‘๐ท=6(โˆ’2,9)=(โˆ’12,54).

Now that we have finished this part, we can follow the order of addition and subtraction as it is written. Since the expression we want to compute is โˆ’4โƒ‘๐ต+2โƒ‘๐ถโˆ’6โƒ‘๐ท, we need to compute the addition โˆ’4โƒ‘๐ต+2โƒ‘๐ถ first. Recall that we can add or subtract 2D vectors by adding or subtracting the corresponding components of the two vectors. Using the vectors we computed above, we have โˆ’4โƒ‘๐ต+2โƒ‘๐ถ=(36,12)+(โˆ’8,โˆ’4)=(36+(โˆ’8),12+(โˆ’4))=(28,8).

Finally, we can compute the subtraction: โˆ’4โƒ‘๐ต+2โƒ‘๐ถโˆ’6โƒ‘๐ท=(28,8)โˆ’(โˆ’12,54)=(28โˆ’(โˆ’12),8โˆ’54)=(40,โˆ’46).

Hence, โƒ‘๐ด=(40,โˆ’46).

In the final example, we will compute a vector operation involving 3 vectors and compute the magnitude of the resulting vector.

Example 5: Finding the Norm of the Result of Operations on Vectors

Given that โƒ‘๐ด=(โˆ’3,5), โƒ‘๐ต=(2,0), and โƒ‘๐ถ=(โˆ’4,5), find โ€–โ€–โˆ’2โƒ‘๐ดโˆ’2โƒ‘๐ต+2โƒ‘๐ถโ€–โ€–.

Answer

In this example, we need to compute the magnitude of a vector resulting from vector operations. Let us first compute the result of the vector operations: โˆ’2โƒ‘๐ดโˆ’2โƒ‘๐ต+2โƒ‘๐ถ.

This expression we need to compute involves scalar multiplication, addition, and subtraction of 2D vectors. Recall that the order of vector operations is given by

  1. parentheses,
  2. scalar multiplication,
  3. addition and subtraction.

Since there is no parentheses in this expression, let us begin by computing the scalar multiplications. Recall that we can multiply a vector by a scalar by multiplying each component of the vector by the scalar. Hence, โˆ’2โƒ‘๐ด=โˆ’2(โˆ’3,5)=(6,โˆ’10),2โƒ‘๐ต=2(2,0)=(4,0),2โƒ‘๐ถ=2(โˆ’4,5)=(โˆ’8,10).

Now that we have finished this part, we can proceed with the addition and subtraction of the vectors in the given order. Since our expression is โˆ’2โƒ‘๐ดโˆ’2โƒ‘๐ต+2โƒ‘๐ถ, we need to compute the subtraction โˆ’2โƒ‘๐ดโˆ’2โƒ‘๐ต first. Recall that we can add or subtract 2D vectors by adding or subtracting the corresponding components of the two vectors. Using the vectors we computed above, we have โˆ’2โƒ‘๐ดโˆ’2โƒ‘๐ต=(6,โˆ’10)โˆ’(4,0)=(6โˆ’4,โˆ’10โˆ’0)=(2,โˆ’10).

Finally, we can compute the addition: โˆ’2โƒ‘๐ดโˆ’2โƒ‘๐ต+2โƒ‘๐ถ=(2,โˆ’10)+(โˆ’8,10)=(2+(โˆ’8),โˆ’10+10)=(โˆ’6,0).

We have obtained that the vector resulting from the vector operations is (โˆ’6,0). Now, we need to find the magnitude of this vector. Recall that the magnitude of a 2D vector is given by โ€–(๐‘Ž,๐‘)โ€–=โˆš๐‘Ž+๐‘.๏Šจ๏Šจ

Since our vector is (โˆ’6,0), we can apply this formula with ๐‘Ž=โˆ’6 and ๐‘=0 to obtain โ€–(โˆ’6,0)โ€–=๏„(โˆ’6)+0=6.๏Šจ๏Šจ

Hence, โ€–โ€–โˆ’2โƒ‘๐ดโˆ’2โƒ‘๐ต+2โƒ‘๐ถโ€–โ€–=6.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • Vector operations of scalar multiplication, addition, and subtraction resemble the corresponding operations of real numbers.
  • When we compute an algebraic expression involving 2D vectors, which includes scalar multiplication, addition, and subtraction, we must follow the order of operations as given below:
    • Parentheses
    • Scalar multiplication
    • Addition and subtraction
  • Given 2D vectors โƒ‘๐ด and โƒ‘๐ต, we express another 2D vector โƒ‘๐ถ in terms of โƒ‘๐ด and โƒ‘๐ต if we find scalars ๐‘Ž and ๐‘ satisfying โƒ‘๐ถ=๐‘Žโƒ‘๐ด+๐‘โƒ‘๐ต.

ุงู†ุถู… ุฅู„ู‰ ู†ุฌูˆู‰ ูƒู„ุงุณูŠุฒ

ุดุงุฑูƒ ููŠ ุงู„ุญุตุต ุงู„ู…ุจุงุดุฑุฉ ุนู„ู‰ ู†ุฌูˆู‰ ูƒู„ุงุณูŠุฒ ูˆุญู‚ู‚ ุงู„ุชู…ูŠุฒ ุงู„ุฏุฑุงุณูŠ ุจุฅุฑุดุงุฏ ูˆุชูˆุฌูŠู‡ ู…ู† ู…ุฏุฑุณ ุฎุจูŠุฑ!

  • ุญุตุต ุชูุงุนู„ูŠุฉ
  • ุฏุฑุฏุดุฉ ูˆุฑุณุงุฆู„
  • ุฃุณุฆู„ุฉ ุงู…ุชุญุงู†ุงุช ูˆุงู‚ุนูŠุฉ

ุชุณุชุฎุฏู… «ู†ุฌูˆู‰» ู…ู„ูุงุช ุชุนุฑูŠู ุงู„ุงุฑุชุจุงุท ู„ุถู…ุงู† ุญุตูˆู„ูƒ ุนู„ู‰ ุฃูุถู„ ุชุฌุฑุจุฉ ุนู„ู‰ ู…ูˆู‚ุนู†ุง. ุงุนุฑู ุงู„ู…ุฒูŠุฏ ุนู† ุณูŠุงุณุฉ ุงู„ุฎุตูˆุตูŠุฉ