VM - Magnitude and Direction of Vectors Lesson

Magnitude and Direction of Vectors

You already know that vectors can be presented in component form or magnitude/direction form. While it is easier to add vectors in component form (add the horizontal and vertical components), vectors will most commonly be given to you by their magnitude and direction. So, let's practice going between forms.

Review

First, let's review the right triangle trig ratios. This will be important when converting component form to magnitude and direction.

right triangle trig ratios including sin, cos, and tan.

Example:

Find all missing angle measures and missing lengths of Triangle ABC.

Triangle problem

Answer:

Since we have a right triangle, we can use the Pythagorean Theorem to find the missing side.

$LaTeX: b^2+2^2=6^2\\b^2+4=36\\b^2=32\\b=\sqrt{32}\;or\;we\;can\;estimate\;this\;as\;5.66$ $b^2+2^2=6^2\\b^2+4=36\\b^2=32\\b=\sqrt{32}\;or\;we\;can\;estimate\;this\;as\;5.66$

Triangle ABC with b

Now let's find the trig ratios. (At this time, we won't reduce our ratios.)

$LaTeX: \sin\;A=\;\frac26\\\cos\;A=\;\frac{5.66}6\\\tan\;A=\;\frac2{5.66}\\\sin\;B=\;\frac{5.66}6\\\cos\;B=\;\frac26\\\tan\;B=\;\frac{5.66}2\\$ $\sin\;A=\;\frac26\\\cos\;A=\;\frac{5.66}6\\\tan\;A=\;\frac2{5.66}\\\sin\;B=\;\frac{5.66}6\\\cos\;B=\;\frac26\\\tan\;B=\;\frac{5.66}2\\$

We can now find the angle measures by using Inverse Trig Functions. Pick any ratio you'd like. I will use sin A.

$LaTeX: \sin\;A=\;\frac26\\$ $\sin\;A=\;\frac26\\$

To find the measure of angle A, we need to find the inverse sine of 2/6.

$LaTeX: A=\sin^{-1}(\;\frac26)\\$ $A=\sin^{-1}(\;\frac26)\\$

Make sure your calculator is in degree mode (not radians). If we find the above value in our calculator, we get

$LaTeX: measure\;of\;angle\;A\approx19.47^\circ\\$ $measure\;of\;angle\;A\approx19.47^\circ\\$

Similarly, if I had picked the ratio tan A,

$LaTeX: \tan\;A=\;\frac2{5.66}\\measure\;of\;angle\;A=\tan^{-1}(\frac2{5.66})\approx19.47\\$ $\tan\;A=\;\frac2{5.66}\\measure\;of\;angle\;A=\tan^{-1}(\frac2{5.66})\approx19.47\\$

We can follow the same process to find the measure of Angle B or we can simply use the fact that all right triangles add up to 180 degrees.

Now we will apply trig ratios and inverse trig functions to switch between component form and the magnitude/direction.

vector a on graph <x1,y1>

Let theta be an angle in standard position, measured from the positive x-axis to vector a <x1,y2>. Then
cos theta=x1/||a|| and sin theta=y1/||a||
x1=||a||cos theta and y1=||a||sin theta

Let's try a few problems. Given the magnitude and direction of each vector below, give the component form. (Give exact answers and approximate answers rounded to two decimal places.)

Problem: $LaTeX: \left|v\right|=8,\:\theta=45°$ $\left|v\right|=8,\:\theta=45°$

Solution: $LaTeX: <4\sqrt[]{2},\:4\sqrt[]{2}\:>\:or\:<5.66,\:5.66>$ $<4\sqrt[]{2},\:4\sqrt[]{2}\:>\:or\:<5.66,\:5.66>$

Problem: $LaTeX: \left|v\right|=24,\:\theta=210°$ $\left|v\right|=24,\:\theta=210°$

Solution: $LaTeX: <-12\sqrt[]{3},\:-12>\:or\:<-20.78,\:-12>$ $<-12\sqrt[]{3},\:-12>\:or\:<-20.78,\:-12>$

Problem: $LaTeX: \left|v\right|=7,\:\theta=300°$ $\left|v\right|=7,\:\theta=300°$

Solution: $LaTeX: <\frac{7}{2},\:-\frac{7\sqrt[]{3}}{2}>\:\:or\:\:<3.5,\:-6.06>$ $<\frac{7}{2},\:-\frac{7\sqrt[]{3}}{2}>\:\:or\:\:<3.5,\:-6.06>$

Problem: $LaTeX: \left|v\right|=11,\:\theta=150°$ $\left|v\right|=11,\:\theta=150°$

Solution: $LaTeX: <-\frac{11\sqrt[]{3}}{2},\:\frac{11}{2}>\:\:or\:<-9.52,\:5.5>$ $<-\frac{11\sqrt[]{3}}{2},\:\frac{11}{2}>\:\:or\:<-9.52,\:5.5>$

Now that we know how to go from the magnitude and direction to component form, let's go the other way.

Magnitude and direction of a vector in component form:
Magnitude: a=<x1,y1>
||a||= sq root ( (x1)sq + (y1)sq

Direction: tan theta=y1/x1
*be sure you chose the appropriate angle for the quadrant your vector is in

Let's try a few problems. Given the component form of each vector below, give the magnitude and direction. Round answers to the nearest tenth. Let 0°≤θ<360°.

Problem: $LaTeX: v=<4,\:-5>$ $v=<4,\:-5>$

Solution: $LaTeX: \left|v\right|=6.4,\:\theta=308.7°$ $\left|v\right|=6.4,\:\theta=308.7°$

Problem: $LaTeX: v=<-6,\:2>$ $v=<-6,\:2>$ $Magnitude and Direction of a Vector in Component Form.$

Solution: $LaTeX: \left|v\right|=6.3,\:\theta=161.6°$ $\left|v\right|=6.3,\:\theta=161.6°$

Problem: $LaTeX: v=<-3,\:-8>$ $v=<-3,\:-8>$

Solution: $LaTeX: \left|v\right|=8.5,\:\theta=249.4°$ $\left|v\right|=8.5,\:\theta=249.4°$

Problem: $LaTeX: v=<2,\:9>$ $v=<2,\:9>$

Solution: $LaTeX: \left|v\right|=9.2,\:\theta=77.5°$ $\left|v\right|=9.2,\:\theta=77.5°$

Example

Let's say you are given vector a with magnitude 10 and direction 30°, and vector b with magnitude 8 and direction 80°. We want to determine magnitude and direction of the resultant vector a + b.

vector b
|b|=8
80°
vector a
|a|=10
30°

Unfortunately, the magnitude of the resultant vector is not the sum of the magnitudes (18 in this case!). We have to find the horizontal and vertical components and add them together. Watch this video to see how.

Given the magnitude and direction of two vectors, find the magnitude and direction of the resultant vector. Round to the nearest tenth.

Solution: $LaTeX: \left|f+g\right|=13.3,\:\theta=125°$ $\left|f+g\right|=13.3,\:\theta=125°$

Solution: $LaTeX: \left|a+b\right|=33.5,\:\theta=260.7°$ $\left|a+b\right|=33.5,\:\theta=260.7°$

Solution: $LaTeX: \left|u+v\right|=22.1,\:\theta=359.8°$ $\left|u+v\right|=22.1,\:\theta=359.8°$

Solution: $LaTeX: \left|c+d\right|=9.3,\:\theta=339.1°$ $\left|c+d\right|=9.3,\:\theta=339.1°$

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