We already know that a vector is a set of numbers that can be represented in a suitable
space by a line segment with a specific length and direction. We can use them to represent
forces or velocity or anything else that’s got an amount and a specific direction associated
In this video, we’re gonna look at the concept of unit vectors, which are basically
line segments with the length of one unit pointing in a specific direction. Here we’ve got
vector 𝐴𝐵, which is four, three. This means that the journey from 𝐴 to 𝐵 involves of
movement of positive four in the 𝑥-direction and positive three in the 𝑦-direction. So
that’s where these numbers come from here.
So what we’ve ended up with here is a right-angled
triangle because the 𝑥-axis and the 𝑦-axis are at right angles to each other, so this angle
here is ninety degrees. So we’ve got a little triangle and we really want to work out the
length of this vector here, 𝐴𝐵, next, so we’re gonna use the Pythagorean theorem to work out
that length, the hypotenuse of that triangle.
So bit of terminology, we’re gonna work out the magnitude of 𝐴𝐵; that is, the length of the
vector 𝐴𝐵. And the notation that we’re gonna use for the magnitude of vector 𝐴𝐵 is these
vertical lines here, so that’s the standard notation. So writing down the Pythagorean theorem,
the magnitude of 𝐴𝐵 all squared, so the length of the hypotenuse here all squared, is equal
to the sum of the squares of the other side.
So 𝐴𝐵, the magnitude of 𝐴𝐵 squared is equal
to four squared plus three squared. And four squared is 16; three squared is nine; add
those two together, we get 25. So the magnitude of 𝐴𝐵 squared is 25. So if
I now take square roots of both sides, that leaves me with the magnitude of 𝐴𝐵 being the
square root of 25, which is five. So we’ve just worked out that the length of vector 𝐴𝐵 is
So if I wanted to generate a unit vector in the same direction as vector 𝐴𝐵, I can take
vector 𝐴𝐵 and just divide it by the length of itself. And the length of itself is the
magnitude of the vector, so we’ve got vector 𝐴𝐵 divided by the magnitude of vector 𝐴𝐵, and
we just worked that out to be five. So the unit vector in the direction of 𝐴𝐵 is the vector
𝐴𝐵 divided by five.
Now what that means is we’ve got to take the 𝑥-component and divide it
by five, and we’ve gotta take the 𝑦-component and divide it by five. The 𝑥-component was
four; the 𝑦-component was three. So the unit vector in the direction of 𝐴𝐵 is this thing
here: four-fifths is the 𝑥-component; three-fifths is the 𝑦-component. That will have a
length of one.
So this red vector here, going from here to here, is the unit vector in the
direction 𝐴𝐵. It’s a fifth of the length of vector 𝐴𝐵. It’s a vector which has an
𝑥-component of four-fifths and a 𝑦-component of three-fifths. Obviously I haven’t drawn that
very accurately on here, but that’s basically what the vector would look like. So the
unit vector in the direction 𝐴𝐵, if 𝐴𝐵 was four, three, is basically a vector which starts
at 𝐴 and has a length of one, but goes in the same direction that 𝐴𝐵 would have gone
So in this case, we’ve ended up with vector four-fifths, three-fifths. So let’s just
check that in fact the length of this side here is one. So we can do the Pythagorean theorem
again. So that length squared is four-fifths squared plus three-fifths squared, which is
sixteen twenty-fifths plus nine twenty-fifths, which is twenty-five twenty-fifths which is
one. So if I take the square root of that, that will give me the actual length, and the square
root of one is equal to one. So that length is definitely equal to one. This is where the unit
vector idea comes in.
So here’s what the method is in general, and then we’ll look at some more examples afterwards
to make sure that you understand it. So let’s say we’ve got a vector called 𝑉, which has an
𝑥-component of a and a 𝑦-component of 𝑏. Then the unit vector in the direction of vector 𝑉
is the vector 𝑉 divided by the magnitude of vector 𝑉. So whatever length 𝑉 has got, if I
divide that length by itself, I’m gonna get an answer of one. So this bit is forcing the length
of that unit vector to have a magnitude of one.
And obviously it’s gonna still be in the same
direction; the 𝑥- and the 𝑦-components will be in the same proportion as the original
vector, so it’s gonna be heading in the same direction. So to work out the 𝑥-component,
I’ve got the original 𝑥-component, 𝐴, and I’m dividing that by the magnitude of the square
root of 𝐴 squared plus 𝐵 squared, from the Pythagorean theorem. And to work out the
𝑦-component correspondingly, I take the original 𝑦-component and I divide that by the
magnitude of the vector, the square root of 𝐴 squared plus 𝐵 squared. That’s gonna give me my
unit vector in the direction of vector 𝑉. Now that might look a little bit intimidating when you write it all out like that, but it’s
actually really easy to apply. So let’s have a look at an example.
So we’ve gotta find the
unit vector in the direction 𝐴𝐵, where 𝐴𝐵 is the vector five, negative six. So the first
thing is I’m gonna do a little sketch to get this clear in my head. So it doesn’t have to be a
hundred-percent-accurate diagram, but a quick sketch here gives you a sort of a flavour of
what’s going on. So we’ve got the vector 𝐴𝐵; we’re going positive five in the 𝑥-direction,
negative six in the 𝑦-direction.
The first thing we need to do is to work out the length of
this vector 𝐴𝐵 here. And then if we divide that vector by that length, it’s gonna create a
little vector here, which starts at 𝐴, which goes in this direction, same direction as 𝐴𝐵,
but it’ll only have a length of one.
So I’ve got a right-angled triangle. So using the
Pythagorean theorem, the magnitude of 𝐴𝐵, the length of a vector 𝐴𝐵, all squared is equal
to the 𝑥-component squared, that’s five squared, plus the 𝑦-component squared, that’s
negative six squared. So that is five squared is 25, negative six squared is
36. So we add them together, we get 61.
Now remember that was the magnitude
squared. So taking square roots of both sides, I’ve got the magnitude of vector 𝐴𝐵 is equal
to the square root of 61. So we haven’t got quite such nice numbers this time, so we’ll
just leave that in that accurate format, root 61. And I can use that to work out what
my unit vector in the direction of 𝐴𝐵 is. Remember that would be the original vector 𝐴𝐵
divided by the magnitude of that vector.
So our original vector was five, negative six. And
we’ve now got to divide each component by the magnitude of the vector. So five divided by the
square root of 61 and negative six divided by the square root of 61. And that’s
it. This vector here — five over root 61, negative six over root 61 — has got a
length of one. And the 𝑥- and the 𝑦-components are in the same proportion as the original
vector 𝐴𝐵. In other words, it’s pointing in exactly the same direction, so it’s the unit
vector in the direction 𝐴𝐵.
Okay let’s just do one more question then. Find the unit vector in the direction of 𝑃𝑄. So
the difference here is we’ve been given the point 𝑃, we’ve been given the point 𝑄, and we’ve
got to calculate what the vector’s gonna be.
So the first thing then is to sketch in our
vector 𝑃𝑄, so it goes in that direction there, and we’ve got to work out what the original
vector 𝑃𝑄 is. So in the 𝑥-direction, we’re starting off with an 𝑥-coordinate of three, and
we’re going down to an 𝑥-coordinate of negative five. So we’re going three here plus another
five here. That’s negative eight in the 𝑥-direction. In the 𝑦-direction, we’re starting off
at negative four and we’re moving. So we’re going four up to here and then another one up to here, so
that’s positive five in the 𝑦-direction.
So just summarising that, 𝑃𝑄 is negative eight,
five. And now we can tackle this just like we did the previous question: use that triangle, do
a bit of Pythagorean theorem to work out what the length of the vector 𝑃𝑄 is, and divide
vector 𝑃𝑄 by that length. So the magnitude of 𝑃𝑄 squared is negative eight squared plus
five squared, which is 64 plus 25. So remember that’s 𝑃𝑄 squared; the
magnitude of 𝑃𝑄 squared is equal to 89.
So taking the square root of both sides,
and we’ve ended up with another horrible number, so the magnitude of 𝑃𝑄 is the square root
of 89. So I’ll just go through, divide the components of 𝑃𝑄 by root 89. And
the unit vector in the direction 𝑃𝑄 is simply the original vector 𝑃𝑄 divided by the length
of itself, so that’s negative eight is the 𝑥-component divided by root 89, and five
was the 𝑦-component, and divide that by root 89 as well.
And there we have it, the
unit vector in the direction 𝑃𝑄. So we’ve got some nice magic tricks we can use now, but why
do we use unit vectors at all in the first place? Well, they encapsulate the information about
the direction, but their length is just one. They’re like ready-made direction descriptors
that you can multiply by a number to tell people how far to go in that direction. And it also
turns out that they’re useful in finding the angle between two different vectors, but that’s a
story for another day.
Okay we can’t resist it. Let’s just do one more example on unit vectors. Now you may recall,
or hopefully you do recall, that we’ve got special vectors 𝑖 and 𝑗, which are unit vectors in
the direction of the positive 𝑥-axis and the positive 𝑦-axis. So 𝑖 is going to the right,
and 𝑗 is moving up on the axes. So 𝑖 is one, zero and 𝑗 is zero, one. So if we’ve got this
vector here two 𝑖 minus three 𝑗, it’s just two of these 𝑖s strung together followed by the
negative of three of the 𝑗s strung together.
In other words, that’s two in the 𝑥-direction
and negative three in the 𝑦-direction, but we’re not gonna worry too much about that format
for this question because we’re gonna do it in terms of the 𝑖 and 𝑗 unit vectors. Now
although we’ve got 𝑖s and 𝑗s all over the place, the process is exactly what we’ve just done
before. So we want to work out the magnitude of the vector. And to do that, we’re gonna use the
Now two, positive two, is the 𝑥-component and negative three is the
𝑦-component. So we can just write that as the magnitude of this vector two 𝑖 minus three 𝑗
all squared is equal to the sum of the squares of the other side of the
triangle. So that’s two squared plus negative three squared, which is obviously four plus
nine, which is 13. So remember that was the magnitude squared, so taking square roots of
both sides says the magnitude of this line is the square root of 13.
And again, all we have to
do to work out the unit vector in that direction is divide each of those components, the 𝑥-
and 𝑦-components, by the magnitude of the resultant vector, so that was root 13. So two
divided by root 13𝑖 and negative three divided by root 13𝑗. So even in the 𝑖,
𝑗-format of our vectors, the process is still just the same. To work out the unit vector in
the direction of a particular vector, we take that vector and we divide by the magnitude of
the vector. We basically divide each of the components in turn by the magnitude of the
original vector, which we use Pythagorean theorem to calculate. So hopefully that gives you a
basic insight into unit vectors.