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
00:19to obviously visualize. Imagine I
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.