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Video: Working with Vectors in Component Form

Tim Burnham

Learn how to express vectors in i and j component form and convert between that form and the magnitude and direction of a given vector.

12:38

Video Transcript

In this video, we’re gonna use the 𝑖 and 𝑗 form of vectors, we’ll add and subtract vectors in that format, we’ll find the vector components when given the magnitude and direction, and we’ll find the magnitude and direction when we’re given vector components.

Firstly, just a quick recap of the 𝑖 and 𝑗 vector format and a couple of quick examples of how to use them. 𝑖 is a unit vector, that is, it’s got length one in the direction of the positive π‘₯-axis; we can write 𝑖 like this: the π‘₯-component is one, the 𝑦-component is zero. And 𝑗 is a unit vector in the direction of the positive 𝑦-axis, and we can write it like this: the π‘₯-component is zero, the 𝑦-component is positive one. So, let’s go on and use them.

You can string together any number of 𝑖 and 𝑗 vectors. So here, we have strung three 𝑖’s together and two 𝑗’s together, and that takes us from position 𝐴 to position 𝐡; it gives us a resultant vector 𝐴𝐡. So, because we’ve done three 𝑖’s and two 𝑗’s, 𝐴𝐡 is just three 𝑖 plus two 𝑗, which of course we can also write in this format: three two.

Now, we can also add and subtract vectors in the 𝑖 and 𝑗 format just by treating the 𝑖 and 𝑗 components separately, so let’s have a couple of examples. So, vector 𝐴𝐡 is three 𝑖’s followed by negative four 𝑗’s. So, we can write that as 𝐴𝐡 equals three 𝑖 take away four 𝑗. And 𝐢𝐷 is negative five 𝑖’s followed by negative five 𝑗’s, and we can write that 𝐢𝐷 equals negative five 𝑖 take away five 𝑗.

Right! Let’s say we wanted to find the result of 𝐴𝐡 plus 𝐢𝐷. So, we’re gonna write out 𝐴𝐡 and then we’re gonna add 𝐢𝐷. So first, just write out those vectors: three 𝑖 minus four 𝑗 plus negative five 𝑖 take away five 𝑗. Then we’re gonna pick the 𝑖 components and sort those out, and then we’ll pick the 𝑗 components separately and deal with those. So first of all, with the 𝑖 components, we’ve got three 𝑖’s and then we’re adding negative five 𝑖’s, so that’s the same as taking away five 𝑖. And then for the 𝑗’s, we’ve got negative four 𝑗 and then we’re adding negative five 𝑗, so that’s just taking away five 𝑗. So now, we’ve got 𝑖’s over here, 𝑗’s over here. We can just simply work out what those terms are. And three 𝑖 take away five 𝑖 is negative two 𝑖, and negative four 𝑗 take away another five 𝑗 is negative nine 𝑗. So there’s our answer: 𝐴𝐡 plus 𝐢𝐷 is equal to negative two 𝑖 take away nine 𝑗.

Okay, so now let’s look at 𝐴𝐡 take away 𝐢𝐷. Again, we’re just gonna write the vectors out. So that’s three 𝑖 take away four 𝑗 take away negative five 𝑖 take away five 𝑗. So, we can now deal with these. So, we’re gonna look at the 𝑖 terms first. Now we’ve gotta be very careful cause we’re taking away a negative five 𝑖. So three 𝑖 take away negative five 𝑖 means we’re adding five 𝑖. And now, we look at the 𝑗 terms, negative four 𝑗 take away negative five 𝑗. And, obviously, taking away negative five 𝑗 again means we are adding five 𝑗. So now, we can just work out the two parts of that. three 𝑖 plus 5 𝑖 is eight 𝑖, and negative four 𝑗 add five 𝑗 is plus one 𝑗, so we don’t need to write the one because we’re going to write plus 𝑗. So there’s our answer: 𝐴𝐡 take away 𝐢𝐷 is eight 𝑖 plus 𝑗.

Right! Now to slightly more interesting stuff. If we’re given a vector in its magnitude and direction format, we can change it into its 𝑖 and 𝑗 format. Here, we’ve got a vector 𝐴𝐡, which has got a length or a magnitude of ten units and a direction of a hundred and twenty degrees. Now, we define the direction as how far we have to rotate counterclockwise from the positive axis in order to line up with the vector. So that’s this rotation here. So imagine we started off with a vector pointing in that direction. How far would we have to rotate it about this sort of origin in order to get it to line up here? So that, in this case, that’s a hundred and twenty degrees. So, what we trying to work out then it is this 𝑖 component here. Obviously to go from 𝐴 to 𝐡, we’re actually kind of moving left in this diagram. So this direction here, this this arrow from β€” this green arrow here is the 𝑖 component moving in that negative direction. So it’s gonna be a negative number. We’ve gotta work out the width of that. And the height of that is going to be the 𝑗 component. So that’s what we’re looking for there. Now what you should have spotted here is that we’ve actually got a right-angled triangle, and the hypotenuse of that triangle is ten, and the height is the 𝑗 component of our vector, and the width is the 𝑖 component of our vector.

So let’s start off and work out the 𝑖 component. Now, the 𝑖 component, the width of that, is the projection of that hypotenuse onto the bottom side of the triangle. So the way that we work that out is to take that magnitude of ten and multiply it by cos of the angle; in this case πœƒ is a hundred and twenty degrees, and the fact that is a hundred and twenty degrees means that cos of a hundred and twenty degrees is gonna give us a negative number. So, that will automatically work out that the direction of the 𝑖 component for us is gonna be negative direction. So that’s going to be ten times, and cos of a hundred and twenty is negative half, and that gives us an answer of negative five. So our 𝑖 component is negative five; it’s gonna be negative five 𝑖.

Now the 𝑗 component is the projection of the hypotenuse of ten onto the height of the triangle. So, that’s gonna be the magnitude ten times sin of πœƒ, so sin of a hundred and twenty. And sin of a hundred and twenty is is root three over two. So the 𝑗 component, the height of that triangle, is going to be ten times root three over two which is five root three.

So when we convert that from the magnitude and direction, so we had 𝐴𝐡 had a magnitude of ten units and a direction of a hundred and twenty degrees, that equates to this negative five 𝑖 plus five root three 𝑗. So just using these formula here, that takes the sine and the cosine based on the angle, takes care of the positive and the negative signs for us. So relatively straightforward, just plug this into the formulae and out pop the components for us.

So let’s just look at the general format for that then. If we’ve got a vector 𝑉 with a magnitude of 𝑉, like we see here, and the direction of πœƒ, then in its 𝑖 and 𝑗 form we can just write this: vector 𝑉 is the magnitude of vector 𝑉 times cos πœƒ 𝑖’s and then the magnitude of vector 𝑉 times sin πœƒ 𝑗’s.

Now things get slightly more complicated when we go in the other direction. If we’re given the 𝑖 and 𝑗 format, we want to convert them back to the magnitude and direction. So let’s take a look at an example of that now. So let’s say we have vector 𝑉 is equal to eight 𝑖 minus six 𝑗 β€” not drawn a hundred percent accurately on there but we get the general gist of it, and we’ve got to find the magnitude and direction of vector 𝑉. Now, we can work out using Pythagoras’s theorem quite easily what the magnitude of vector 𝑉 is. But to work out that angle πœƒ, I’m going to be a little bit careful because the answers that we’re gonna get from my calculator are slightly different.

So our journey from the start of the vector here to the end of the vector here can be summed up by saying we’ve got a positive eight in the π‘₯ direction and a negative six in the 𝑦 direction. And that’s created a nice little right-angled triangle here, and we want to work out the magnitude of 𝑉, which is the hypotenuse of that triangle. So using the Pythagorean theorem, the hypotenuse squared is equal to the sum of the squares of the other sides. So, if we square root both sides there we’ve got 𝑉, the magnitude of 𝑉, is equal to the square root of eight squared plus negative six squared. And that basically, sixty-four plus thirty-six, is a hundred; the square root of a hundred is ten. So the length of that line is ten.

Now to find out the angle, we’re gonna use this formula here: πœƒ is the inverse tan of the 𝑗 component divided by the 𝑖 component, which gives us tan to the minus one of minus six over eight, and when we put that into the calculator we get negative thirty-six point nine degrees to one decimal place. Now that’s not quite the answer we’re looking for, because if you think about it, positive is counterclockwise, negative is clockwise. So what this is actually giving us is this angle in here okay? So that angle there is thirty-six point nine. We’ve called it negative because it’s travelling in the clockwise direction, but it just means that physically that angle there is thirty-six point nine. So what we can do (remember a full circle is three hundred and sixty) if we take that away, thirty-six point nine, from three hundred and sixty, we’re gonna get the πœƒ angle: the blue bit all the way round the outside here. And when we work that out, three hundred and sixty take away thirty-six point nine is three hundred and twenty-three point one degrees to one decimal place.

So there we have our final answers: ten for the magnitude and three hundred and twenty-three point one degrees to one decimal place for the direction. And it’s absolutely crucial, I think, in these questions to draw yourself little diagram as you go, because otherwise you’ll get very confused about the angle you get at the end. Do we have to add it to something? Do we have to subtract it from something? But if you look at that diagram, it’s pretty clear just by looking at it what the answer we’ve got from the tan (the inverse tan calculation), what that is on the diagram, and then what do we have to do with that to work out our actual answer in terms of the direction in a counterclockwise direction. So hopefully that’s fairly clear.

So let’s just summarise that again then. So in general, if I’ve got a vector 𝑉, which is in the format π‘Ž 𝑖 plus 𝑏 𝑗, and I’m trying to work out the angle, the direction theta, and the magnitude 𝑉 with the two lines by it, so the magnitude of 𝑉, this is the process I’ve got to go through. The magnitude of 𝑉 will use the Pythagorean theorem with the components of 𝑖 and 𝑗, so π‘Ž squared plus 𝑏 squared all square rooted. That gives us our magnitude. And to work out the direction, it’s just the inverse tan of the 𝑗 component divided by the 𝑖 component, but remember you must check your diagrams to see what version of the angle your calculator has given you. So in this case here, this angle, it would give you the right answer; everything will be quite simple. But if we had a vector which goes down and to the right, it’s gonna give you this angle here, so you’re going to need to make this adjustment to make sure you give it as a positive angle and a counterclockwise rotation. So that about wraps it up for working with vectors in terms of 𝑖 and 𝑗 in their component format.