Explainer: Vector Operations in 2D

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.

We are going to look at the i and j form of vectors; we will add and subtract vectors in this form and practice finding vector components given a magnitude and direction and vice versa.

First, let us recap the i and j vector format. The vector i, as seen in the figure, is a unit vector, so it has length one, in the direction of the positive π‘₯-axis.

The vector j, as seen in the figure, is a unit vector in the direction of the positive 𝑦-axis.

We can write the vector i as i=⟨1,0⟩ and the vector j similarly as j=⟨0,1⟩.

We can string together any number of i and j vectors. For example, the vector 𝐴𝐡=⟨3,2⟩ can be expressed using i and j notation. The vector takes us from 𝐴 to 𝐡 by traveling three units right and two units up. So 𝐴𝐡=⟨3,2⟩ is the same as 3i+2j. This can be seen in the figure.

We can add and subtract vectors in i and j form by treating the i and j components separately. For example, given a vector v=⟨π‘₯1,𝑦1⟩ and a vector u=⟨π‘₯2,𝑦2⟩, the sum v+u=⟨π‘₯1+π‘₯2,𝑦1+𝑦2⟩. We will demonstrate this in examples 1 and 2.

Example 1: Adding Vectors

Find the sum of the vectors 𝐴𝐡=3iβˆ’4j and 𝐢𝐷=βˆ’5iβˆ’5j.

Answer

Firstly, let us draw a sketch of the two vectors.

If we write this out as a sum, we have 𝐴𝐡+𝐢𝐷=(3iβˆ’4j)+(βˆ’5iβˆ’5j).

If we then group the horizontal and vertical components, we will be able to simplify the expression: 𝐴𝐡+𝐢𝐷=3iβˆ’5iβˆ’4jβˆ’5j=βˆ’2iβˆ’9j.

This process can be expressed visually by placing the vectors together, as seen in the figure, where the pink line represents the vector 𝐴𝐡+𝐢𝐷.

Example 2: Subtracting Vectors

Given the vectors 𝐴𝐡=3iβˆ’4j and 𝐢𝐷=βˆ’5iβˆ’5j, calculate οƒ π΄π΅βˆ’οƒ πΆπ·.

Answer

Firstly, we will write out the difference of the two vectors: οƒ π΄π΅βˆ’οƒ πΆπ·=(3iβˆ’4j)βˆ’(βˆ’5iβˆ’5j).

This time we need to be a little bit more careful as we are subtracting negatives. Remember, if you subtract a negative, this is equivalent to adding. So if we group the horizontal and vertical components as before, we have that οƒ π΄π΅βˆ’οƒ πΆπ·=3i+5iβˆ’4j+5j.

Simplifying, we find that οƒ π΄π΅βˆ’οƒ πΆπ·=8i+j.

What we have done by subtracting 𝐢𝐷 is reverse its direction; this can be seen in the following figure.

If we are given a vector’s magnitude and its direction, we can convert this to find the vector’s component form using trigonometry. We will demonstrate this in the next couple of examples.

Example 3: Finding the Component Form of a Vector given Its Magnitude and Direction

The vector 𝐴𝐡 has a magnitude of 10 units and a direction of 120∘ counterclockwise from the positive π‘₯-axis. Write the vector in component form.

Answer

Let us start by drawing a sketch.

In order to convert 𝐴𝐡 to component form, we need to find its horizontal and vertical displacement, which can be calculated using trigonometry. First, it can be helpful to draw a right triangle and note the direction of each component and the value of the hypotenuse. Then, the i component can be calculated by multiplying the magnitude (or hypotenuse) by cosπœƒ and similarly the j component can be calculated by multiplying the magnitude by sinπœƒ, πœƒ being the angle counterclockwise from the positive π‘₯-axis.

In this example, the magnitude (or hypotenuse) is 10 and the value of πœƒ is 120∘ so the i component is 𝐴𝐡cosπœƒ=10cos(120)=βˆ’5 and the j component is 𝐴𝐡sinπœƒ=10sin(120)=5√3.

Notice here that the the sign of the horizontal component is negative which reflects what we established when drawing our diagram. So, our vector 𝐴𝐡 can be written in component form as follows: 𝐴𝐡=βˆ’5i+5√3j.

An alternative approach here would be to work out the acute angle in the right triangle and use standard right triangle trigonometry. If we do this, however, we need to be particularly careful in checking the direction of our vectors.

Example 4: Finding the Component Form of a Vector given Its Magnitude and Direction

The vector 𝐴𝐡 has a magnitude of 5 units and a direction of 5∘ counterclockwise from the positive π‘₯-axis. Write the vector in component form. Give the coefficients of i and j to two decimal places.

Answer

In order to answer this question, we need to first determine the quadrant in which the vector lies. We are told that the vector is positioned in a position 5∘ above the positive real axis, which puts it in the first quadrant. If we then sketch out a right triangle to represent the vector, we get the following triangle.

Using standard right trigonometry, we can form an equation for π‘₯. We have π‘₯=5cos(5), which gives π‘₯=4.98(2d.p.).

We can also form an equation for 𝑦: 𝑦=5sin(5), which gives 𝑦=0.44(2d.p.).

Therefore, the component form of the vector 𝐴𝐡 is 𝐴𝐡=4.98i+0.44j.

We can generalize the method used in examples 3 and 4 to find the component form of a vector. Given a vector v with magnitude |v| and direction πœƒβˆ˜ counterclockwise from the positive π‘₯-axis, the component form of v can be calculated using the formula v=|v|cosπœƒi+|v|sinπœƒj.

Now let us look at this process in reverse. If we are given a vector in component form, how do we find the vectors’ magnitude and direction? It is easier to approach this problem geometrically. We will explain this in example 5.

Example 5: Finding the Magnitude and Direction of a Vector in Component Form

Given the vector v=8iβˆ’6j, find its magnitude and direction, giving the direction as an angle, πœƒ, to two decimal places in the range βˆ’180∘<πœƒβ‰€180∘.

Answer

Our first step in answering this question is to draw a diagram. This will help us identify the quadrant in which the vector lies.

We can see from our diagram that the vector lies in the fourth quadrant. We can calculate the magnitude, |v|, of the vector by finding the length of the hypotenuse using the Pythagorean theorem: |v|=82+(βˆ’6)2=√64+36=10.

We can work out the angle in the right triangle using trigonometry. Let π‘₯ be the angle of the vector below the π‘₯-axis, if we consider the opposite to be βˆ’6 and the adjacent to be 8, we have that π‘₯=tanβˆ’1ο€Όβˆ’68=βˆ’36.87∘.

Notice here that π‘₯ is negative; the negative is informing us of a direction. A negative angle is considered to be measured in the clockwise direction and a positive angle is considered to be measured in the counterclockwise direction. Here, we have worked out the angle to v in the clockwise direction from the positive π‘₯-axis. However, the convention is to give a vector’s direction in a counterclockwise direction from the positive π‘₯-axis. So, πœƒ in our diagram will be equal to 360βˆ˜βˆ’36.87∘=323.13∘. In summary, our vector v has a magnitude of 10 units and a direction of 323.13∘.

As before, we can generalize this result. For a vector v=π‘Ži+𝑏j, the magnitude will be equal to |v|=βˆšπ‘Ž2+𝑏2, and the direction is calculated using the formula πœƒ=tanβˆ’1ο€½π‘π‘Žο‰.

With the direction, however, remember to be pay attention to the quadrant in which the vector lies, as the convention is to give the direction as an angle, counterclockwise from the positive π‘₯-axis.

Key Points

To convert vectors into component form, remember the following steps:

  1. It is often useful to draw a diagram of the vector to determine the quadrant in which it lies. You can then use standard trigonometry to determine the angle.
  2. Make sure you take note of the direction and reference point of the angle of the vector, for example, 40∘ counterclockwise from the positive π‘₯ axis.
  3. You can use the general formula to find the component form of a vector:
    Given a vector v with magnitude |v| and direction πœƒβˆ˜ counterclockwise from the positive π‘₯-axis, the component form of v can be calculated using the formula v=|v|cosπœƒi+|v|sinπœƒj.
  4. The unit vector in the positive π‘₯ direction is i and the unit vector in the positive 𝑦 direction is j. Once the vectors are written in component form, you can easily find the sum or difference of two vectors by adding or subtracting the corresponding coefficients of the unit vectors.

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