IB · Mathematics Analysis and Approaches (AA)
AAHL 3.13.2 Properties of the scalar product
Learn AAHL 3.13.2 Properties of the scalar product in this free IB Mathematics Analysis and Approaches (AA) video lesson for AHL 3.13—Scalar (dot) product.
Learn AAHL 3.13.2 Properties of the scalar product in this free IB Mathematics Analysis and Approaches (AA) video lesson for AHL 3.13—Scalar (dot) product.
Learn AAHL 3.13.2 Properties of the scalar product in this free IB Mathematics Analysis and Approaches (AA) video lesson for AHL 3.13—Scalar (dot) product.
The lesson focuses on the properties of the scalar product (dot product) of vectors, highlighting key concepts that are not included in the formula booklet. Important properties discussed include:
Examples are provided to illustrate these properties, including calculations involving unit vectors and angles between vectors, leading to conclusions about the relationships between them. The lesson concludes with a discussion on the implications of these properties in geometric contexts, such as the perpendicularity of diagonals in a parallelogram formed by vectors.
00:00Okay guys, so in this
00:02lesson we're gonna look at
00:03the properties of the scalar
00:04product. Now I've written here
00:06in red different properties. As
00:11you know, that means they
00:12are not in the formula
00:13booklet. This is what you
00:14get in the formula booklet.
00:15You don't get this. But
00:17a lot of this is
00:18actually quite intuitive. If you
00:19think of like a scalar
00:21product is just this times
00:24times this, then kind of
00:26obviously V dot W.
00:28equals w dot v. And
00:30then u dot v plus
00:32w would equal u plus,
00:34now u dot v plus
00:36u dot w. If you
00:37just want to try and
00:39work out a few of
00:39these, you'll see that it's
00:41very clear. K in this
00:43case is a scalar. So
00:46a scalar times vector dot
00:48another vector is equal to
00:49the scalar times the dot
00:51portrait of the two vectors.
00:52We've come across this one
00:53before v dot w equals
00:55zero
00:56For, this is my perpendicular
00:57science. So when they're perpendicular,
00:59the dot product equals zero.
01:01When they're parallel, this is
01:03parallel. The dot product of
01:06V and W equals the
01:07product of the magnitude of
01:09the vectors. Now I'll just
01:11explain very quickly why that's
01:12the case. If class of
01:14theta equals a dot b,
01:18a dot b in the
01:19middle, over magnitude of a
01:22times magnitude of b,
01:24Well, if you've two vectors
01:27that are parallel, what's the
01:28angle between them? Like what's
01:30the angle between this and
01:31this? If you put them
01:32right beside each other. Well,
01:34it's zero. There is no
01:36angle because they're basically the
01:38same line or the same
01:40direction. So if the angle
01:44is zero, what's the cosine
01:45of zero? Or the cos
01:47of zero is one. So
01:48you left with one equals
01:51a
01:52that B over magnitude of
01:55eight times magnitude of B.
01:56And then that gives us
01:58the magnitude of eight times
01:59magnitude of B when I
01:59multiply across is equal to
02:01A dot B. So this
02:03is for parallel vectors. So
02:08that's where this comes from.
02:10And then this comes from
02:12this basically is the same
02:13rule because it's V dot
02:15V. That means the same
02:17vector, the dot product of
02:19the vector
02:20and itself is the magnitude
02:23squared because if you have
02:25the two vectors that are
02:26the same like this and
02:27this, these are the exact
02:29same vectors, well, they're parallel
02:31also so this applies to
02:32that too. Okay, that's basically
02:35it. I think this is
02:37all pretty straightforward. It works
02:39out how you would expect.
02:44However, they can obviously give
02:45you quite complicated questions. This
02:47one is fairly
02:48It's not that straightforward. Let's
02:51put it that way. Okay,
02:52so part a, I'm going
02:53to do two, I think
02:54I have two examples for
02:55you to add on this.
02:57So if a is a
02:58unit vector perpendicular to b,
03:00so I see perpendicular to
03:01strader, strader, I'm like, okay,
03:02that's important. A is perpendicular
03:04to b. Find the value
03:05of a dot two a
03:07minus b, right? So I'm
03:08going to write that out.
03:09A dot two a minus
03:11three b. So I know
03:14using my properties of
03:16of scalar products and look,
03:18I don't even feel the
03:19need to go and look
03:20at this, but this multiplies
03:23out as you would expect
03:24it to do. So you
03:25get two a dot a
03:29minus three a dot b.
03:32That's it. Now it says
03:35if a is a unit
03:36vector perpendicular to b, find
03:39the value of this. So
03:41if they are perpendicular
03:44if a and b are
03:46perpendicular, the dot product equals
03:48zero. So as soon as
03:50I see that I'm like
03:50dot product equals zero. So
03:51this is zero. And it
03:53says a unit vector. So
03:56a dot a is going
03:57to be the magnitude of
03:59a squared that comes from
04:01this one minus three times
04:03zero. This is just zero
04:05because it's their perpendicular a
04:06and b are perpendicular. And
04:08then a unit vector means
04:09it has length one. So
04:11it's just two times one
04:12square
04:12which is two. So that's
04:13it. So arguably that wasn't
04:16that difficult. B, if p
04:18is a unit vector, making
04:20a 45 degree angle with
04:22vector q and p dot
04:24q is three root two,
04:26find the magnitude of q,
04:27right? That seems somewhat confusing.
04:32However, we're looking at two
04:36vectors and an angle between
04:38them. So I'm definitely going
04:39to need this formula. The
04:40cost of
04:40Theta is equal to, let's
04:42go with p dot q
04:44over magnitude of p, magnitude
04:47of q. It says p
04:49is a unit vector, and
04:51q, it doesn't say is
04:52a unit vector. It says
04:54the angle between them is
04:5545 degrees. Now note that
04:58they're using degrees here. So
04:59sometimes for vectors, you'll see
05:01degrees. Sometimes you see ratings
05:03for often its degrees. So
05:04this is equal to p
05:05dot q. Now they take
05:08LSP .q is three root
05:12two. So that's easy. And
05:15then, they say P is
05:18a unit vector. So the
05:19length of P is just
05:20one, and the magnitude of
05:21P is one. And so
05:23it's just one times one,
05:29and put that on a
05:30bracket. One times the magnitude
05:33of Q. So look, I
05:34actually have an equation with
05:36only one.
05:36unknown, which is the thing
05:38I'm trying to find the
05:39magnitude of Q. So this
05:40is actually not too bad
05:42either. So this is going
05:43to be, let's just go
05:44over here. Cost of 45
05:46is 1 over root 2
05:48equals 3 root 2 over
05:53the magnitude of Q. Cost
05:54multiply the magnitude of Q
05:56is equal to 3 root
05:582 times root 2, which
05:59is 3 times 2, which
06:00is 6. There we go.
06:04Yeah, it's not not necessarily
06:08that that difficult, but a
06:11little bit tricky confusing the
06:13first time you see it
06:14Okay, this is a pass
06:16paper question Consider the vectors
06:19o a equals a ob
06:20was b and o c
06:21equals a plus b show
06:23that if a Okay, I
06:26like to when I win
06:28you can I like to
06:29draw out something so our
06:32draw out a
06:33the diagram. So this is
06:35a, this is O A,
06:37so this is O, this
06:38is A, and this is,
06:42O A is equal to
06:44little A. And then I
06:47have a, I have B.
06:51Now the length of A,
06:52the magnitude of A is
06:53equal to the magnitude of
06:54B. So let's go with
06:57that, but they're the same.
06:59So these are the same
07:00magnitude.
07:01This is point B. It
07:03then says, oh, C is
07:08A plus B. So it's
07:10A plus B. So what
07:14we get when this is
07:16going to give us a
07:19parallelogram. This is a parallelogram
07:22because A plus B, A
07:25plus B is the same
07:27as B plus A.
07:29This is going to give
07:29us a parallel diagram, but
07:30this is point C. So
07:32that's point B. That's point
07:34A, that's point A, B,
07:36C, and zero. Find, show
07:39that if magnitude of A
07:40and magnitude of B are
07:41the same, then this equals
07:43zero. Let's do that first.
07:46So obviously one of the
07:49things they're testing here is
07:50that you are okay with
07:52the fact that you
07:57in the scalar product, using
07:59the properties of the scalar
08:00product, this becomes a .a.
08:02So just like you'd multiply
08:03out to, to, um, binomials,
08:09like you'd do where you'd
08:09expand the brackets. So it's
08:10a, this times this plus
08:12this times this, so minus
08:14a dot b plus this
08:17times this plus b dot
08:19a minus this times this
08:22so minus b dot b.
08:25equals a dot a is
08:28the magnitude of a squared.
08:31a dot b equals b
08:32dot a. So minus a
08:33dot b plus b dot
08:35a minus a dot b
08:37plus b dot a equals
08:39zero. They cancel if you
08:40like. So I'm going to
08:41put this minus the magnitude
08:44of b squared. And because
08:46b dot b is the
08:48magnitude of b squared, it
08:50then says if these are
08:52equal,
08:53show that this equals zero,
08:55well this equals zero, as
08:57I'm just going to write
08:58here as the magnitude of
09:00a equals the magnitude of
09:02b, just to be clear
09:03why you're doing that. Because
09:05it says show, if they
09:05say show, make sure you
09:07show. What does this tell
09:10us about the parallelogram OACB?
09:14So, okay, this is an
09:17interesting one. This, this vector
09:20here,
09:21is A, this is A
09:26plus B, this vector. Now
09:29what's the vector? Because what
09:31it's telling us is A
09:32plus B and A minus
09:36B, this vector and this
09:37vector are perpendicular. Because when
09:40the dot product is zero,
09:41they're perpendicular. So A plus
09:43B and A minus B
09:45are perpendicular. What's A minus
09:47B? Well, A minus B
09:49minus B, A minus B
09:52is B, A. If you
09:56remember, B, A, remember this
09:58rule that I once told
09:59you was super, super important.
10:01B, A is A minus
10:02B, when these are position
10:04vectors. So this is A
10:13minus B, that's the A
10:14minus B vector. So basically
10:17what is
10:17telling us is what does
10:18it tell us about the
10:19parallelogram? Let me complete the
10:21parallelogram. This is your parallelogram.
10:25What it's telling us is
10:26that the diagonals of the
10:29parallelogram are perpendicular. So it
10:32tells us, let me call
10:33that part A, that part
10:36B, I like. It tells
10:39us diagonals, diagonals of the
10:45Power a L diagram are
10:51perpendicular. Per pen. Decor. Okay,
10:59that's it. As I say,
11:03you will be able to
11:04find difficult challenging questions on
11:08the properties of skater products.
11:11But
11:13certainly the main thing you
11:14need to know is all
11:16of these. So they work
11:20as you would expect them
11:22to do, which is not
11:23the case for the cross
11:23product, which we're going to
11:24do in an upcoming lesson.
11:28I'll be giving you a
11:30lot more warnings about properties
11:31of the cross product, because
11:33it's sometimes it gets a
11:35bit strange, but the properties
11:36of the scalar product work
11:37as you would expect them
11:38to work. These three here,
11:41and come up all the
11:44time so make sure you
11:44understand that v dot w
11:46is zero when the perpendicular
11:49and v dot v is
11:51equal to the magnitude of
11:52v squared. Okay, that's it.
11:56See you in the next
11:57lesson.