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## Precalculus

### Unit 6: Lesson 8

Adding vectors in magnitude and direction form# Adding vectors in magnitude and direction form

In this example, Sal takes two vectors given by magnitude and direction, and finds the magnitude and direction of their sum. Created by Sal Khan.

## Want to join the conversation?

- Is there no formula to do this more easily?(6 votes)
- Sal incorporated the equation into the problem. If you have a vector with magnitude |a| and at an angle α from positive x axis and a vector with magnitude of |b| at an angle of ß from the positive x axis,

the x coordinate of the vector sum is |a|cos(α)+|b|cos(ß) and the y coordinate is |a|sin(α)+|b|sin(ß).(7 votes)

- how does Sal know its in the 3rd quadrant?(1 vote)

## Video transcript

- [Instructor] We're told that
vector a has magnitude four and direction 170 degrees
from the positive x-axis. Vector b has magnitude three
and direction 240 degrees from the positive x-axis. Find the magnitude and direction
of vector a plus vector b. So pause this video and see
if you can have a go at that. All right, now let's work
through this together. And the way that I'm going to approach it, I'm going to represent each
vector in component form. And then I'm going to add
the corresponding components. And from that, I'll try to
figure out the magnitude and the direction of the sum. So vector a, what is its x-component? Well, the change in x
here, there's multiple ways that you could try to do
this using trigonometry. But we've reviewed this or
gone over this in other videos. The simplest way to think about it is our change in x here is
going to be the length. And we know vector a has
magnitude four times the cosine of the angle that the vector makes with the positive x-axis,
cosine of 170 degrees. And so that's our
x-component right over here, four times cosine of 170 degrees. And then what's our y-component? Well, our y-component is going
to be this change in y here. And as we've reviewed in other videos, that's going to be the
length times the sine of the angle we make
with a positive x-axis, sine of 170 degrees. And we can maybe use a calculator in a bit to get approximations for these values. But then we can do the exact
same thing for vector b. Vector b here is going
to be, by the same logic, it's x-component is going to
be the length of the vector, and it would be three. They tell us that. So it's going to be three times the cosine of this angle, 240 degrees. And then the y-component
is going to be the length of our vector three times the sine of 240 degrees. Now, when we wanna take
the sum of the two vectors, let me write it here, vector a plus vector b, I can just add the
corresponding components. This is going to be equal to four cosine of 170 degrees plus three cosine of 240 degrees. And then the y-component
is going to be four sine of 170 degrees plus three sine of 240 degrees. And so let me get my calculator
out to evaluate these. We say 170 degrees. We take the cosine times
four, that equals this. And then we're going to add to that. I'll open parentheses. We'll take the cosine of 240. 240 cosine times three, close parentheses, is equal to this, negative, approximately negative 5.44. So this is approximately negative 5.44. And then if we were to take 170 degrees, take the sine of it, multiply it by four. And then to that, I'm
going to open parentheses. I'm gonna take 240 degrees, take the sine, multiply that times three,
close my parentheses. That is going to be equal to
approximately negative 1.90. So this is approximately negative 1.90. And this is consistent with our intuition. If the sum has both negative components, that means it's going to
be in the third quadrant. And if I were to do
the head to tail method of adding vectors, if
I were to take vector b and I were to put it right over here, we see that the resulting vector, the sum will sit in the third quadrant. It makes sense that our x and y-components would indeed be negative. Now, the question didn't ask just to find the components of the sum. It asked to find the
magnitude and the direction of the resulting sum. And so to do that, we just have to use a little
bit more of our trigonometry and actually a little bit of our geometry. For example, our change in x
is this value right over here as we go from the tail to the tip. It's negative 5.44. If we're just thinking in terms
of length right over here, the absolute value, this
side would have length 5.44. And then same way, you are
changing y, its negative, we're going down in y. But if we were just thinking
in terms of a triangle, the length on this side
of a triangle is 1.90. And we can see from
the Pythagorean theorem that the length of our hypotenuse, which is the same thing as
the magnitude of this vector, squared is going to be equal
to the sum of the squares of these two sides. Or another way of thinking about it is, the length of this vector,
the magnitude of this vector, which we can write as
a magnitude of vector a plus vector b is going to be equal to, or I should say approximately equal to since we're already
approximating these values, the principal root of 5.44 squared. And that's 'cause I'm just thinking about the absolute length of the side. I could also think about a change in x. But if I had a negative
5.44 and I square that, that would still become positive. And then I'll have plus 1.90 squared. And I can get our calculator out for that. This is going to be approximately equal to 5.44 squared plus 1.9 squared, is equal to that. Take the square root of that. It's approximately equal to 5.76, 5.76, which is going to be our magnitude. And then to figure out the direction, so we essentially want to figure out this angle right over here. You might recognize that
the tangent of this angle, theta right over here, should be equal to, and I'll
do approximately equal to since we're using these approximations, our change in y over our change in x. So negative 1.90 over negative 5.44, or we could say that theta is going to be approximately
equal to the inverse tangent of negative 1.90 over negative 5.44. And we're gonna see in a second whether this is actually
going to get us the answer that we want. So let's try this out. If we were to take 1.9 negative divided by 5.44 negative, that gets us that, which makes sense. Negative divided by a
negative is a positive. And now let's try to take
the inverse tangent of that. So here I press second, and
then I'll do inverse tangent. So I'm getting 19.25 degrees, approximately. So this is saying that this is approximately 19.25 degrees. And my question to you
is, does that seem right? Well, 19.25 degrees would
put us in the first quadrant. It would get us a vector that looks something like this. This would be 19.25 degrees. But clearly, that's not the
vector we're talking about. We're talking about a vector
in the third quadrant. And the reason why we got this result, is that when you take the inverse tangent on most calculators, it's going to give you an angle that's between negative 90 degrees and positive 90 degrees. While here we are at an angle that puts us out in the third quadrant. So we have to adjust. And to adjust here, we just
have to add 180 degrees to get to the actual angle
that we are talking about. So in our situation, the magnitude here is going to be approximately 5.76 and then the direction is going to be approximately 19.25 plus 180 degrees, which is going to be 199.25 degrees. And now we are done.