Main content

## Magnitude and direction form of vectors

Current time:0:00Total duration:6:42

# Direction of vectors from components: 3rd & 4th quadrants

## Video transcript

- [Voiceover] Let's get some more practice finding the angle, in these
cases the positive angle, between the positive X axis and a vector drawn in standard form
where it's initial point, or it's tail, is sitting at the origin. Here for vector A we can write
it in two different ways. In engineering notation it would be -2 times a unit vector I, that's the unit vector in the X direction, minus four times the unit
vector in the Y direction, or we could just say
it's X component is -2, it's Y component is -4. And we see that here. If we're starting at the
origin we go two to the left and we go four down to
get to the terminal point or the head of the vector. Also figure out what theta is. What we've seen before when
we're thinking about vectors drawn in standard form,
we could say the tangent of this angle is going to
be equal to the Y component over the X component. So the Y component is -4 and the X component is -2. And in the previous video
we explained why this is, it really comes straight out
of the unit circle definition of trig functions, tangent of theta is equal to the Y coordinate
over the X coordinate of where a line that
defines an angle intersects the unit circle. And I encourage you to watch that video if that doesn't make much sense. But so we could say tangent
of theta is equal to two. And so we might want to say,
if we want to solve for theta, we could say theta is equal to the inverse tangent function of two. And I'm gonna put a question mark, and I think you might know why I'm putting that question mark. So let's see what that gets us. So if we were to take two, and I wanna take the inverse
tangent not just the tangent. I wanna figure out what angle
gives me a tangent of two. So inverse tangent, it's about 63.4 degrees roughly. So this gives me theta is approximately 63.4 degrees. And I'm gonna put a question mark here. And I think you might sense why that is. If you don't, pause the video and think about why am I
putting a question mark here? Why does this angle look fishy? Well, it looks fishy because
an angle of 63.4 degrees would put us squarely in the first quadrant. If our vector looked like this, let me see if I can draw it. If our vector looked like this, so if our vector's
components were positive two and positive four then that
looks like a 63-degree angle. This looks like a 63-degree angle. But we're not in the first quadrant. Our vector A that we care
about is in the third quadrant. So it's clear that it's in
the exact opposite direction, and I think you see why. When we take the inverse tangent
function on our calculator it assumes that the angle
is between -90 degrees and positive 90 degrees. Well, here we have an angle
that's over 180 degrees. It's between 180 and 270 degrees. And so to find this angle, and this is why if you're
ever using the inverse tangent function on your calculator
it's very, very important, whether you're doing
vectors or anything else, to think about where does
your angle actually sit? What quadrant does it actually put you in because you might have
to adjust those figures. So, it's not going to be 63.4
degrees it's going to be that plus another 180 degrees to
go all the way over here. So, theta is going to be 180,
and I should say approximately 'cause I still rounded,
180 plus 63.4 degrees is going to be 200 and, what is that? 180 plus 60 is 240, so 243.4 degrees. So let's do one more. So here I have a vector
sitting in the fourth quadrant like we just did. Pause the video and see
if you can figure out the positive angle that it
forms with the positive X axis. Well, we could do the same drill and maybe we could skip a few steps here now that we've done it many times. We might wanna say that
the inverse tangent of, let me write it this way,
we might want to write, I'll do the same color. We might wanna say that
theta is equal to the inverse tangent of my Y component
over my X component of -6 over four, and we know what that is but let me just actually
not skip too many steps. So the inverse tangent of -1.5. - 1, -1.5 and once again, I get to get my calculator out and so 1.5 negative, and I wanna find the
inverse tangent of it, I get roughly -56.3 degrees. So this is approximately equal to - 53.6 degrees. Did I do that right, 56.3, 56.3 degrees. And once again, I'm gonna
put the question marks here. And why did I do that? Because the angle that it's giving, and this isn't wrong
actually in this case, it's just not giving
us the positive angle. Because if you start the positive X axis and you were to go clockwise, well now your angle is
going to be negative, and that is -56.3 degrees. But we wanna figure out the
positive angle right over here. So, there's a couple of ways that you could think about doing it. One way to think about it is well to go from this negative angle to the positive version of it we have to go completely around once. So we have to add 360 degrees. So if it's really approximately -56.3 degrees plus 360 degrees, which is going to be, what is that? So it's going to be, so it's going to be approximately, see if I subtracted 50 degrees
I would get to 310 degrees, I subtract another six
degrees, so it's 304 degrees, and then .3, so 303.7 degrees. Did I do that right? Let's see, if I add this .3 to the seven, that's gonna get to
304, then at 310 to 360. So there ya go. That is our positive angle that we form. So always really think about
what they're asking from you, or what a question is asking from you. Anyway, you get the idea.