𝐴𝐵𝐶 is a triangle with the point 𝐴 having coordinates two, three, one; the point 𝐵 having coordinates three, five, four; and the vector 𝐵𝐶 having components negative one, four, zero. Find the measure of the angle 𝐴𝐵𝐶 and the component of the vector 𝐴𝐶 in the direction of the vector 𝐴𝐵.
Let’s sketch the scenario to help us. We have the point 𝐴 with coordinates two, three, one; and 𝐵 with coordinates three, five, four. And we’re given that the vector 𝐵𝐶 has components negative one, four, zero. So, drawing this vector starting from 𝐵, the endpoint will be 𝐶. And we’re interested in the triangle 𝐴𝐵𝐶. So, we draw in the sides 𝐴𝐵 and 𝐴𝐶.
We are required to find two things, firstly the measure of angle 𝐴𝐵𝐶, that’s this angle here, which we’ll call 𝜃. Note, that this is the angle between the vectors 𝐵𝐶, which is already drawn, and 𝐵𝐴. And so, the cos of this angle 𝜃 that we want to find is just the dot product of the vectors 𝐵𝐶 and 𝐵𝐴 divided by the product of the magnitudes of 𝐵𝐶 and 𝐵𝐴. This follows from the fact that the dot product of two vectors is the product of their magnitudes times the cos of the angle 𝜃 between them. All we’ve done here is to divide by the product of the magnitude of the vectors to get cos 𝜃 as the subject.
Okay, so, let’s get going. We know the components of the vector 𝐵𝐶. We’re given that in the question. And so, we can substitute that in. However, we don’t know the components of 𝐵𝐴 yet. We’re going to have to work those out. The vector 𝐵𝐴 is the position vector of 𝐴 minus the position vector of 𝐵. And the components of the position vector of 𝐴 are just the coordinates of 𝐴. So, that’s two, three, one, similarly for 𝐵.
And to subtract two vectors in component form, we just subtract their components. So, the difference of the first components, two minus three is negative one. Difference of the second components, three minus five is negative two. And the difference of the third components, one minus four is negative three. Having now found the components of 𝐵𝐴, we can substitute them in.
Now we just have to worry about the denominator of our fraction on the right-hand side where we have the magnitudes of 𝐵𝐶 and 𝐵𝐴. What are these magnitudes? Well, the magnitude of the vector with components 𝐴𝐵 and 𝐶 is the square root of 𝐴 squared plus 𝐵 squared plus 𝐶 squared. So, to find the magnitude of the vector 𝐵𝐶, we write the vector in its component form.
We were told these components in the question. Now we apply the formula for the magnitudes to get the square root of negative one squared plus four squared plus zero squared. Negative one squared is one. Four squared is 16. And zero squared is zero. So, the magnitude of the vector 𝐵𝐶 is the square root of 17. So, we substitute this value in.
Now all we need to do is to find the magnitude of the vector 𝐵𝐴. And as we know its components, this is again straightforward. Following the same procedure as for the magnitude of 𝐵𝐶, we find that the magnitude of 𝐵𝐴 is the square root of 14. And we substitute this value as well. Now let’s clear some room so we can continue the calculations.
We have to compute the dot product in the numerator of our fraction. We get the products of the first components, that’s negative one times negative one, plus the product of the second components, that’s four times negative two, plus the product of the third component, that’s zero times negative three. In the denominator, we still have the square root of 17 times the square root of 14. Now negative one times negative one is one. Four times negative two is negative eight. And zero times negative three is zero. So, we get negative seven in the numerator.
And applying the inverse cosine function to both sides, using calculator, we get an approximate value of 116.984 and so on degrees, which to the nearest degree is 117 degrees. Remember that 𝜃 was the measure of the angle 𝐴𝐵𝐶, which we were required to find. And so, this is our answer. Now we can move on to finding the component of 𝐴𝐶 in the direction of 𝐴𝐵. But first, let’s clear some more space.
What does it mean to talk about the component of 𝐴𝐶 in the direction of 𝐴𝐵? The idea is that we can write our vector 𝐴𝐶 as the sum of two component vectors. One, which as a multiple of 𝐴𝐵, points in the direction of 𝐴𝐵, and the other which is perpendicular to 𝐴𝐵. The vector that we’re looking for is the vector 𝑘 times 𝐴𝐵, which is the component of 𝐴𝐶, which points in the direction of 𝐴𝐵.
But how do we find this? What we can do is we can take this equation 𝐴𝐶 equals 𝑘 times 𝐴𝐵 plus 𝑉 and take the dot product with 𝐴𝐵 on both sides. As the vectors on the left- and right-hand sides are equal, their dot products with 𝐴𝐵 will be two. Now the dot product is distributive over vector addition. So, the right-hand side becomes 𝑘 times 𝐴𝐵 dot 𝐴𝐵 plus 𝑉 dot 𝐴𝐵. And as 𝑉 was defined to be perpendicular to 𝐴𝐵, its dot product with 𝐴𝐵 will be zero. And this term disappears.
Swapping the sides and dividing through by 𝐴𝐵 dot 𝐴𝐵, we get that the value of the constant 𝑘 is 𝐴𝐶 dot 𝐴𝐵 divided by 𝐴𝐵 dot 𝐴𝐵. And this is a quantity we can compute. So, let’s substitute it for 𝑘. And so, we see that the component of 𝐴𝐶 in the direction of 𝐴𝐵 is 𝐴𝐶 dot 𝐴𝐵 over 𝐴𝐵 dot 𝐴𝐵 times the vector 𝐴𝐵. Let’s clear some room then and compute.
To compute the right-hand side, we’re going to need the vector 𝐴𝐶, which we don’t know yet. But we can work it out. The vector 𝐴𝐶 is the vector 𝐶 minus the vector 𝐴. Now the question is what is the vector 𝐶? Again, we don’t know. We’re going to have to work it out.
Looking at the diagram, we can see that we know the vector 𝐵 and the vector 𝐵𝐶. And so, it’s natural to write 𝐶 as 𝐵 plus 𝐵𝐶. The components of the vector 𝐵 are just the coordinates of the point 𝐵. So, that’s three, five, four. The vector 𝐵𝐶, we were told in the question, has components negative one, four, zero. And the vector 𝐴 has components, which are the coordinates of the point 𝐴, two, three, one.
We add the first two vectors together, adding their components. Three plus negative one is two. Five plus four is nine. And four plus zero is four. Now from this, we have to subtract the vector with components two, three, one. So, again, subtracting components, two minus two is zero, nine minus three is six, and four minus one is three. And we’ve found our vector 𝐴𝐶.
How about the vector 𝐴𝐵? Well, we know the coordinates of 𝐴 and 𝐵. So, we could work it out. But, of course, it’s also the additive inverse of the vector 𝐵𝐴. And so, its components will be the additive inverses of the components of 𝐵𝐴. That’s one, two, three. If you don’t believe me, you can work out the components of 𝐴𝐵 using the coordinates of 𝐴 and 𝐵. So, we can also substitute this value. And having made these substitutions, we get something that we can calculate, but only if we have space to do so.
Let’s compute the dot product in the numerator, zero times one is zero, six times two is 12, and three times three is nine. And in the denominator, one times one is one, two times two is four, and three times three is nine. So, we get 21 over 14 times the vector one, two, three. And 21 over 14 is not in its simplest form. We can divide both numerator and denominator by seven, giving three over two.
And to multiply the vector one, two, three by three over two, we multiply each of the components by three over two. Three over two times one is three over two. Three over two times two is three. And three over two times three is nine over two. This is the component of the vector 𝐴𝐶, which is in the direction of the vector 𝐴𝐵. In this question, we’ve used the dot product in two ways. Firstly, to find the angle between two vectors and then to find the projection of one vector onto another.