If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Algebra (all content)>Unit 19

Lesson 8: Component form of vectors

# Converting between vector components and magnitude & direction review

Review how to find a vector's magnitude and direction from its components and vice versa.

## Cheat sheet

### Vector magnitude from components

The magnitude of $\left(a,b\right)$ is $||\left(a,b\right)||=\sqrt{{a}^{2}+{b}^{2}}$.

### Vector direction from components

The direction angle of $\left(a,b\right)$ is $\theta ={\mathrm{tan}}^{-1}\left(\frac{b}{a}\right)$ plus a correction based on the quadrant, according to this table:
Q1${\mathrm{tan}}^{-1}\left(\frac{b}{a}\right)$
Q2${\mathrm{tan}}^{-1}\left(\frac{b}{a}\right)+180\mathrm{°}$
Q3${\mathrm{tan}}^{-1}\left(\frac{b}{a}\right)+180\mathrm{°}$
Q4${\mathrm{tan}}^{-1}\left(\frac{b}{a}\right)+360\mathrm{°}$

### Vector components from magnitude & direction

The components of a vector with magnitude $||\stackrel{\to }{u}||$ and direction $\theta$ are $\left(||\stackrel{\to }{u}||\mathrm{cos}\left(\theta \right),||\stackrel{\to }{u}||\mathrm{sin}\left(\theta \right)\right)$.

## What are vector magnitude and direction?

We are used to describing vectors in component form. For example, $\left(3,4\right)$. We can plot vectors in the coordinate plane by drawing a directed line segment from the origin to the point that corresponds to the vector's components:
Considered graphically, there's another way to uniquely describe vectors — their $\text{magnitude}$ and $\text{direction}$:
The $\text{magnitude}$ of a vector gives the length of the line segment, while the $\text{direction}$ gives the angle the line forms with the positive $x$-axis.
The magnitude of vector $\stackrel{\to }{v}$ is usually written as $||\stackrel{\to }{v}||$.

## Practice set 1: Magnitude from components

To find the magnitude of a vector from its components, we take the square root of the sum of the components' squares (this is a direct result of the Pythagorean theorem):
$||\left(a,b\right)||=\sqrt{{a}^{2}+{b}^{2}}$
For example, the magnitude of $\left(3,4\right)$ is $\sqrt{{3}^{2}+{4}^{2}}=\sqrt{25}=5$.
Problem 1.1
$\stackrel{\to }{u}=\left(1,7\right)$
$||\stackrel{\to }{u}||=$

Either enter an expression with a square root symbol or a decimal rounded to the nearest hundredth.

Want to try more problems like this? Check out this exercise.

## Practice set 2: Direction from components

To find the direction of a vector from its components, we take the inverse tangent of the ratio of the components:
$\theta ={\mathrm{tan}}^{-1}\left(\frac{b}{a}\right)$
This results from using trigonometry in the right triangle formed by the vector and the $x$-axis.

### Example 1: Quadrant $\text{I}$‍

Let's find the direction of $\left(3,4\right)$:
${\mathrm{tan}}^{-1}\left(\frac{4}{3}\right)\approx {53}^{\circ }$

### Example 2: Quadrant $\text{IV}$‍

Let's find the direction of $\left(3,-4\right)$:
${\mathrm{tan}}^{-1}\left(\frac{-4}{3}\right)\approx -{53}^{\circ }$
The calculator returned a negative angle, but it's common to use positive values for the direction of a vector, so we must add ${360}^{\circ }$:
$-{53}^{\circ }+{360}^{\circ }={307}^{\circ }$

### Example 3: Quadrant $\text{II}$‍

Let's find the direction of $\left(-3,4\right)$. First, notice that $\left(-3,4\right)$ is in Quadrant $\text{II}$.
${\mathrm{tan}}^{-1}\left(\frac{4}{-3}\right)\approx -{53}^{\circ }$
$-{53}^{\circ }$ is in Quadrant $\text{IV}$, not $\text{II}$. We must add ${180}^{\circ }$ to obtain the opposite angle:
$-{53}^{\circ }+{180}^{\circ }={127}^{\circ }$
Problem 2.1
$\stackrel{\to }{u}=\left(5,8\right)$
$\theta =$
${}^{\circ }$
Enter your answer as an angle in degrees between ${0}^{\circ }$ and ${360}^{\circ }$ rounded to the nearest hundredth.

Want to try more problems like this? Check out this exercise.

## Practice set 3: Components from magnitude and direction

To find the components of a vector from its magnitude and direction, we multiply the magnitude by the sine or cosine of the angle:
$\stackrel{\to }{u}=\left(||\stackrel{\to }{u}||\mathrm{cos}\left(\theta \right),||\stackrel{\to }{u}||\mathrm{sin}\left(\theta \right)\right)$
This results from using trigonometry in the right triangle formed by the vector and the $x$-axis.
For example, this is the component form of the vector with magnitude $2$ and angle ${30}^{\circ }$:
$\left(2\mathrm{cos}\left({30}^{\circ }\right),2\mathrm{sin}\left({30}^{\circ }\right)\right)=\left(\sqrt{3},1\right)$
Problem 3.1
$\right)$

Want to try more problems like this? Check out this exercise.

## Want to join the conversation?

• problem2.2 has two answers,am i right?
325.01 & 145.01
(1 vote)
• I think there's only one answer that fits the question. You are right that arctan(7/-10) yields two answers in the range 0-360 degrees, but the vector u is in the second quadrant (u=-10i+7j), and so its angle cannot be 325 degrees, even though a vector with that angle has the same slope/tangent value. A vector with a direction of 325 degrees would be in the fourth quadrant.
• What is the pedagogical reason for using parenthesis vector notation? I find that for a first introduction to vectors, it is hard for students to understand how vectors are different from ordered pairs. Using the identical notation, just makes the situation worse. Why not use matrix notation (x above y) or angle vector brackets?
• It may be to underline the fact that there really is no difference between a vector and a standard ordered pair. The only real difference is that we've defined addition between vectors.
• How do you calculate the component form of a vector in 3D (x,y,z)?
• If you have all three coordinates it is super simple. xi + jy + zk. If you only have the magnitude and some angle, you would actually need two angles, usually from two of the three planes. the xy plane, xz plane or yz plane.
• If vector A = 12i – 16j and vector B = –24i + 10j, what is the direction of the vector C = 2A – B?
• Vector addition, or subtraction, is just combining steps in the various directions. then finding the direction is taking the inverse tangent of the ratio of the combined j steps over the combined i steps.

Your question says 2A - B. if this were just A - B you would do 12i - 16j -(-24i + 10i). Here A is multiplied by 2, so let's get that done first.

2A
2(12i-16j)
24i - 32j

There, now we can do the subtraction part

2A - B
24i - 32j - (-24i + 10j)
24i - 32j + 24i - 10j
48i - 42j

So now we have C but we want the direction. First it's important to note which quadrant it will be in. The i direction is positive, so we know it will be right of the y axis, and the j term is negative so we know it will be below the x axis. this is quadrant 4, so we know the answer will be between 270 and 360

When you want the direction of a vector you take the i term as a and j term as b then take arctan(b/a)

here a = 48 and b = -42 so arctan(-42/48) = -41.19. just to double check this is in quadrant 4 so it is the right answer. If you started with a = -48 and b = 42 you would get the same answer but -48i + 42j is in quadrant 2, so you would take -41.19 and add 180 to get 138.81, which is in quadrant 2.

Let me know if this didn't help
• What about 3 Dimensional Vectors? How would I calculate and express the angle it builds with the planes of the origin?
• How do I find which quadrant the vector would be in? I think i missed something along the way.