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.