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## Algebra (all content)

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

Lesson 9: Adding vectors in magnitude & direction form# Adding vectors in magnitude & direction form (1 of 2)

Watch Sal add two vectors given in magnitude and direction form by breaking them down into components first. Created by Sal Khan.

## Want to join the conversation?

- At2:46, he says the side is 3(sqrt3)/2. Why? Where did the sqrt of 3 come from?(8 votes)
- The sqrt(3) comes from the fact that this is a 30/60/90 triangle. You can search google for "30 60 90 triangles" and see more clearly (with pictures) why this is such.(13 votes)

- Wouldn't it be easier to just convert vectors a and b to their component forms and then add the components, getting a decimal approx. for the sum of a and b (call it c) and then use the magnitude formula to solve?

a = (3 cos 30, 3 sin 30)

b = (2 cos 135, 2 sin 135)

a + b = c = (3 cos 30 + 2 cos 135, 3 sin 30 + 2 sin 135)

c ~= (-1.53, -2,79)

|| c || = sqrt((-1.53)^2 + (-2.79)^2) ~= 3.18 (I know it's slightly off from the answer in the part 2)

I guess this video just took the component form of the vectors and changed it into exact representations of the trig functions?(5 votes)- 𝒄 ≈ (1.18, 2.91)

You were in radian mode when you calculated the coordinates.

Nonetheless,

|𝒄| ≈ √((3 cos 30° + 2 cos 135°)² + (3 sin 30° + 2 sin 135°)²) ≈ 3.146

Let 𝜃 be the direction angle of 𝒄 ⇒ (since 𝒄 ∈ 𝐈) ⇒

⇒ 𝜃 = arctan((3 sin 30° + 2 sin 135°) ∕ (3 cos 30° + 2 cos 135°)) ≈ 67.89°

These are, of course, the same values that Sal got, so you're right, we don't need to worry about what the values of cos 30°, sin 30°, cos 135° and sin 135° are, as long as we don't care about the exact values of the magnitude and direction angle of 𝒄.

If we simplify the above expressions for |𝒄| and 𝜃, we get:

|𝒄| = √(13 + 3(√2 −√6))

𝜃 = arctan((9√3 + 6√2 + 6√6 + 8) ∕ 19)

Now, these expressions are hardly any easier to work with,*but*this doesn't mean that there aren't times when the simplified expressions actually are simpler, so it is still a good idea to at least try and simplify before taking the easy way out.(7 votes)

- I was surprised by how fast the video went (at 2.11) to assuming we know about 30/60/90 triangles and their proportions. I tried to work this out for myself using my memory of sine (soh), cos (cah) and to check on these calculations tan (toa). But where do we learn these pre trigonometrical theses?(5 votes)
- It is a big time-saver to use the triangle facts (side ratios) for 30-60-90 triangles and 45-45-90 triangles. The 30-60-90 triangle is half of an equilateral triangle, while the 45-45-90 triangle is half of a square cut on the diagonal. It is worthwhile to review them because so many examples are based on these triangles and angles. In case you have forgotten, it is easy to use Pythagoras to rediscover. This is covered in 8th grade geometry and Geometry Basics and in the Geometry section (with practice exercises) in case you want a refresher.

Meanwhile:

30-60-90

short side = half of long side (hypotenuse or vector)

the third side is length of short side times √3

If vector is 60 units, short side is 30 units and third side is 30√3 Boom! Done just like that.

45-45-90

two sides are equal (it is half of a square)

the long side (hypotenuse or vector) will equal the side length times √2

If a vector is 12 units long, divide by √2 to get the side length, which gives 6√2

Example with Pythagoras: If the triangle is 30-60-90 and the vector is 13 units

a² + b² = c²

If the short side is a, we know it is 13/2 units (this could be either the x or the y, depending upon the orientation of the triangle.

(13/2)² + b² = 13²

b² = 13² - (13/2)²

b = √13² - (13/2)²

b = √169 - 169/4

b = √676/4 - 169/4

b = √507/4

b = ½√169 ∙ 3

b = ½13 √3

So, b = 13/2 √3

Since the short side was 13/2, the result for the third side is 13/2`√3`

which is exactly what is expected using the shortcut ratios.

Hope that helps.(5 votes)

- But 3/2 and 3sqrt3/2 dont give 3(2 votes)
- Using the Pythagorean theorem, we know that

a^2 +b^2 =c^2

This means that the the length of the hypotenuse squared is equal to the sum of the two other sides squared.

If we find (3/2)^2 + (3sqrt3/2)^2, we get 9, which is the hypotenuse squared, therefore the actual length of the hypotenuse is 3.(6 votes)

- I don't know about others but in India we don't get calculator in exam, so how will I solve degree/radian problems...(5 votes)
- 2 pi radians equals 360 degrees. As a result, you can multiply your degrees by pi/180 to convert to radians. To go from radians to degrees, simply multiply by 180/pi(2 votes)

- what does sin stand for?(0 votes)
- That is the "sine function" from Trigonometry, typically used to describe the ratio of vertical motion to diagonal motion.(13 votes)

- Hi. How do you know that it is 30-60-90 triangle, not 45-45-90 ?(2 votes)
- Hello,

It will be a 30-60-90 angle triangle as one given angle is 30degree and other is a right angle. Applying the angle sum property of a triangle gives you the remaining angle, i.e. 60 degree.(2 votes)

- Wait...at1:10, Sal says that the horizontal vector for the triangle (I hope I'm using that term correctly) is something multiplied by an imaginary number. However, I remember doing skills that had graphs featuring the y-axis as the imaginary axis. I don't really understand why he's doing that and I'm not sure if I missed something(2 votes)
- At5:52how would we know that sine or cosine of 45 degrees is equal square-root of 2 over 2 (if Sal did not tell us) if we get decimal numbers on our calculators? How do we figure this out on our own for similar problems?(2 votes)
- Sines and cosines of certain degrees should be memorized. The cosine of 45 degrees in particular can be easily found to be √(2)/2 with just elementary geometry if you know the side ratios of a 45-45-90 triangle (which are 1 : 1 : √2).(1 vote)

- Okay, given that vector 𝑎+ vector 𝑏=0, how is the magnitude of 𝑎 related to the magnitude of 𝑏? Using the example Sal used, I am still unable to answer this question.(1 vote)
- When we add vectors, we arrange them tip-to-tail, then draw a new vector from 0 to the endpoint.

In our case, the new vector has length 0, so if we travel along a, b must then take us right back to the origin.

After we travel the length of a, we are a distance |a| from the origin (where |a| is the magnitude of a). We then travel another distance |b| and get back to the origin. So |a|=|b|. a and b have equal magnitudes.(3 votes)

## Video transcript

Voiceover:We have two vectors here. Vector A, it has a magnitude of three so the length of this blue arrow is three. Its direction, it forms a 33
degree angle with the positive, I guess you could say the positive x axis. I haven't drawn that here and vector B has a magnitude of two, the length of this arrow is two and it forms a 135 degree
angle with a positive x axis. What I want to think about in this video and maybe the next one depends
on how we do with the time is what is the magnitude
and direction of the sum of these two vectors. What is the magnitude and
direction of the vector A plus B? I encourage you to pause this video and try to work this through on your own before I work through it. Well to think about this, I'm going to decompose
each of these vectors into their horizontal and
their vertical components. For example vector A could be viewed as the sum of this
horizontal pointing vector plus this vertical pointing vector. Another way of thinking
about this horizontal vector, this right over here, this is going to be a scaled up version of the unit vector I and so this is going
to be something times I and this vector right over
here is going to be something times the unit vector in
the vertical direction. The vertical direction might
look something like that or let's see if this is
three then the unit vector, yeah, actually would
look something like that. That's what the J unit vector looks like and this is what the I unit
vector would look like. This is a scaled up version of it. It's something times I and
this is something times J. The same exact argument right over here. This is going to be something
times the I unit vector, its horizontal component and its vertical component
is going to be something times the J unit vector. What we really need to do is figure out the magnitudes
of these horizontal and vertical components and then we know how much to scale up I and how much to scale up J. Let's think through this a little bit. This one might jump out at you immediately is this is a 30, 60, 90 triangle. This is a 30, 60, 90 triangle
then this side right over here is going to be half the
length of the hypotenuse. This is going to be half of three, so it's three over two and
this one right over here, let me do that in same color. This is going to be three,
that's not the same, that's a different color. This is going to be three over two and then this length right over here. This is going to be square root of three times the shorter side. This is three times the
square root of three over two. I just took this and multiplied
it by the square root of three and once again that just came
out of 30, 60, 90 triangles. Now another way that you could tackle this is to use your, what would you know about your trig functions? You say, "Okay I know
this angle right here" "is a 30 degree angle,
this is the opposite side." You could say, "Well,
the opposite over three" "is going to be equal to
the sine of 30 degrees," let me right this down. Soh cah toa. Sine of 30 degrees is going to
be equal to the opposite side over the length of the hypotenuse. Over three or we could
say that the opposite side is equal to three times
the sine of 30 degrees. If you put this in your calculator, sine of 30 degrees is one half and so you're going to
get three halves here. Similarly for this side, this side is adjacent
to the 30 degree angle. You could say cosine, cosine
is adjacent over hypotenuse. You could say that cosine of 30 degrees is equal to adjacent over the hypotenuse or multiplying both sides by three. The adjacent is going to be equal to three times the cosine of 30 degrees and the cosine of 30 degrees
if you type in your calculator, you're actually going to
get some type of a decimal but it's square root of three over two. Three square root of three over two and you can verify this if you like. We see here, sine of 30 is indeed one half and cosine of 30 well you get this kind of crazy decimal but notice that is the same thing as square root of three divided by two. The exact same value right over here but we were able to figure that out just using what we know
about 30, 60, 90 triangles. Now this triangle, you might
say, "Well, what's the angle?" What angles do we know about it? Well this angle right over here is this is supplementary
to this 135 degree angle. This is a 45 degree angle. This is 90 so this is 45. This is a 45, 45, 90 degree triangle. Now you could say that
using the trig identities and what we did right
over here that this length is going to be the hypotenuse times the cosine of 45 degrees and you could say that
this length right over here is going to be the hypotenuse
times the sine of 45 degrees. Using the exact same logic here and as you get more practice
with finding the components, you'll realize, okay,
you take the hypotenuse times the cosine of this angle. You're going to get the adjacent side. If you do the hypotenuse
times the sine of this angle, you're going to get the opposite side but we could use this sine of 45 degrees. Once again if you put in your calculator, you get a crazy decimal but we can figure that out,
we know 45, 45, 90 triangles. Sine of 45 degrees is
square root of two over two. Cosine of 45 degrees is also
square root of two over two. I know my writing is
getting a little bit messy and actually I should have
done all of these in green, square root of two over two. What's the length of that green vector? What's two times square
root of two over two, that's going to be square root of two. What's the length of this orange vector? It's two times square root of two over two so it's going to be square root of two. Now we know the magnitudes
of the component vector is a horizontal and vertical
components of each of these so now we can write these out as the sum of these horizontal and vertical vectors. Vector A, we can write
as square root of three or three square roots of
three over two times I. That's this vector right over here, we scaled up the I unit vector by three squares root of three over two plus three halves times J. This right over here is
this vector right over here. Scaled up version of the J unit vector and if you add this orange
vector to this green vector, you get vector A. Similarly, vector B is equal to the length of the horizontal component and we got to be very careful. It's length is square root of two but it's going in the leftward direction. We're going to put a
negative on it times I. If we just squared it, I is
doing something like this, square root of two times
I would look like that. Negative square root of two
would point it to the left. This is negative square
root of two times I and then we're going to have plus square root of two times J. Now that we've broken them
up in their components, we're ready to figure out what, at least broken up into its components what A plus B is. A plus B is equal to, well
it's going to be the sum of all of these things. Let me just write that down. It's going to be that, copy and paste. That plus this, let me
just copy and paste it and you're going to get that but of course we can simplify this. We can add the I unit
vectors to each other. If I have three times
the square root of three over two Is and then I have another negative square root of two I, I can add that together. This is going to be equal to three times the square root of three over two minus square root of two times I and then I can add this to this and I'm going to get plus three halves plus square root of two times J. It looks a little bit complicated but we could type it in to our calculator and get approximations of
each of these two values and we essentially have
a at least a broken down into its components
representation of A plus B. In the next video, we're
now going to take this and figure out the actual
magnitude and direction of A plus B.