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
00:23this plus this times this
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.