Video Transcript
Find the unit vector in the
direction of the vector ππ, given that vector π is zero, one, negative two and
the vector π is one, one, two.
In this question, weβre given two
vectors: the vector π and the vector π. We need to determine the unit
vector which has the same direction as the vector of ππ. And to do this, letβs start by
recalling what we mean by a unit vector. We say that any vector which has
magnitude one is a unit vector, and we often represent this with a hat notation. For example, π hat would be the
vector in the same direction as vector π; however, its magnitude would be equal to
one. So when the question asks us to
find the unit vector in the direction of ππ, we could have also written this as
ππ hat. Itβs the vector in the same
direction as ππ. However, its magnitude will be
equal to one.
We also know a formula for finding
the unit directional vector pointing in the same direction as the vector π―. π― hat will be equal to one over
the magnitude of π― multiplied by the vector π―. And in our case, weβre going to
want to apply this to the vector ππ. So first, weβre going to need to
find the vector ππ. First, recall the vector from π to
π is going to be the vector π minus the vector π. Thatβs the vector one, one, two
minus the vector zero, one, negative two. And remember, to subtract two
vectors with the same number of components, we need to subtract each of our
corresponding components separately. This gives us the vector one minus
zero, one minus one, and two minus negative two.
And now we can just calculate each
of these components separately. We get the vector ππ is the
vector one, zero, four. Now weβre going to want to apply
our formula to find the unit directional vector which points in the same direction
as ππ. We get that ππ hat is going to be
equal to one over the magnitude of ππ multiplied by our vector ππ. But we canβt evaluate this yet
because we donβt know the magnitude of our vector ππ. So before we do this, weβre going
to need to find the magnitude of vector ππ. And to do this, weβre going to need
to recall how we find the magnitude of a vector. The magnitude of a vector is the
square root of the sums of the squares of its components. So the magnitude of the vector π,
π, π will be the square root of π squared plus π squared plus π squared.
We can then use this to find the
magnitude of our vector ππ. Our value of π is one, our value
of π is zero, and our value of π is four. We need to take the square root of
the sums of the squares of these components. This gives us the square root of
one squared plus zero squared plus four squared. And if we calculate this, we see
itβs equal to the square root of 17. Now, we can substitute both of
these into our equation for ππ hat. We get that ππ hat is equal to
one over root 17 multiplied by the vector one, zero, four. Now, we could leave our answer like
this. However, we can simplify this
because we can notice weβre multiplying a constant by a vector. So we can simplify this by using
scalar multiplication of a vector.
We recall to multiply a scalar by a
vector, we need to multiply all of our components by our scalar. In other words, π multiplied by
the vector π, π, π is equal to the vector ππ, ππ, ππ. So we need to multiply all of the
components of our vector by our scalar one over root 17. This gives us the vector one over
root 17 times one, one over root 17 times zero, one over root 17 times four. And we can evaluate each of these
components. We get one over root 17, zero, four
over root 17. And we could just leave our answer
like this. However, we can also simplify the
components of our vector by rationalizing their denominators.
In our first and third component,
we want to multiply by the vector root 17 divided by root 17. And then evaluating this and
simplifying, we get our final answer which is the vector root 17 over 17, zero, four
root 17 over 17. Therefore, we were able to show,
given the vector π is zero, one, negative two and the vector π is one, one, two,
we were able to find the vector ππ by subtracting the vector π from our vector of
π. Then we were able to find the unit
vector in the same direction as ππ by multiplying ππ by one over its
magnitude. This gave us ππ hat is equal to
the vector root 17 over 17, zero, four root 17 over 17.