Vector magnitude and direction review

Review your knowledge of vector magnitude and direction, and use them to solve problems.
Magnitude of (a,b)(a,b)
(a,b)=a2+b2\mid\mid\!\!\!(a,b)\!\!\mid\mid=\sqrt{a^2+b^2}
Direction of (a,b)(a,b)
θ=tan1(ba)\theta=\tan^{-1}\left(\dfrac{b}{a}\right)
Components from magnitude u\mid\mid\!\!\!\vec u\!\!\mid\mid and direction θ\theta
(ucos(θ),usin(θ))\bigg(\mid\mid\!\!\!\vec u\!\!\mid\mid\cos(\theta),\mid\mid\!\!\!\vec u\!\!\mid\mid\sin(\theta)\bigg)

What are vector magnitude and direction?

We are used to describing vectors in component form. For example, (3,4)(3,4). 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 magnitude\blueD{\text{magnitude}} and direction\greenD{\text{direction}}:
The magnitude\blueD{\text{magnitude}} of a vector gives the length of the line segment, while the direction\greenD{\text{direction}} gives the angle the line forms with the positive xx-axis.
The magnitude of vector v\vec v is usually written as v||\vec v||.
Want to learn more about vector magnitude? Check out this video.
Want to learn more about vector direction? Check out this video.

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):
(a,b)=a2+b2||(a,b)||=\sqrt{a^2+b^2}
For example, the magnitude of (3,4)(3,4) is 32+42=25=5\sqrt{3^2+4^2}=\sqrt{25}=5.
Problem 1.1
u=(1,7)\vec u = (1,7)
u=||\vec 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:
θ=tan1(ba)\theta=\tan^{-1}\left(\dfrac{b}{a}\right)
This results from using trigonometry in the right triangle formed by the vector and the xx-axis.

Example 1: Quadrant I\text{I}

Let's find the direction of (3,4)(3,4):
tan1(43)53\tan^{-1}\left(\dfrac{4}{3}\right)\approx 53^\circ

Example 2: Quadrant IV\text{IV}

Let's find the direction of (3,4)(3,-4):
tan1(43)53\tan^{-1}\left(\dfrac{-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 360360^\circ:
53+360=307-53^\circ+360^\circ=307^\circ

Example 3: Quadrant II\text{II}

Let's find the direction of (3,4)(-3,4). First, notice that (3,4)(-3,4) is in Quadrant II\text{II}.
tan1(43)53\tan^{-1}\left(\dfrac{4}{-3}\right)\approx -53^\circ
53-53^\circ is in Quadrant IV\text{IV}, not II\text{II}. We must add 180180^\circ to obtain the opposite angle:
53+180=127-53^\circ+180^\circ=127^\circ
Problem 2.1
u=5i^+8j^\vec u = 5\hat i +8\hat j
θ=\theta =
^\circ
Enter your answer as an angle in degrees between 00 ^\circ and 360360^\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:
u=(ucos(θ),usin(θ))\vec u=\bigg(||\vec u||\cos(\theta),||\vec u||\sin(\theta)\bigg)
This results from using trigonometry in the right triangle formed by the vector and the xx-axis.
For example, this is the component form of the vector with magnitude 2\blueD 2 and angle 30\greenD{30^\circ}:
(2cos(30),2sin(30))=(3,1)\bigg(\blueD 2\cos(\greenD{30^\circ}),\blueD 2\sin(\greenD{30^\circ})\bigg)=(\sqrt 3,1)
Problem 3.1
u( \vec u \approx (~
 ,~,
))
Round your final answers to the nearest hundredth.
Want to try more problems like this? Check out this exercise.
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