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## Precalculus

### Course: Precalculus > Unit 6

Lesson 7: Vector components from magnitude and direction- Vector components from magnitude & direction
- Vector components from magnitude & direction
- Vector components from magnitude & direction: word problem
- Vector components from magnitude & direction (advanced)
- Converting between vector components and magnitude & direction review

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# Vector components from magnitude & direction

Sal finds the components of a couple of vectors given in magnitude and direction form.

## Want to join the conversation?

- At9:15, how did Sal come up with 5√2?

I didn't quite understand that part, can someone help me on this please?

Thanks!(8 votes)- It comes from knowing the unit circle and trigonometric functions.

The cosine of 45 degrees is √2/2, therefore 10(√2/2) = 5√2.

You should familiarize yourself with the unit circle, as these types of trig questions are more frequent in calculus.

Print out this image and have it handy when doing your work. There are videos on Khan that talk about it as well.

https://lelandmath.files.wordpress.com/2013/10/screen-shot-2013-10-14-at-7-19-46-pm.png(24 votes)

- If x component is 4 cos50 then its y component can also be written as 4 cos(90-θ) so its gonna be 4 sin40 right?? Just a little bit doubt in it?(6 votes)
- In my textbook the <?,?> signs are used. What is the difference between <?,?> and just (?,?)?(4 votes)
- Typically, <x,y> is a vector and (x,y) is a point.(10 votes)

- Hi, quick question, is there a way to find the components of a vector with just the angle and nothing else? Thanks!(3 votes)
- No, because there are infinitely many vectors with just that angle.(10 votes)

- If the x and y vector components of Vector A are 11 and 7,

what is the magnitude of Vector A? Round your answer to the nearest hundredth.(2 votes)- ||A||=sqrt(11^2+7^2)=sqrt(121+49)=sqrt(170)=13.05

Hope it helps..(3 votes)

- http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w02_qp_3.pdf can someone please help me with Q10 (ii) &(iii) from this link?

and Q9 (ii) form this link http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s03_qp_3.pdf

Please!(1 vote)- position vector: OA=4i+k, OB=5i-2j-2k, OC=i+j, OD=-i-4k

point: a=(4,0,1) b=(5,-2,-2) c=(1,1,0) d=(1,0,-4)

vector: AB =(5-4,-2-0,-2-1)=(1,-2,-3) DC=(1--1,1-0,0--4)=(2,1,-4)

symmetric form : line AB : (x-4)/1=(y-0)/-2=(z-1)/-3

line CD : (x-1)/2=(y-1)/1=(z-0)/4

solve for x,z (x-4)=(z-1)/-3 , (x-1)/2=z/4

-3x+13=z and 2x-2=z

x=3 z=4

apply to line formula : (3-4)/1=(y-0)/-2=(4-1)/-3 , (3-1)/2=(y-1)/1=(4-0)/4

get y/-2=-1 , y-1=1 y=2

so point (3,2,4) satisfied both line AB and CD ,therefore two line intersect.

point p=(1,5,6) line AB= (x-4)/1=(y-0)/-2=(z-1)/-3 , so point x=4+n , y=-2n , z=1-3n on line AB.

let q on AB , vector PQ =(4+n-1 , -2n-5 , 1-3n-6), if PQ perpendicular AB , then (n+3)*1+(-2n-5)*-2+(-3n-5)*-3=0 (point product of perpendicular vector=0)

n+3+4n+10+9n+15=0 , n=-2 . PQ=(4-2-1 , 4-5 , 1+6-6)=(1,-1,1) magnitude =squart(1^2+1^2+1^2)=squart(3)(4 votes)

- Can you please tell me what will happen if angle is 180 degrees.

Can ever component will be greater than vector itself?(2 votes)- When an angle is at 180 degrees, we won't have a y component because the vector will lie along the negative x-axis.(1 vote)

- I have done it with cos45 degree and the x-axis answer comes positive 7.07(2 votes)
- I'm using the calculator and it's not giving me the correct answer. Why is this happening?(2 votes)
- If there were to be a question that says the vector is 50m/s, and 0 degrees from horizontal, how would you go about solving? Would it just be a straight line? How could you apply SOH CAH TOAH..(2 votes)
- 50 units on the x axis.(1 vote)

## Video transcript

- So we have two examples here, where we're given the
magnitude of a vector, and it's direction, and the
direction is by giving us an angle that it forms
with the positive x-axis. What we need to do is go from having this magnitude and this
angle, this direction, to figuring out what
the x and y components of this vector actually are. So, like always, pause this video and see if you can work
through this on your own. Alright, now let's work
through this together, and it's really just gonna involve a little bit of trigonometry. So, we wanna break this vector
down into it's horizontal and it's vertical components,
or it's x and y components. So I could draw it's, I could draw it's horizontal component just like this. So it's gonna look something like-- no, let me draw it, I can
do a better job than that. So, it's gonna look something like that. That's it's horizontal component, or it's x component. And then it's vertical component is going to look something like this. It's going to look like that. That's it's vertical
component, and notice, if you add the horizontal component and the vertical component, you are going to get your original vector. Now, I've just constructed
a right triangle, and in this type of a right triangle, I could just use, actually,
some of my most basic trig definitions, or the simplest form, which is the Soh-Cah-Toa definition of the trig functions. I might wanna break into the unit circle definition later on, but if I wanna figure out
the magnitude of this base, right over here, we see
that that is adjacent to this 50 degree angle. It's not the hypotenuse, it's the other side that forms the angle. And so what trig function deals with adjacent and hypotenuse? Well, we could just put
a little reminder here, Soh-Cah-Toah. Well, cosine deals with
adjacent over hypotenuse. So we could say that if I call, if I call this x the
length of our x component, we could say that the cosine, the cosine of 50 degrees of our angle is going to be equal to the
length of the adjacent side, x, over the length of our hypotenuse, which is the magnitude of the vector, over four. And so, if I wanna solve for x, I just multiply both sides by four. So I could get four times cosine of 50 degrees, four times cosine of 50 degrees, is equal to x. Now what about the y component? The y component, what is that going to be? Well, what, that side, the length of that side,
is opposite to the angle, the 50 degree angle. So what trig function deals
with opposite and hypotenuse? Well, that is sine. So we know that the sine of 50 degrees is going to be equal to y over
the length of the hypotenuse. Y over four, and so we can multiply both sides by four and we get, we get four times sine of 50 degrees, sin of 50 degrees, is equal to y. And so if I don't have a calculator, I could just write that this, this vector, I could write as, if I write it in the component form, it's x component if four
cosine of 50 degrees. And it's y component is
four sine of 50 degrees. And you might notice
something interesting here. I have cosine of the angle, the angle we formed with
the positive x axis, I have that for the x coordinate. I have sine on the y coordinate, and then I just multiply
it by the magnitude of the vector. Can I always do that? Well, it turns out you can, and this comes out of the
unit circle definition of the trig functions. The unit circle definition
of trig functions, cosine, if you have a unit circle, so unit circle looks like this. A unit circle has a radius of one. Cosine is the x coordinate of where you intersected the unit circle, and sine is the y coordinate. Or if you had a vector of magnitude one, it would be cosine of that angle, would be the x component, for the, if we had a unit
vector there in that direction. And then sine would be the y component. Well, we don't have a unit vector, our vector has a magnitude of four, it's four times bigger than a unit vector, so each of the components are going to be four times bigger. So that's why we multiply the x, the cosine of fifty by four
to get the x component, and we take the sine of 50, and we multiply it by four
to get the y component. And that's gonna come in handy when we think about this one over here. But we can get our calculator out, and approximate what
these are going to be. So let me get it out. Where's my calculator? Alright, there we go. So, 50, I'm in degree mode, gotta make sure, so 50 degrees. I'm gonna take the cosine, and I'm gonna multiply it times four. So times 4, is equal to two, approximately 2.57. So this is approximately equal to 2.57, is our x component, and
then our y component. Our y component, if I take 50 degrees, and if I take the sine of it, and then multiply it by four, I get approximately 3.06. 3.06, and we see that over here, this x component looks
like it's a little more than two and a half, and this y component looks like it's slightly more than three. So it all worked out, even
though this is kind of a sloppy, hand drawn graph. Alright, now let's tackle this one. This is interesting cause this, the terminal point, when we
draw it in the standard form, is in the second quadrant. So what would be the x
and y components here, and you can immediately tell, because we're in the second quadrant, our x component is gonna be negative, and our y component is going to be, is going to be positive. Well, we could just resort exactly to what we did just there. We could say this vector is going to be the magnitude, it's x component is going
to be it's magnitude times the cosine of the angle that forms at the positive x axis. Cosine of 135 degrees. And it's y component is going to be the magnitude times the sine of the angle that forms with the positive x axis, and we'd be done. And we can evaluate each of these things, so we could get, let me get
my calculator out again. So if we take 135 degrees, and if I take the cosine of that, and then multiply it by ten, times ten, I get approximately negative 7.07. So that's approximately negative 7.07, and then if I take the sine, we're going to see something very similar. So 135 degrees, and I take the sine of it, I get positive .707, and so let me multiply
that times the magnitude, times ten, is equal to 7.07. So 7.07, and we see that over here. This is roughly, looks like
it's a little more than seven, looks like it's a little
bit more than seven in that direction. And we'd be done, and you might say well-- once again, how did this work? Well, if I had a unit circle right over here. And if I had a unit vector, so it's terminal point would
sit on the unit circle, that went in the exact same direction, it still formed 135 degrees, this point right over here, it would have the coordinates cosine of 135 degrees, sine of 135 degrees, let me make it a little bit more visible, this point right over here. Cosine 135, sine of 135 would be it's coordinates. Or for this unit vector right over here, that going in that direction, it's x component would be cosine of 135, and it's y component would be sine of 135. Well the vector that we care about has ten times the magnitude of a unit vector in that direction. So it's x component is going to have ten times the magnitude, and so is it's y component. And you could even resort to your Soh-Cah-Toa definitions
to figure this out. We could construct a
right triangle if we like, we could say here is, we could say here is my x component, that is my x component, it's gonna be something like that, and then there-- and then my y component is going to look something like, I'm drawing
it a little bit imprecise, but I could draw that, let me
draw that a little bit better. My y component is gonna
look something like that, right triangle, and so
if I want this magnitude, the magnitude of this bottom side, I would say well, look, I know if this is 135 degrees, this angle right over here
is supplementary to that, so 45 degree angle, and so what trig function deals with adjacent and hypotenuse, Well cosine. So I could say cosine of 45 degrees is equal to the length of this, so let me just call the length of that, oh I don't know, let me just call it x, so the length of it, x, over the hypotenuse, over ten. Multiply both sides by ten, you get ten times cosine of 45 degrees is equal to x. And cosine of 45 degrees is square root of two over two, so this is going to be
five square roots of two is equal to x, and you might say, wait, wait, I thought this
was supposed to be negative. Well, the way we just solved it right now, with our right triangle, we just figured out our
magnitude of this side. And then we would just have to reason, ok, we're not going five square
roots of two to the right, we're going five square
roots of two to the left. So, the magnitude in the x direction, or not the magnitude, our x component, I should say, is going to be negative five square roots of two. And you could do the exact
same reasoning to say, well ten cosine of f-- it's ten sine of 45 degrees is going to give us the
magnitude of the y component, and that's actually going
to be our y component. So, same exact thing, this is going to be
ten sine of 45 degrees, which is five square roots of two, and if we took our calculator out, they would evaluate to approximately these things right over there.