# Video: Finding a Unit Vector in a Given Direction

What is the unit vector in the same direction as 𝐀 = 〈3, 4〉?

02:30

### Video Transcript

What is the unit vector in the same direction as the vector 𝐀 which has components three and four?

So we’ve been asked to find a unit vector in the same direction as the given vector 𝐀. Now, this means that two things must be true. Firstly, our vector must be parallel to the vector 𝐀. But it must have a magnitude of one. Diagrammatically, that will look a little like this. The two vectors are parallel. They’re in the same direction. But the vector 𝐮 will have a magnitude of one. Now, the way that we can achieve this is to take the vector 𝐀 and divide it by its own magnitude, which we’ll first need to calculate.

We recall that for the general two-dimensional vector with components 𝑎 and 𝑏, its magnitude is equal to the square root of 𝑎 squared plus 𝑏 squared. And this is simply an application of the Pythagorean theorem. The magnitude of our vector 𝐀 then is the square root of three squared plus four squared. That’s the square root of nine plus 16, which is the square root of 25, which is simply equal to five. In fact, we should recognise this as one of our Pythagorean triples. That is, right triangles in which all three sides have integer lengths.

To find our unit vector 𝐮, which is in the same direction as 𝐀 then, we take the vector 𝐀 and we divide it by its magnitude. So that’s the same as multiplying by one-fifth. We have that 𝐮 is equal to one-fifth of the vector three, four. In order to multiply a vector by a scalar, we just multiply each component by that scalar. So the first component of our vector 𝐮 will be one-fifth multiplied by three, which is three-fifths. And the second component will be one-fifth multiplied by four, which is four-fifths. We’ve, therefore, found a unit vector in the same direction as 𝐀. It’s the vector with components three-fifths and four-fifths.

We can, of course, check that the magnitude of this vector is equal to one. We have the square root of three-fifths squared plus four-fifths squared, which is the square root of nine over 25 plus 16 over 25. That’s the square root of 25 over 25 or the square root of one which is equal to one. So we’ve confirmed that the magnitude of our vector is indeed equal to one. And so, it is a unit vector.