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

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 𝑣=(𝑎,𝑏) and 𝑤=(𝑐,𝑑), then 𝑣+𝑤=(𝑎+𝑐,𝑏+𝑑),𝑣𝑤=(𝑎𝑐,𝑏𝑑).

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 𝑣=(𝑎,𝑏) and a scalar 𝑐, 𝑐𝑣=(𝑐𝑎,𝑐𝑏).

We also recall that a 2D vector can be expressed in terms of fundamental unit vectors 𝑖 and 𝑗, which are defined by 𝑖=(1,0),𝑗=(0,1).

Starting from the component form of a vector, we can apply vector addition and scalar multiplication to write (𝑎,𝑏)=(𝑎,0)+(0,𝑏)=𝑎(1,0)+𝑏(0,1)=𝑎𝑖+𝑏𝑗.

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 𝑖 and 𝑗 as indeterminates and collect the like terms together. For instance, the addition of two vectors can be written as 𝑎𝑖+𝑏𝑗+𝑐𝑖+𝑑𝑗=𝑎𝑖+𝑏𝑗+𝑐𝑖+𝑑𝑗=𝑎𝑖+𝑐𝑖+𝑏𝑗+𝑑𝑗=(𝑎+𝑐)𝑖+(𝑏+𝑑)𝑗.

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 𝑐𝑎𝑖+𝑏𝑗=𝑐𝑎𝑖+𝑐𝑏𝑗, 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 92+4, we would first compute the subtraction 92=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 𝐴=(2,4) and 𝐵=(7,6), find 𝐴4𝐵.

Answer

In this example, we need to compute 𝐴4𝐵, 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 4𝐵. Recall that given a vector 𝑣=(𝑎,𝑏) and a scalar 𝑐, the scalar multiplication is defined by 𝑐𝑣=(𝑐𝑎,𝑐𝑏).

Hence, 4𝐵=4(7,6)=(28,24).

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 𝐴4𝐵=(2,4)(28,24)=(2(28),4(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 𝑖 and 𝑗, where we write (𝑎,𝑏)=𝑎𝑖+𝑏𝑗.

Given two 2D vectors 𝐴 and 𝐵, we can sometimes write a third 2D vector 𝐶 in terms of 𝐴 and 𝐵 if we find scalars 𝑎 and 𝑏 satisfying 𝐶=𝑎𝐴+𝑏𝐵.

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 𝑎 and 𝑏. We should keep in mind that this system of equations does not always have a solution, in which case we cannot express vector 𝐶 in terms of 𝐴 and 𝐵. But when this system of equations leads to solutions 𝑎 and 𝑏, then we can express 𝐶 in terms of 𝐴 and 𝐵.

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 𝐴, 𝐵, and 𝐶 given below.

Let us consider what these operations mean graphically to express 𝐶 as 𝑎𝐴+𝑏𝐵=𝐶.

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 𝑎𝐴 and 𝑏𝐵 are scaled (or reversed) versions of vectors 𝐴 and 𝐵 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 𝐴 and 𝐵 to draw a parallelogram having 𝐶 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 𝐴=(4,1) and 𝐵=(2,1), express 𝐶=(8,1) in terms of 𝐴 and 𝐵.

Answer

In this example, we need to express vector 𝐶 in terms of two vectors 𝐴 and 𝐵. This means that we want to find scalars 𝑎 and 𝑏 such that 𝑎𝐴+𝑏𝐵=𝐶.

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 𝑣=(𝑎,𝑏) and a scalar 𝑐, the scalar multiplication is defined by 𝑐𝑣=(𝑐𝑎,𝑐𝑏).

This leads to 𝑎𝐴=𝑎(4,1)=(4𝑎,𝑎),𝑏𝐵=𝑏(2,1)=(2𝑏,𝑏).

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, 𝑎𝐴+𝑏𝐵=(4𝑎,𝑎)+(2𝑏,𝑏)=(4𝑎2𝑏,𝑎𝑏).

Since this needs to equal vector 𝐶, we obtain (4𝑎2𝑏,𝑎𝑏)=(8,1).

We know that two vectors are equal if the corresponding components of the vectors are equal. This leads to a pair of simultaneous equations: 4𝑎2𝑏=8,𝑎𝑏=1.

To solve this pair of simultaneous equations, let us first divide both sides of the first equation by 2: 2𝑎+𝑏=4,𝑎𝑏=1.

Then, adding these two equations leads to 𝑎=3. Substituting this value into the second equation leads to 3𝑏=1.

Rearranging this equation so 𝑏 is the subject, we obtain 𝑏=2. Hence, 𝐶 can be written as 3𝐴2𝐵.

Next, we will consider an example involving vector operations where vectors are defined by their endpoints. If we are given points 𝐴(𝑥,𝑦) and 𝐵(𝑥,𝑦), vector 𝐴𝐵 is defined by the difference of the position vectors 𝑂𝐵 and 𝑂𝐴. This leads to 𝐴𝐵=𝑂𝐵𝑂𝐴=(𝑥,𝑦)(𝑥,𝑦)=(𝑥𝑥,𝑦𝑦).

We also recall that 𝐴𝐵+𝐵𝐶=𝐴𝐶.

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 𝐴𝐶=(3,3), 𝐵𝐶=(13,7), and 2𝑂𝐶+2𝐴𝐵=(4,4), find the coordinates of the point 𝐶.

Answer

In this example, we need to find the coordinates of point 𝐶 from the given vector equation. On the left-hand side of the given equation, we see two vectors 𝑂𝐶 and 𝐴𝐵. We recall 𝑂𝐶 is the position vector of point 𝐶, which means that the components of this vector is given by the corresponding coordinates of 𝐶. Since we do not know the coordinates of 𝐶, let us denote the coordinates of 𝐶 by (𝑥,𝑦). Then, 𝑂𝐶=(𝑥,𝑦).

Next, consider the other vector in the given equation, 𝐴𝐵. 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 𝐴𝐵+𝐵𝐶=𝐴𝐶.

Let us denote 𝐴𝐵=(𝑎,𝑏). Substituting 𝐵𝐶=(13,7) and 𝐴𝐶=(3,3) into the equation above leads to (𝑎,𝑏)+(13,7)=(3,3).

We recall that we can add a pair of vectors by adding the corresponding components of the two vectors. This leads to (𝑎+13,𝑏7)=(3,3).

Hence, we must have 𝑎+13=3,𝑏7=3.

Rearranging these two equations leads to 𝑎=10 and 𝑏=10. This gives us 𝐴𝐵=(10,10).

Then, the given equation can be written as 2(𝑥,𝑦)+2(10,10)=(4,4).

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(𝑥,𝑦)=(2𝑥,2𝑦),2(10,10)=(20,20).

Then, the equation can be written as (2𝑥,2𝑦)+(20,20)=(4,4).

Adding the vectors on the left-hand side, (2𝑥20,2𝑦+20)=(4,4).

Setting the corresponding components of the two vectors equal leads to a pair of equations: 2𝑥20=4,2𝑦+20=4.

Rearranging these equations leads to 𝑥=8,𝑦=12, which gives the 𝑥- and 𝑦-coordinates of 𝐶. Hence, the coordinates of 𝐶 are (8,12).

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

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

Given that 𝐵=(9,3), 𝐶=(4,2), and 𝐷=(2,9), determine the vector 𝐴 that satisfies the equation 𝐴=4𝐵+2𝐶6𝐷.

Answer

In this example, we need to compute the right-hand side of the equation to find vector 𝐴. 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, 4𝐵=4(9,3)=(36,12),2𝐶=2(4,2)=(8,4),6𝐷=6(2,9)=(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 4𝐵+2𝐶6𝐷, we need to compute the addition 4𝐵+2𝐶 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 4𝐵+2𝐶=(36,12)+(8,4)=(36+(8),12+(4))=(28,8).

Finally, we can compute the subtraction: 4𝐵+2𝐶6𝐷=(28,8)(12,54)=(28(12),854)=(40,46).

Hence, 𝐴=(40,46).

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 𝐴=(3,5), 𝐵=(2,0), and 𝐶=(4,5), find 2𝐴2𝐵+2𝐶.

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: 2𝐴2𝐵+2𝐶.

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, 2𝐴=2(3,5)=(6,10),2𝐵=2(2,0)=(4,0),2𝐶=2(4,5)=(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 2𝐴2𝐵+2𝐶, we need to compute the subtraction 2𝐴2𝐵 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 2𝐴2𝐵=(6,10)(4,0)=(64,100)=(2,10).

Finally, we can compute the addition: 2𝐴2𝐵+2𝐶=(2,10)+(8,10)=(2+(8),10+10)=(6,0).

We have obtained that the vector resulting from the vector operations is (6,0). Now, we need to find the magnitude of this vector. Recall that the magnitude of a 2D vector is given by (𝑎,𝑏)=𝑎+𝑏.

Since our vector is (6,0), we can apply this formula with 𝑎=6 and 𝑏=0 to obtain (6,0)=(6)+0=6.

Hence, 2𝐴2𝐵+2𝐶=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 𝐴 and 𝐵, we express another 2D vector 𝐶 in terms of 𝐴 and 𝐵 if we find scalars 𝑎 and 𝑏 satisfying 𝐶=𝑎𝐴+𝑏𝐵.

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