00:00Hi guys, so the next
00:01lesson we're going to do
00:02here is find the intersection
00:04between two planes. So how
00:07do they meet? Well, look,
00:08here's a nice picture, two
00:09planes. They meet at a
00:13straight line. So this, this
00:15black line is a straight
00:17line that goes through with
00:18these two vectors meet. A
00:20nice simple way to think
00:21of it is a book.
00:23Imagine you have a book
00:23and you put it when
00:24you open it up and
00:26you kind of open up
00:27two pages.
00:28Well the two pages come
00:29down and they meet together
00:30at the kind of whatever
00:32the name of that part
00:32of the book is and
00:34the middle of the book
00:34and it's like a straight
00:35line. So we're trying to
00:37find, when it says find
00:39the vector equation of the
00:40line of intersection of the
00:41planes with equations this and
00:42this. We've got two planes
00:44and we're trying to find
00:45the equation of this line
00:46where they intersect. That's it.
00:50Can two planes not intersect?
00:52Well yes, they can. If
00:54they are parallel and a
00:55nice easy way to
00:56see if they are parallel
00:57is to look at their
00:59normals. And if their normals
01:00are parallel, then the vectors
01:01are parallel. These two vectors
01:03are these two normals are
01:05not parallel because you can
01:07quickly see it. This guy
01:08is this three, one negative
01:09one. And this guy is
01:10one negative two, four. So
01:12they're definitely not multiples of
01:15each other and therefore not
01:16parallel. OK, I'm going to
01:19show you two ways to
01:20do this. The first way
01:21is the easy way with
01:23a calculator.
01:24So we can solve, we
01:27have two equations. So this
01:28is like a system of
01:29linear equations. So I'm going
01:30to go to algebra, solve
01:31system of linear equations. I
01:33have two equations, but I
01:34have three unknowns. So you
01:36may ask, well, how on
01:37earth are we going to
01:39do this? I thought we
01:41needed three equations to solve
01:43three unknowns. Well, yes, but
01:46we are going to, we're
01:49not going to find a
01:51point where
01:52going to find a line.
01:54So let me just show
01:55you what the calculator gives
01:56us. So x minus 2i
01:58plus 4z equals negative 5
02:04enter. OK. What on earth
02:07is this? Well, this is
02:09actually, it's given this c2
02:12in bold. Usually it's c1.
02:13I guess if you see
02:15one already for some reason.
02:17So this c2 is their
02:19parameter. So remember
02:20Remember, we have the parametric
02:22form of the vector equation
02:23of a line. This is
02:25the parametric form. It is
02:28essentially, so a seventh, one
02:31seventh minus one seventh minus
02:34two seventh T. So, X
02:37is equal to one seventh
02:40minus two seventh T. I
02:43certainly prefer T to C2.
02:46So, use T or lambda.
02:47This guy is,
02:48They've swapped it around only
02:50because it's positive and minus
02:51but this is 18 over
02:547 plus 13 over 7
02:57T. So y is equal
02:59to 18 over 7 plus
03:0313 over 7 T and
03:08then z is just equal
03:10to T. So that's you'll
03:14see why z is equal
03:15to t and
03:16Second, we actually choose it,
03:18let it equal to t,
03:20because we need to bring
03:21in a parameter, because we
03:22can't find a solution. We
03:25can't find a single solution,
03:27because we have three unknowns
03:29and two equations. Okay, so
03:32that's the easy way to
03:33do it with a calculator.
03:34Without a calculator, what do
03:37we do? Well, we basically
03:39need to start trying to
03:41eliminate variables. So let's try
03:43and eliminate variables.
03:44The nicest thing I can
03:46see to do is let's
03:48multiply this one by two,
03:51leave this one and then
03:51our y's will cancel nicest.
03:52I'm going to multiply this
03:53one by two. Give me
03:56six x plus two y
03:58minus two z equals six.
04:03And then this one, x
04:04minus two y plus four
04:07z equals negative five. Then
04:11I'm going to eliminate
04:12the y's. So I'm going
04:15to add the two equations.
04:16So this becomes 7x. Add
04:18this plus 2z at this
04:22equals 1. So I have
04:247x plus 2z equal 1.
04:27Now I'm going to write
04:28x in terms of z.
04:29So I'm going to say
04:317x is equal to 1
04:33minus 2z. And then x
04:37is actually equal to 1
04:39over 7.
04:40minus two over seven. Z.
04:46Okay, at this stage I'm
04:48gonna say let, and now
04:52remember this guys, is nothing
04:55to do with this one
04:56because I've done this with
04:58my calculator. I'm now assuming
05:00I don't have a calculator.
05:02So I'm gonna say let
05:03Z equal T. We have
05:08to bring in this
05:08parameter at this point because
05:11we're not going to be
05:13able to solve it. We
05:14have to let's say it's
05:16a bit like saying, okay,
05:17let's z equal anything. So
05:19z is just something and
05:21we'll get x and y
05:23in terms of whatever that's
05:24something is. That's what a
05:25parametric equation is. So I'm
05:27going to say let's say
05:28it equal t. So x
05:31is now equal to a
05:32seventh minus two seventh t.
05:36And that's, look, actually have
05:38it already. That's x, a
05:40seventh minus two seventh t,
05:42and z is equal to
05:43t. All I need now
05:44is to get all I
05:45need to find is y.
05:47So let's choose this equation,
05:49I think. I'm just going
05:51to put a line down
05:52here, so there's no confusion.
05:56Okay, so I can now
05:58say, I'm going to, let's
05:59say, sub into equation y.
06:04So I can say 3x,
06:07which is 3 times this
06:091 7th minus 2 7th
06:12t plus y plus y
06:17minus z, which is t
06:19minus t is equal to
06:223. Now I just need
06:24to find y in terms
06:26of t. So this is
06:273 7ths minus 6 7ths.
06:33t plus y minus t
06:38equals 3. And then finally
06:41let's just put y over
06:42here. Y is equal to
06:44I've 3 minus 3 7th.
06:49So that's 21 7th minus
06:503 7th is 18 7th.
06:54And then I have so
06:57on the left hand side
06:58I've negative 6 7th t
07:00plus 7th.
07:017th T which is 13,
07:037th T are negative, negative
07:0513, 7th T and then
07:07add it to the, our
07:09report size, giving me plus
07:1113, 7th T. And I'm
07:16left with what I have
07:17my X, I have my
07:18Y, I have my Z,
07:19we wanted in, we wanted
07:22to find the vector equation
07:23of the line. So actually
07:24I probably should have set
07:25it at this point. The
07:27final answer we want to
07:28write it in the
07:29form or therefore, or equals
07:34x is, we're gonna write
07:38it like this, it's 1
07:391 7th, y is 18
07:427th, and z is 0.
07:49I don't like what I'm
07:50about to do here, so
07:51let me move, I have
07:52to move all this down.
07:55And this is in a
07:56vector.
07:57like this, this is my
07:59position vector of the point
08:01on the line. And then
08:03this is plus t times,
08:07and I have negative 2
08:10-7th t, negative 2 -7th
08:13t, that's, there's my t,
08:1613 -7th and then z
08:21is just t, so this
08:22is just 1. So this
08:25is my vector equation of
08:27the line of intersection of
08:30these two planes. And that's
08:33it. Without a calculator, with
08:35a calculator. There is one
08:37other nice way to do
08:38this, an easier way to
08:41do it if you know
08:42a point on the line.
08:44Now you could arguably find
08:46a point on this line
08:47by just if you let
08:48z equals zero and then
08:50find x and y.
08:53by solving the two simultaneous
08:56equations with two unknowns, you
08:57could actually find a point.
09:01And then, so that's the
09:03point on the line, and
09:05then the direction of the
09:07line is actually equal to,
09:09and this is something you
09:10need to know anyway. The
09:11direction of that line is
09:14equal to the cross product
09:17of the normals of the
09:19two planes. Now this is
09:21a little bit hard to
09:23see, but imagine this plane
09:25has a normal coming straight
09:27up out of the ground
09:28like this, and this guy's
09:30normal is coming straight out
09:32like this. Now imagine this
09:34in three dimensions. What's happening
09:37is that, let's say this
09:40is my straight line here.
09:45So, if you can think
09:47about it, this guy's going,
09:49Actually, let me get rid
09:50of that. This guy's going
09:52straight up. This guy's going
09:53straight across. The line is
09:55actually perpendicular to both of
09:57these. And again, if you
09:59take out a book, get
10:01two pencils and make the
10:03pencils perpendicular to the two
10:04pages, you'll see that the
10:06line goes. It's perpendicular to
10:11this normal and this normal
10:13hence, the direction of the
10:16line, the direction
10:17the direction is equal to
10:24the cross product of n1,
10:27n1 cross n2 gives me
10:31the direction of the line.
10:32So if you have a
10:33point on the line or
10:34if you can easily find
10:34one, for example if this
10:36is 0 and this is
10:370, you know the point
10:390, 0, 0 is on
10:40the line, then you can
10:42easily get the direction of
10:44the line by getting the
10:45cross product
10:45the two normals and then
10:47you have the line. It's
10:48point plus a t times
10:49direction. Okay, that's it. Hope
10:54that made sense. That's one
10:55of the more confusing lessons
10:56for sure. But yeah, I
10:59hope you got it and
11:01see you in the next
11:02video.