WEBVTT

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JOEL LEWIS: Hi.

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Welcome back to recitation.

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In lecture, you've learned a
little bit about the dot

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product, and how the dot product
can be used to compute

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angles between vectors, or the
relationship between the dot

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product of vectors and the
angle between them.

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So, here I have two questions,
two related questions for you.

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So the first one, you've
got two vectors.

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i hat plus j hat plus 2k hat.

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And a second vector, 2i hat
minus j hat plus k hat.

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I'm probably going to stop
saying the hats in a minute,

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just warning you.

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So you've got these two
different vectors, and what

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I'd like to know is what the
angle between them is.

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And the second one is that
we have three points.

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So the P has coordinates
a, 1, and minus 1.

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Q has coordinates 0, 1, and 1.

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And R has coordinates
a, minus 1, and 3.

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So when you've given
three points,

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that define some angles.

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So what I'd like to know is for
which values of a is the

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angle PQR a right angle?

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So why don't you take some time,
work out these problems,

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come back, and we can work
them out together.

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Hopefully you had some
luck on these

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problems. Let's get started.

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Let's start with
the first one.

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So we have these two vectors,
i plus j plus 2k, and 2i

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minus j plus k.

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And we want to know what the
measure of the angle

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between them is.

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So the thing that we have to use
here, is that for vectors

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v and w, we know that v dot w
can be written in terms of the

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angle between them.

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So in particular, we know that
for vectors v and w, if the

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angle between them is equal to
theta, then cosine of theta is

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equal to v dot w, but I have to
divide by the length of v

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and the length of w.

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So I have this simple formula
for cosine of theta in terms

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of v and w.

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And so then if we want,
we can find theta by

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taking an arc cosine.

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So let's apply this formula
in the case of problem a.

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So we could say that v is equal
to i plus j plus 2k, and

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w is equal to 2i
minus j plus k.

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And so what that means is that
in our case, v dot w-- well,

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we just take, you know, the
products of the coordinates

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and add them up.

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So in that case, this is 1 times
2 plus 1 times minus 1

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plus 2 times 1.

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So this is 1 times 2 plus 1
times minus 1 plus 2 times 1.

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So that's 2 minus 1 plus 2,
so that's equal to 3.

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And so then we also need to know
what the length of v and

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w are in order to apply
this formula.

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So the length of v is equal to,
well it's the square root

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of the sum of the squares
of the coordinates--

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or the square root of v dot v.
So in this case, that's the

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square root of 1 squared plus
1 squared plus 2 squared.

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So that's the square
root of 6.

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And the length of w is equal to
it's the square root of 2

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squared plus minus 1 squared
plus 1 squared, which is also

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equal to the square root of 6.

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So if you put these three things
together, we have that

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cosine of theta is equal--

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so I'm going to write it right
up here next to this general

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formula just to save a
little bit of space--

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so in our case, this is equal
to 3 divided by the square

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root of 6 times the
square root of 6.

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So that's 3 divided by 6,
which is equal to 1/2.

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So if cosine of theta is equal
to 1/2, well we know that that

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means that theta is equal
to pi over 3.

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OK?

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So the angle between these
two vectors is pi over 3.

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OK, great.

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That's part a.

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Now let's talk about part b.

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So we have these three points
and we want to know for what

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values is the angle that they
define a right angle.

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For what values of
this parameter.

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Right?

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Our points depend on
this parameter, a.

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So in order to do that,
we have these

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three points in space.

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Let me draw a picture up here.

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So we have point P somewhere
and point Q somewhere and

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point R somewhere, and they
define this angle, and we want

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to know when this angle
is a right angle.

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Well, we don't have any vectors
yet, but it's easy to

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get some vectors involved
in this problem.

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We can look at the vectors
QP and QR.

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So those two vectors are the two
sides of this angle that

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we're interested in.

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And it's easy when you're given
coordinates for points

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to find the vector
between them.

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So you just subtract, right?

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So QP.

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In order to get from our point
Q over here-- in order to get

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from 0, 1, 1 to a, 1, minus 1--
well, you have to increase

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the first coordinate by a, the
second coordinate has to stay

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the same, and the third
coordinate has

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to decrease by 2.

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So this vector QP is equal
to a, 0, minus 2.

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So now I'm using
angle brackets,

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because this is a vector.

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SPEAKER 1: Positive
2 or negative 2?

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JOEL LEWIS: Negative 2.

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Yes.

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Right.

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Negative 2, yeah?

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Because we're going
from 1 to minus 1.

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OK, good.

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And similarly QR.

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So to get from Q to R--

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I have to go back and look at
the coordinates over here-- so

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I increase the first coordinate
by a, decrease the

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second coordinate by 2, increase
the third coordinate

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by 2, to get from Q to R. So the
vector QR is equal to a--

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what did I say--

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minus 2, 2.

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So these are the vectors QP and
QR, and we want to know

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when the angle between these two
things is a right angle.

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We want to know when these
vectors are perpendicular.

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OK?

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So the two vectors are
perpendicular exactly when

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their dot product is 0.

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Right?

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When the angle is 90 degrees.

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Cosine of 90 degrees.

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Cosine of pi over 2 radians is
equal to 0, so those two

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things are the same.

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The dot product equals 0 and
the angle is 90 degrees.

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So OK.

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So in our case, we just have
to look then at QP dot QR.

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So that's equal to-- well, we
just take this dot product

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here-- so it's equal
to a squared plus--

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0 times minus 2 is 0--

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minus--

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2 times 2, so plus minus 2 times
2-- so that's minus 4.

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So that's a squared minus 4.

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And so this dot product is
equal to 0 exactly when a

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squared minus 4 is equal to 0.

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So a squared minus 4 is equal
to 0 when a is equal to 2 or

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when a is equal to minus 2.

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So there are two values of a for
which those three points

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form a right angle at Q.

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So just to quickly summarize,
we used this general formula

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here for the cosine of an angle
in terms of the two

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vectors that are the
sides of the angle.

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And with that formula, we can
compute the measure of angles

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between two vectors as here--
or, as in the first

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part, in part a.

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Or in a special case when we're
looking for a 90 degree

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angle, we can use the simplified
form when we just

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need to figure out when the
dot product is equal to 0.

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OK.

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So I'll end there.