Video Transcript
Given the point 𝐴 negative three, negative five, six; the point 𝐵 zero, three,
negative seven; the point 𝐶 negative eight, 10, negative two; and the point 𝐷
negative three, nine, negative six, determine the measure of the angle between
vectors 𝐀𝐁 and 𝐂𝐃 rounded to the nearest hundredth.
In this question, we’re given the coordinates of four points, the coordinates of the
points 𝐴, 𝐵, 𝐶, and 𝐷. We need to use this to determine the measure of the angle between the two vectors:
the vector from 𝐴 to 𝐵 and the vector from 𝐶 to 𝐷. We need to give our answer rounded to the nearest hundredth.
To do this, we can start by recalling how we find the measure of the angle between
two vectors. We recall if 𝜃 is the angle between two vectors 𝐮 and 𝐯, then the cos of 𝜃 will
be equal to the dot product of 𝐮 and 𝐯 divided by the magnitude of vector 𝐮
multiplied by the magnitude of vector 𝐯. It’s also worth pointing out the same is true in reverse. If 𝜃 is an angle between these two vectors, then it will satisfy this equation. Therefore, we can find one angle between these two vectors by taking the inverse
cosine of both sides of this equation. In fact, this will give us the smallest nonnegative angle between these two
vectors. This is of course the same as the measure of the angle between these two vectors. Therefore, we just need to find the dot product of our two vectors and the magnitude
of both of these vectors.
But before we can do this, we’re going to need to find expressions for our two
vectors. Let’s start with the vector from 𝐴 to 𝐵.
Let’s start with the vector from 𝐴 to 𝐵. Remember, this is going to be the vector from the origin to 𝐵 minus the vector from
the origin to 𝐴. The vectors 𝐎𝐁 and 𝐎𝐀 will just be the position vectors of the points given to us
in the question. The position vector of point 𝐵 is the vector zero, three, negative seven. And the position vector of the point 𝐴 is the vector negative three, negative five,
six.
To subtract two vectors of the same dimension, all we need to do is subtract the
corresponding components from each of our two vectors. Doing this, we get the vector zero minus negative three, three minus negative five,
negative seven minus six. Evaluating the expressions in each of our components, we get the vector three, eight,
negative 13. We can do exactly the same to find the vector from 𝐶 to 𝐷. That’s the vector negative three, nine, negative six minus the vector negative eight,
10, negative two. Evaluating the subtraction component-wise, we get the vector five, negative one,
negative four.
Now that we found expressions for the vector from 𝐴 to 𝐵 and the vector from 𝐶 to
𝐷, we’re ready to find the dot product of these two vectors and the magnitude of
each of these two vectors. We’ll do this by clearing some space and finding the dot product between these two
vectors. Remember, to find the dot product between two vectors with an equal number of
dimensions, we do this by finding the sum of the products of the corresponding
components. In our case, we get three multiplied by five plus eight multiplied by negative one
plus negative 13 multiplied by negative four, which if we evaluate, we can see it’s
equal to 59.
Next, we’re going to need to find the magnitude of the vector from 𝐴 to 𝐵 and the
magnitude of the vector from 𝐶 to 𝐷. Remember, to find the magnitude of a vector, we need to find the square root of the
sum of the square of its components. In our case, we get the magnitude of the vector from 𝐴 to 𝐵 is equal to the square
root of three squared plus eight squared plus negative 13 all squared, which if we
calculate the expression inside of the square root symbol, we get the square root of
242.
We can do exactly the same to find the magnitude of the vector from 𝐶 to 𝐷. It’s equal to the square root of five squared plus negative one all squared plus
negative four all squared, which we can calculate is equal to the square root of
42.
We’re now ready to form an equation through our angle 𝜃 between the vector from 𝐴
to 𝐵 and the vector from 𝐶 to 𝐷. We know the cos of 𝜃 will be equal to the dot products of the vector from 𝐴 to 𝐵
and the vector from 𝐶 to 𝐷 divided by the magnitude of the vector from 𝐴 to 𝐵
multiplied by the magnitude of the vector from 𝐶 to 𝐷. We can then substitute the values we found for the three expressions on the
right-hand side of this equation. Doing this, we get the cos of 𝜃 will be equal to 59 divided by the square root of
242 multiplied by the square root of 42.
We could simplify the right-hand side of this equation. However, it’s not necessary. We just need to find the measure of angle 𝜃 and we can do this by taking the inverse
cosine of both sides of this equation. We get that 𝜃 is the inverse cos of 59 divided by root 242 multiplied by root 42,
which if we calculate in degrees, we get that this is equal to 54.181, and this
expansion continues degrees.
However, the question wants us to give our answer to the nearest hundredth. To do this, we need to look at the third decimal place, which is one. This means we need to round this value down. And this gives us our final answer. The measure of the angle between the vector from 𝐴 to 𝐵 and the vector from 𝐶 to
𝐷 rounded to the nearest hundredth is 54.18 degrees.
