Learn AAHL 3.13.1 Scalar product and angle between vectors in this free IB Mathematics Analysis and Approaches (AA) video lesson for AHL 3.13—Scalar (dot) product.
Learn AAHL 3.13.1 Scalar product and angle between vectors in this free IB Mathematics Analysis and Approaches (AA) video lesson for AHL 3.13—Scalar (dot) product.
The video introduces the scalar product, also known as the dot product, which is a method for multiplying two vectors to find the angle between them. The scalar product is defined mathematically as:
where and are the magnitudes of the vectors, and is the angle between them.
Key points covered include:
00:00Hi guys, so in this
00:01lesson I'm going to introduce
00:02you to the scalar product
00:04and we're going to use
00:04it to find the angle
00:06between two vectors. So I'm
00:10going to do this example
00:11in a second, but first
00:12I want to just kind
00:13of talk to you about
00:14the scalar product using this
00:15example here. So imagine I
00:17have, I will just look
00:18in 2D because it's easier
00:19
00:21have two vectors, so this
00:22vector, this vector, this vector
00:23is length 6, this guy's
00:26length or magnitude
00:28is five. That's a nice
00:30three, four, five triangle. So
00:31he is a magnitude of
00:32five. I think I mentioned
00:35in the previous operations of
00:38vectors lesson that multiplying vectors
00:40is not straightforward because how
00:43can it be straightforward? What
00:44does it mean to multiply
00:45together to vectors? It's easy
00:48to multiply five times six,
00:50like a 30, but what
00:55if
00:56what if this vector was
00:57going this way? Let's say
00:59it was going like this.
01:01So how could he have
01:03the same? How could this
01:05times this be the same
01:07as this times this? It
01:08just wouldn't make any sense.
01:10So we define we've two
01:11definition, the two ways to
01:13multiply vectors, the scalar product
01:14and the cross product. We'll
01:16get to the cross product
01:17later. They're all soon, but
01:21just in this lesson, I
01:21want to talk about the
01:22scalar product. So the scalar
01:24product is defined as this
01:26we multiply. So it's it's
01:29this it's the length. It's
01:31the length of V and
01:32we say V dot W
01:33or a dot B that's
01:35what it's called. So it's
01:35called the dot product. I
01:37actually call it the dot
01:38product. So the dot product
01:39and because we've a dot
01:40in the middle, that's the
01:42multiplication sign. So we're going
01:44to multiply this time. So
01:45it's the it's the length
01:47of the magnitude of one
01:49vector times the magnitude of
01:51the other vector times the
01:52the cost of the angle
01:53between them. So kind of
01:54what you're doing is, like
01:56if this was, let's say
01:59this is, let's say this
02:03is V and this is
02:05W. Then this red line
02:09here, there, that's actually the,
02:15that is actually W. So
02:19the red line here,
02:20going this direction is, well,
02:23I don't even need the
02:24direction, but it's the W
02:27cos theta. We know that
02:30from trigonometry. So this is
02:32actually the magnitude of W
02:35times cos theta where theta
02:38is obviously this angle here
02:39between them. So we end
02:41up just doing six times,
02:45and in this case, it's
02:46in this case, it is
02:48three. So it's just six
02:50times three gives me, six
02:55times three gives me eighteen.
02:57Now if you were to
02:58actually, so this vector here
03:00is, let's say it's six
03:03zero, because it's six along
03:05zero up. And if I
03:07do six zero dot, so
03:08I'm going to multiply it
03:09by this vector, which is
03:12three four, three four. The
03:16way
03:16I can get the dot
03:17product is this. Look, it's
03:19v1 times v1, w1 plus
03:22v2, w2 plus v3, w3.
03:25So quite simply, this times,
03:27this plus this times, this,
03:28plus this times, this if
03:29there was a k component.
03:31But there isn't. So it's
03:32just six times 13 is
03:3318 plus zero, which is
03:36just 18. And that's kind
03:38of shown it there. So
03:39the scalar product is when
03:42you multiply the two vectors
03:44using the
03:44get a product or the
03:45top product, you get a
03:46scalar, which is why it's
03:47called a scalar product. So
03:48you don't get a vector.
03:49When we do the cross
03:50product, you're going to get
03:51a vector. But the scalar
03:53product or dot product, we
03:56get a scalar. OK, so
04:00let's do this. It's very,
04:02very straightforward. Once you understand
04:04this or even if you
04:06don't understand this, it's very
04:08simple. It is this dot
04:12This, which is 2 times
04:164, 8 plus negative 1
04:19times 2 is minus 2
04:20plus 6 times 1 is
04:226 equals 8 minus 2
04:24is 6 plus 6 is
04:2512. So the dot product
04:27of A and B, A
04:29dot B equals 12. Simple.
04:32So there's two kind of
04:33two ways to get the
04:34dot product this way and
04:37this way. And we're going
04:37to use this formula, this
04:39formula is what we're going
04:40to use.
04:40to find the angle between
04:42two vectors. And you see
04:43it here rearranged. So this
04:46is just, if you do,
04:48if you rearrange this and
04:49say, cost of theta is
04:50v dot w over magnitude
04:52of v times magnitude of
04:53w, you get this because
04:56this v dot w is
04:58this from here. So I'm
05:00not going to make sense.
05:01Okay, example two. Now I
05:04just mix and match whether
05:05I write it like in
05:06this form or
05:08Ijk form just so you're
05:10all getting used to the
05:11two forms that they can
05:12write it in. Okay, so
05:14find the angle between the
05:15vectors a and b. So
05:16the angle between two vectors,
05:19cos of theta, is equal
05:21to a dot b, a
05:24dot b, over the magnitude
05:26of a times the magnitude
05:28of b, simple. Which is
05:31equal to, now at the
05:32side here I'm gonna do
05:33a dot b. a dot
05:35b,
05:36And remember I said when
05:38I first introduced you to
05:40vectors, I said I always
05:41prefer working with like this,
05:45not I chain case. This
05:47dot b is negative two,
05:50five, two. And this equals
05:53negative four minus five plus
05:57six equals negative nine plus
05:59six is negative three. Then
06:02I'm going to do, I'm
06:03going to move this down
06:04guys
06:04because I want space, always
06:07use space. Okay, so this
06:11is a dot b. The
06:12magnitude of a, the magnitude
06:15of a, magnitude of a
06:18is equal to the square
06:20root of the square root
06:23of two squared plus one
06:25squared plus three squared, which
06:27is four, five plus nine
06:30is 14. So that's root
06:3114. And the magnitude
06:33Chude of B is equal
06:34to the square root of
06:382 squared, 2 squared plus
06:415 squared plus 2 squared.
06:43So C now I've started
06:44to, I ignore the negative
06:46because I'm squaring it so
06:48it's gonna be positive anyway.
06:49It's just saves time and
06:52less likely to make mistakes.
06:53So 4 plus 4 is
06:549, plus 25 is 34,
06:56so it's root to 34.
06:59So all I need to
07:00write here
07:01is, let me write this
07:06here. So, cos of theta,
07:12cos of theta equals a
07:16dot b, which is negative
07:18three, over, and you don't
07:22even have to work this
07:23out, just leave it like
07:24this, root 14 times root
07:2634, and then theta
07:29is equal to and you
07:30can just go straight to
07:31your calculator once it's open
07:34and You don't even need
07:37to work this out first.
07:39You just go straight to
07:40inverse costs so Close this
07:45add a calculator. I want
07:48trig inverse costs I'm then
07:51gonna use this function here,
07:53and I'm gonna just do
07:54negative three over
07:57Route 14, out of the
08:02route, multiplied by route 34.
08:08Am I in degrees or
08:09radians? He didn't ask. So,
08:12the vector is one time
08:13you might see degrees come
08:15up more than radians. But
08:17if he doesn't say, I
08:18can certainly give radians 1
08:20.70874, 1 .70874,
08:25eight, seven, four. I can
08:28put radians or I don't
08:29even need to. If they
08:30ask, if the whole question's
08:32in degrees, of course, change
08:33your calculator to degrees and
08:35give you answer in degrees.
08:37But that's it. That's how
08:38we find an angle between
08:40two vectors. Now, one question
08:45I often get is, well,
08:46how is there a vector,
08:48or an angle between two
08:49vectors that aren't even touching?
08:50Well, remember, if you've got
08:52two vectors, one there,
08:53and one there. Remember a
08:55vector, you can pick it
08:57up and move it. It's
08:58still the same vector. So
08:59now you've got an angle
09:01there between two vectors. And
09:04actually that reminds me. When
09:06you find the angle, so
09:07here you've got, if these
09:09two vectors are crossing, you've
09:10got two angles. You've got
09:11one there and one there.
09:12And when we do an
09:13equation of a line, it'll
09:15be very important that you
09:16differentiate between the two. But
09:17when you get the angle
09:18between two vectors, what you're
09:20finding is this angle
09:21angle, it's the angle, not
09:22the acute angle, but the
09:24angle, where the direction of
09:28the vectors is moving away
09:30from. So let me give
09:32you a separate example. Imagine
09:35I has this vector and
09:40I don't know this vector.
09:45The angle is
09:49this angle here. So it's
09:52in between the two, this
09:54moving away from here, it's
09:56in between the two, it's
09:57not this angle. Okay, last
10:00example. Let A equals this,
10:04this is from a page
10:05of the pass paper. Let
10:06A equal two k negative
10:08one and B equals this.
10:10Given that A and B
10:11are perpendicular, find the possible
10:12values of k. Now, this
10:14is a very, very important
10:15concept. The formula is
10:17cos theta equals a dot
10:20b over the magnitude of
10:22a times the magnitude of
10:24b. Now if they're perpendicular,
10:27perpendicular means right angles, 90
10:30degrees, pi over two, radians.
10:33What is the cos of
10:3590 degrees or the cos
10:36of pi over two radians?
10:37Hopefully you know it's zero.
10:39So the only way that
10:41vectors can be perpendicular is
10:43if this is zero,
10:45which means this is zero,
10:47which means this, because the
10:49numerator has to be zero.
10:51So for when you see
10:52perpendicular in vectors, and this
10:54comes up all the time,
10:55check your pass papers, you'll
10:57see, you'll see the word
10:57perpendicular in vectors questions. When
11:00it comes up, you immediately
11:01go ding, ding, a dot
11:04b equals zero. So a
11:05dot b equals zero. So
11:09I have my a now,
11:11there's a k in there,
11:12but whatever.
11:14I'm going to put my
11:16pen. I have A is
11:192k negative 1. So 2k
11:24negative 1 dot b negative
11:283k plus 2k. This has
11:32to equal 0. Now you'll
11:34see 2 times negative 3
11:36is negative 6 plus k
11:38times k plus 2.
11:41minus k equals zero. So
11:44what I've got here is
11:45a quadratic minus six plus
11:47k squared plus two k
11:50minus k equals zero. Let's
11:52bring this up here. Make
11:54it k squared plus k
11:57minus six equals zero. This
12:01is a nice quadratic that
12:02factorizes. I'm gonna go with
12:04k plus three times k
12:09minus
12:09two. Hopefully you guys can
12:11all do this now. It's
12:13just factors of k squared
12:14or k and k, factors
12:15of six are three and
12:16two and has to be
12:17plus and a minus and
12:19the big one has to
12:19be plus. So k, therefore
12:22k plus three equals zero
12:24or k minus two equals
12:26zero. k equals negative three
12:29or k equals two. Just
12:32check, go back to the
12:34question. k has to be
12:36an element of the real
12:37numbers.
12:37is negative three, a real
12:39number, yes, is k equals
12:41two is two a real
12:43number, yes, so they are
12:45our two solutions done. Okay,
12:48that's it. I know I
12:49went through that pretty quickly,
12:50but trust me, once you
12:54do a lot of vector
12:54stuff and we go further
12:56into the into the series
12:58of lessons, this will be
13:00actually very very, this will
13:01become very very familiar to
13:02you, the dot product. It's
13:06Just this times, this times,
13:08this plus this times, this
13:08plus this times this. This
13:10is also the formula, which
13:12comes from this thing. And
13:14it is how we find
13:16the angle between two vectors
13:17and very, very importantly, if
13:19they're perpendicular, the angle is
13:2290 degrees, which means the
13:24cost of the angle is
13:25zero, which means a dot
13:27b equals zero. I'm just
13:28going to write that here,
13:29because it's important. V, I
13:30go to V dot w
13:32equals zero.
13:34If perpendicular to the perpendicular,
13:41straight line. If perpendicular, v
13:46dot w is zero. That's
13:48it. See you in the
13:49next lesson.