Lesson Objectives
  • Learn about the basic terminology of vectors
  • Learn how to find the component form of a vector
  • Learn how to find the magnitude of a vector
  • Learn how to find the direction angle of a vector

How to Find the Component Form, Magnitude, & Direction Angle of a Vector


Basic Terminology of Vectors

In math, the magnitude of a mathematical object is a property that determines whether the object is larger or smaller than other objects of the same kind. The magnitude of any complex number is known as its absolute value, which we have seen throughout our course.
Real Number r: $$|r|=r, r ≥ 0$$ $$|r|=-r, r < 0$$ Complex Number z: $$z=a + bi$$ $$|z|=\sqrt{a^2 + b^2}$$ Quantities that involve magnitudes, such as 50 pounds, 200 miles per hour, or 64° can be represented by real numbers called scalars. A scalar is any measurement that has magnitude, but no direction. When we refer to vector quantities, these involve both magnitude and direction. As an example, traveling 80 miles per hour north represents a vector quantity.
A vector quantity can be represented with a directed line segment called a vector. A directed line segment has both magnitude and direction. The magnitude will refer to the length of the directed line segment, whereas, the direction of the vector, indicated by the arrowhead represents the direction of the quantity.

Naming Vectors

Vectors can be named with one lowercase or uppercase letter or with two uppercase letters. showing the vector OP We can refer to our vector in the image above using an arrow over the letters: $$\overrightarrow{OP}$$ Note: Some textbooks will use boldface type instead of placing an arrow on top of the letter or letters. We will use boldface type when working inline for this tutorial, otherwise, we will stick with the arrow placed on top of the letter or letters that name our vector. When two letters are used to name a vector, the first letter indicates the initial point, whereas, the second indicates the terminal point of the vector. When we know these points, it gives us the direction of the vector. Two vectors will only be equal if and only if they have the same direction and the same magnitude. This means they must be parallel and in the same direction. showing the vector OP is not equal to vector PO As an example, we can see in the image above that vector OP is not equal to vector PO. Although they have the same magnitude, they have opposite directions, therefore, they can't be equal. $$\overrightarrow{OP}≠ \overrightarrow{PO}$$ The magnitude of a vector is written inside of vertical bars. For example, we could show the magnitude of vector OP as: $$\left|\overrightarrow{OP}\right|$$ Note: some books will use two vertical bars on each side. $$\left|\left|\overrightarrow{OP}\right|\right|$$

Magnitude and Direction Angle of a Vector

A vector with its initial point at the origin in a rectangular coordinate system is known as a position vector and is in standard position. A position vector u with an endpoint (a, b) will be written in component form as: $$\overrightarrow{u}=\langle a, b \rangle$$ Every vector in our real coordinate plane will correspond to an ordered pair of real numbers (x,y). Therefore, we can say that algebraically, a vector is an ordered pair, while geometrically, a vector is just a directed line segment. The numbers given as a and b represent the horizontal component and the vertical component, respectively, of vector u. We think of our first letter a as the x-value in our real number coordinate plane, therefore, this gives us a horizontal location for the terminal point. Similarly, we think of our second letter b as the y-value in our real number coordinate plane, therefore, this gives us the vertical location for the terminal point. algebraic interpretation of vectors In the image above, we see the vector u. The positive angle θ between the x-axis and a position vector is the direction angle for the vector. The length of u or the magnitude is notated with vertical bars: |u|.

Magnitude (length) of Vector u

Recall that the distance between two points on the coordinate plane can be found as: $$d=\sqrt{(x_{2}- x_{1})^2 + (y_{2}- y_{1})^2}$$ If one of our points is the origin, and the other is just labeled as (x,y), the formula becomes: $$d=\sqrt{x^{2}+ y^2}$$ Let's now relate this to our study of vectors. The magnitude or length of vector u is given by the following formula: $$\overrightarrow{u}=\langle a, b \rangle$$ $$\left|\overrightarrow{u}\right|=\sqrt{a^2 + b^2}$$

Direction Angle θ

The direction angle θ satisfies the following: $$\text{tan}\hspace{.1em}θ=\frac{b}{a}, a ≠ 0$$ Let's look at a few examples.
Example #1: Find the magnitude and direction angle for each vector. $$\overrightarrow{k}=\langle -2, -5 \rangle$$ Let's begin by finding the magnitude. $$\left|\overrightarrow{k}\right|=\sqrt{(-2)^2 + (-5)^2}$$ $$\left|\overrightarrow{k}\right|=\sqrt{4 + 25}$$ $$\left|\overrightarrow{k}\right|=\sqrt{29}$$ Now, let's find the direction angle θ. It may help to sketch a graph. Graphing our vector k on the coordinate plane $$\text{tan}\hspace{.1em}θ=\frac{b}{a}=\frac{-5}{-2}$$ Our position vector k has a negative horizontal component and a negative vertical component. This places our position vector in quadrant III as we can see from our sketch. What we will do is use our inverse tangent function with 5/2 as the argument. We will then want to find θ as an angle in quadrant III with this reference angle. Let's use our inverse tangent function. $$\text{tan}^{-1}\left(\frac{5}{2}\right) ≈ 68.2°$$ Since we are in quadrant III, we will add 180° to 68.2° to get our answer for θ $$θ ≈ 180° + 68.2° ≈ 248.2°$$ Example #2: Find the magnitude and direction angle for each vector. $$\overrightarrow{v}=\langle -7, -6 \rangle$$ Let's begin by finding the magnitude. $$\left|\overrightarrow{b}\right|=\sqrt{(-7)^2 + (-6)^2}$$ $$\left|\overrightarrow{b}\right|=\sqrt{49 + 36}$$ $$\left|\overrightarrow{b}\right|=\sqrt{85}$$ Now, let's find the direction angle θ. It may help to sketch a graph. Graphing our vector v on the coordinate plane $$\text{tan}\hspace{.1em}θ=\frac{b}{a}=\frac{-6}{-7}$$ Our position vector v has a negative horizontal component and a negative vertical component. This places our position vector in quadrant III as we can see from our sketch. What we will do is use our inverse tangent function with 6/7 as the argument. We will then want to find θ as an angle in quadrant III with this reference angle. Let's use our inverse tangent function. $$\text{tan}^{-1}\left(\frac{6}{7}\right) ≈ 40.6°$$ Since we are in quadrant III, we will add 180° to 40.6° to get our answer for θ $$θ ≈ 180° + 40.6° ≈ 220.6°$$

Finding the Component Form

In some cases, our initial point will not be the origin. If we are given an initial point P (p1, p2) and a terminal point Q (q1, q2), then our component form can be found as: $$\overrightarrow{PQ}=\langle q_{1}- p_{1}, q_{2}- p_{2}\rangle$$ Let's look at an example.
Example #3: Find the magnitude and direction angle for each vector. $$\overrightarrow{PQ}$$ $$P=(2, -9)$$ $$Q=(1, -4)$$ Since the letter P comes first, this will be our initial point and Q will be the terminal point. First, let's find the component form: $$\overrightarrow{PQ}=\langle 1 - 2, -4 - (-9) \rangle$$ $$\overrightarrow{PQ}=\langle -1, 5 \rangle$$ We can look at a sketch of the two and see they are equivalent vectors. Sketching vector PQ and PQ in standard position Now, let's use our formula to find the magnitude: $$\left|\overrightarrow{PQ}\right|=\sqrt{(-1)^2 + (5)^2}$$ $$\left|\overrightarrow{PQ}\right|=\sqrt{26}$$ Now, let's find the direction angle θ. It may help to sketch a graph. Graphing our vector PQ on the coordinate plane in standard position $$\text{tan}\hspace{.1em}θ=\frac{5}{-1}$$ Since the horizontal component is negative and the vertical component is positive, this places us in quadrant II. Let's use our inverse tangent function. $$θ=\text{tan}^{-1}(-5) ≈ -78.69°$$ This tells us we want to find an angle in quadrant II with a reference angle of 78.69° $$180° - 78.69°=101.31°$$ $$θ ≈ 101.31°$$

Horizontal and Vertical Components

$$a=\left|\overrightarrow{u}\right| \hspace{.1em}\text{cos}\hspace{.1em}θ$$ $$b=\left|\overrightarrow{u}\right| \hspace{.1em}\text{sin}\hspace{.1em}θ$$ $$\overrightarrow{u}=\langle a, b \rangle=\left\langle \left|\overrightarrow{u}\right| \hspace{.1em}\text{cos}\hspace{.1em}θ, \left|\overrightarrow{u}\right| \hspace{.1em}\text{sin}\hspace{.1em}θ \right\rangle$$ In some cases, we will be given the magnitude and direction angle and be asked to find the vector component form. Let's look at a few examples.
Example #4: Write each vector in component form. $$\left|\overrightarrow{k}\right|=32$$ $$θ=120°$$ Let's find our a, the horizontal component and b, the vertical component. $$a=\left|\overrightarrow{u}\right| \hspace{.1em}\text{cos}\hspace{.1em}θ$$ Let's replace this formula with what is given: $$a=32 \hspace{.1em}\text{cos}\hspace{.1em}120°$$ $$a=32 \hspace{.1em}\cdot -\frac{1}{2}=-16$$ Similarly, let's find b. $$b=\left|\overrightarrow{u}\right| \hspace{.1em}\text{sin}\hspace{.1em}θ$$ Let's replace this formula with what is given: $$b=32 \hspace{.1em}\text{sin}\hspace{.1em}120°$$ $$b=32 \hspace{.1em}\cdot \frac{\sqrt{3}}{2}$$ $$b=16 \sqrt{3}$$ $$\overrightarrow{k}=\langle -16, 16\sqrt{3}\rangle$$ Example #5: Write each vector in component form. Round your answer to the nearest hundredth. $$\left|\overrightarrow{k}\right|=8$$ $$θ=293°$$ Let's find our a, the horizontal component and b, the vertical component. $$a=\left|\overrightarrow{u}\right| \hspace{.1em}\text{cos}\hspace{.1em}θ$$ Let's replace this formula with what is given: $$a=8 \hspace{.1em}\cdot \text{cos}\hspace{.1em}293° ≈ 3.13$$ Similarly, let's find b. $$b=\left|\overrightarrow{u}\right| \hspace{.1em}\text{sin}\hspace{.1em}θ$$ Let's replace this formula with what is given: $$b=8 \hspace{.1em}\cdot \text{sin}\hspace{.1em}293° ≈ -7.36$$ $$\overrightarrow{k}=\langle 3.13, -7.36 \rangle$$

Skills Check:

Example #1

Find the magnitude. $$\overrightarrow{b}=\langle 15, -36 \rangle$$

Please choose the best answer.

A
$$\left|\overrightarrow{b}\right|=36$$
B
$$\left|\overrightarrow{b}\right|=15$$
C
$$\left|\overrightarrow{b}\right|=39$$
D
$$\left|\overrightarrow{b}\right|=30$$
E
$$\left|\overrightarrow{b}\right|=2\sqrt{15}$$

Example #2

Find the direction angle θ. $$\overrightarrow{v}=\langle -12, 18 \rangle$$

Please choose the best answer.

A
$$θ ≈ 123.69°$$
B
$$θ ≈ 146.31°$$
C
$$θ ≈ 213.69°$$
D
$$θ ≈ 33.69°$$
E
$$θ ≈ 22.97°$$

Example #3

Write the vector k in component form. $$\left| \overrightarrow{k}\right|=68$$ $$θ=300°$$

Please choose the best answer.

A
$$\overrightarrow{k}=\left\langle -34\sqrt{3}, 34\right\rangle$$
B
$$\overrightarrow{k}=\left\langle 34, -34\sqrt{3}\right\rangle$$
C
$$\overrightarrow{k}=\left\langle 34\sqrt{3}, 34\right\rangle$$
D
$$\overrightarrow{k}=\left\langle \sqrt{17}, 2\sqrt{2}\right\rangle$$
E
$$\overrightarrow{k}=\left\langle 2\sqrt{2}, \sqrt{17}\right\rangle$$
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