Calculating components of a vector

In this video, we are going to look at finding the components of the vector. So for example, if we wanted to write the following vector as a column vector where we know the length is 7, or the magnitude 7, and the angle between the horizontal and the vector is 36 degrees, then we can find this as a column vector. And by thinking of trigonometry, we’ve got a right angled triangle with an angle of 36 and hypotenuse of 7. And we can find the opposite and the adjacent using these rules that you should know from trigonometry.

So the adjacent, using trigonometry, would be cosine of the angle multiplied by the hypotenuse, which if we remember to change our calculator to degree mode, gets us an answer of 5.66 to two decimal places. And similarly, for the opposite, the opposite would be sine of the angle multiplied by the hypotenuse, which would get us 4.11 to two decimal places. So we can see that the vector goes 5.66 to the right and 4.11 up, which we can represent the column vector a as the following.

So more generally, what we can write down is a definition. Using what we’ve just got using trigonometry, is that given a 2D vector of length r with an angle theta in degrees from the horizontal, the corresponding vector in column form can be found as r cosine theta or sine theta as a vector. And we’re now going to use that to find the components of the following vector. So we can see a length of 6, and the angle from the horizontal there is 13 degrees. So r cosine theta will be 6 cosine 13. And r sine theta will be 6 sine 13.

Which when we type that in a calculator, that gets us 5.85 for the first component, 1.35 for the second component. So we’re just going to label them. And that makes sense for the lengths, but we’ve got to make sure that the column vector makes sense. So 5.85 saying it’s going 5.85 to the right and 1.35 up, which is clearly wrong. We can see it’s going 5.85 to left and 1.35 down. So we just check that and make sure that these are negative in the final answer. We’re also going to introduce that the magnitude of a vector, ab, the magnitude of the vector, which we denote by these two lines is the square root of a squared plus b squared.

So if we wanted to find the magnitude of the vector that goes from a to b, which is 2, 3, then the magnitude of a to b would be the square root of 2 squared plus 3 squared, which in said form is the square root of 13. Similarly if we have a vector with three components, abc, then the magnitude of that vector, v, will be the square root of a squared plus b squared plus c squared. Such that if we wanted to find the magnitude of the vector that goes from a to b, which is negative 2, 3, negative 1, then the magnitude of ab would be negative 2 squared add 3 squared add negative 1 squared.

An important tip, remember if you square a negative number, then the answer is positive. Such that negative 2 squared would be 4, 3 squared would be 9, and negative 1 squared would be 1. Such that the answer is the square root of 14 in said form.

We also have to know the direction of a vector. So if we have a 2D vector with components a, b, then we can find the direction of a vector using the formula, the angle is the tan inverse of b divided by a. And this can be derived from the rules you know from trigonometry. So we want to find the magnitude and direction of the vector that goes from a to b, which is negative 5, negative 4. So we’re first going to draw this, and this would be what the vector would look like. And we want to find this angle here.

So theta, using the formula, would be the tan inverse of negative 4 divided by negative 5, which is 38.66 degrees. Well, if we draw 38.66 degrees from the horizontal, that would be that angle, which we see is not the full angle. But we can get the full angle by doing 38.66 degrees plus 180 degrees, to get us that the direction is 218.66 degrees to two decimal places. And then we can also find the magnitude of that vector by getting the square root of the first component squared plus the second component squared, which would be the square root of 41. So we’ve wrote a column vector in the form of its direction and its magnitude.

So we’re next going to do a similar thing for the vector 5, negative 4. So for the direction, we draw it. This would be the angle that we want to find. And we find that the angle, using the formula, is the tan inverse of negative 4 divided by 5, which is negative 38.66 degrees, which we can see is this here. It’s 38.66 degrees below the horizontal. If we wanted it as an angle between 0 and 360 degrees, we just have to do 360 degrees take away the 38.66 degrees is 321.34 degrees. So tip, just remember, in the question look what they want the angle to be between.

And finally, for the magnitude, it’s the square root of the first component squared plus the second component squared, which is the square root of 41.

Finally, in the comments below, you might want to discuss, can you give advice? How do you ensure, when you’re doing vector questions, that the direction is correct? Thank you for watching.