# Video: MATH-ALG+GEO-2018-S1-Q8A

MATH-ALG+GEO-2018-S1-Q8A

09:31

### Video Transcript

π΄π΅πΆ is a triangle with the point π΄ having coordinates two, three, one; the point π΅ having coordinates three, five, four; and the vector π΅πΆ having components negative one, four, zero. Find the measure of the angle π΄π΅πΆ and the component of the vector π΄πΆ in the direction of the vector π΄π΅.

Letβs sketch the scenario to help us. We have the point π΄ with coordinates two, three, one; and π΅ with coordinates three, five, four. And weβre given that the vector π΅πΆ has components negative one, four, zero. So, drawing this vector starting from π΅, the endpoint will be πΆ. And weβre interested in the triangle π΄π΅πΆ. So, we draw in the sides π΄π΅ and π΄πΆ.

We are required to find two things, firstly the measure of angle π΄π΅πΆ, thatβs this angle here, which weβll call π. Note, that this is the angle between the vectors π΅πΆ, which is already drawn, and π΅π΄. And so, the cos of this angle π that we want to find is just the dot product of the vectors π΅πΆ and π΅π΄ divided by the product of the magnitudes of π΅πΆ and π΅π΄. This follows from the fact that the dot product of two vectors is the product of their magnitudes times the cos of the angle π between them. All weβve done here is to divide by the product of the magnitude of the vectors to get cos π as the subject.

Okay, so, letβs get going. We know the components of the vector π΅πΆ. Weβre given that in the question. And so, we can substitute that in. However, we donβt know the components of π΅π΄ yet. Weβre going to have to work those out. The vector π΅π΄ is the position vector of π΄ minus the position vector of π΅. And the components of the position vector of π΄ are just the coordinates of π΄. So, thatβs two, three, one, similarly for π΅.

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 π΅π΄, 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 π΅πΆ and π΅π΄. What are these magnitudes? Well, the magnitude of the vector with components π΄π΅ and πΆ is the square root of π΄ squared plus π΅ squared plus πΆ squared. So, to find the magnitude of the vector π΅πΆ, 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 π΅πΆ 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 π΅π΄. And as we know its components, this is again straightforward. Following the same procedure as for the magnitude of π΅πΆ, we find that the magnitude of π΅π΄ is the square root of 14. And we substitute this value as well. Now letβ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, thatβs negative one times negative one, plus the product of the second components, thatβs four times negative two, plus the product of the third component, thatβ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 π was the measure of the angle π΄π΅πΆ, which we were required to find. And so, this is our answer. Now we can move on to finding the component of π΄πΆ in the direction of π΄π΅. But first, letβs clear some more space.

What does it mean to talk about the component of π΄πΆ in the direction of π΄π΅? The idea is that we can write our vector π΄πΆ as the sum of two component vectors. One, which as a multiple of π΄π΅, points in the direction of π΄π΅, and the other which is perpendicular to π΄π΅. The vector that weβre looking for is the vector π times π΄π΅, which is the component of π΄πΆ, which points in the direction of π΄π΅.

But how do we find this? What we can do is we can take this equation π΄πΆ equals π times π΄π΅ plus π and take the dot product with π΄π΅ on both sides. As the vectors on the left- and right-hand sides are equal, their dot products with π΄π΅ will be two. Now the dot product is distributive over vector addition. So, the right-hand side becomes π times π΄π΅ dot π΄π΅ plus π dot π΄π΅. And as π was defined to be perpendicular to π΄π΅, its dot product with π΄π΅ will be zero. And this term disappears.

Swapping the sides and dividing through by π΄π΅ dot π΄π΅, we get that the value of the constant π is π΄πΆ dot π΄π΅ divided by π΄π΅ dot π΄π΅. And this is a quantity we can compute. So, letβs substitute it for π. And so, we see that the component of π΄πΆ in the direction of π΄π΅ is π΄πΆ dot π΄π΅ over π΄π΅ dot π΄π΅ times the vector π΄π΅. Letβs clear some room then and compute.

To compute the right-hand side, weβre going to need the vector π΄πΆ, which we donβt know yet. But we can work it out. The vector π΄πΆ is the vector πΆ minus the vector π΄. Now the question is what is the vector πΆ? Again, we donβt know. Weβre going to have to work it out.

Looking at the diagram, we can see that we know the vector π΅ and the vector π΅πΆ. And so, itβs natural to write πΆ as π΅ plus π΅πΆ. The components of the vector π΅ are just the coordinates of the point π΅. So, thatβs three, five, four. The vector π΅πΆ, we were told in the question, has components negative one, four, zero. And the vector π΄ has components, which are the coordinates of the point π΄, 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 weβve found our vector π΄πΆ.

How about the vector π΄π΅? Well, we know the coordinates of π΄ and π΅. So, we could work it out. But, of course, itβs also the additive inverse of the vector π΅π΄. And so, its components will be the additive inverses of the components of π΅π΄. Thatβs one, two, three. If you donβt believe me, you can work out the components of π΄π΅ using the coordinates of π΄ and π΅. 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.

Letβ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 π΄πΆ, which is in the direction of the vector π΄π΅. In this question, weβ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.