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## www.mathcentre.ac.uk/.../Cartesian%20components%20of%20vectors.mp4

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The natural way to describe the
position of any point is to use
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Cartesian coordinates. In two
dimensions it's quite easy.
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We just have. Picture like
this and so we have an X
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axis and Y Axis.
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Origin oh where they cross and
if we want to have vectors in
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that arrangement, what we would
have is a vector I associated
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with the X axis and a vector Jay
associated with the Y axis.
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What all these vectors I&J?
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Well, they have to be unit
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vectors. A unit vector I under
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unit vector. J.
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In order to make sure that
we do know that they are
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unit vectors, we can put
little hat on the top.
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So if we have a point P.
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And let's say the coordinates of
that point are three, 4, then
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the position vector of P which
remember is that line segment
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joining oh to pee is 3 I.
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Plus four jazz.
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Notice the crucial difference.
That's a set of coordinates
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which refers to the point
that's the vector which refers
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to the position vector. So
point and position vector are
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not the same thing.
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We can write this as
a column vector 34.
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And sometimes. This is
used sometimes that one
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is used, just depends.
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What about moving then into 3
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dimensions? We've got XY
and of course the tradition
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is to use zed.
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So let's have a look. Let's draw
in our three axes.
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So then we've got XY.
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And zed.
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I'll always write zed with a bar
through it that so it doesn't
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get mixed up with two. I don't
want the letters Ed being
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confused with the number 2.
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So I've got these three axes or
at right angles to each other
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and meeting at this origin. Oh,
and of course I'm going to
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describe any point P by three
coordinates XY and Z.
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Now when I drew up this set of
axes, I indicated them.
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Quite easily.
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I could of course Interchange
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X&Y. I might choose to
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interchange Y&Z. But this is the
standard way. Why is it the
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standard way? What is it about
this that makes it the standard
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way? It's standard because it's
what we call a right?
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Hand.
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System.
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Now. How can we describe
workout? What is a right hand
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system? Take your right hand and
hold it like this.
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Middle finger. Full finger and
thumb at right angles.
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This is the X axis.
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The middle finger. This is the Y
axis, the thumb.
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Now rotate as though we were
turning in a right handed screw.
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And we rotate like that.
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And so the direction in which
we're moving this direction
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becomes zed axis. So we rotate
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from X. Why?
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And we move in the direction of
the Z axis. So right hander
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rotation as those screwing in a
screw right? Handedly notice
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that it works whatever access we
choose. So if we take this to be
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Y again, the thumb and we take
this to be zed then if we make a
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right handed rotation from why
route to zed, we will move along
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the X axis. So let's do that.
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You can see that as we rotate
it, we are moving right handedly
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along the X axis and you can try
the same for yourself in terms
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of rotating from X to zed and
moving along the Y axis. So
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that's our right handed system.
So let's have a look at that in
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terms of having a point P that's
got its three coordinates XY.
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And said
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X.
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And why?
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And said now origin, oh.
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Will take a point P anywhere
there in space. What we're
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interested in is this point P.
It's got coordinates, XY and
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zed. And its position
vector is that line segment
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OP. And so we can write down,
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Oh, P. Bar is
equal to XI.
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Plus YJ.
Plus, Zed and the unit vector
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that is in the direction of the
Zed Axis is taken to be K.
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So again, notice the
difference. These are the
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coordinates XY, zed. This is
the position vector
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coordinates and position
vector are different.
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Coordinates signify
appoint, position vector
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signifies a line segment.
We sometimes write again
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as we did with two
dimensions. We sometimes
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write this as a column
vector XY zed.
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Now there are various things we
would like to know and certain
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notation that we want to
introduce for start. What's the
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magnitude of Opie bar? What's
the length of OP? Well, let's
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drop a perpendicular down into
the XY plane there and then.
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Let's join this up.
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The axes there and
across there.
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Now let's just think what this
means this length here.
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Is the distance of the point
above the XY plane, so it must
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be of length zed.
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This length, here and here is
the same length. It's the
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distance along the X coordinate,
so that must be X.
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And that's also X. Similarly,
This is why and so that must be
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why as well.
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So if we join up from here out
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to here. What we have here is a
right angle triangle, and of
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course we've got a right angle
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triangle here as well. So this
length here. There's are right
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angle this length using
Pythagoras must be the square
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root of X squared plus Y
squared, and so because we've
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got a right angle here, if we
use Pythagoras in this triangle
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then we end up with the fact
that opie, the modulus of Opie
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Bar is the square root of. We've
got to square that and add it to
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the square of that. So that's
just X squared plus.
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Y squared plus Zed Square.
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Now I'm going to draw this
diagram again, but I'm going to
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try and miss out some of the
extra lines that we've added.
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So XY.
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Zedd.
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We'll take our point P with
position vector OP bar.
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Again.
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Drop that perpendicular down on
to the XY plane.
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Draw this in across here.
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And that in there.
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Now. This line OP
makes an angle with this
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axis here.
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It makes an angle Alpha.
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And if I draw it out so that we
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can see it. Let me call this a.
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If we draw out the triangle
so that we can actually see
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what we've got, then we've
got the line.
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From O to a.
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There. Oh, to A and we've got
this line going out here from A
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to pee and that's going to be at
right angles there like that.
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And so if we now join P2O, we
can see the angle here, Alpha.
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Now we know the length of this
line. We know that it is the
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square root of X squared plus Y
squared plus zed squared and we
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also know the length of this
line, it's X.
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And that is a right angle, and
so therefore we can write down
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cause of Alpha is equal to X
over square root of X squared
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plus Y squared plus zed squared.
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Why have we chosen this? Well,
cause Alpha is what is known
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as a direction.
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Cosine
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be cause. It is the cosine of
an angle that in some way helps
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to specify the direction of P.
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An Alpha is the angle that Opie
makes with the X axis. So of
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course what we can do for the X
axis we can do for the Y axis
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and for the Z Axis.
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So we have calls Alpha which
will be X over the square
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root of X squared plus Y
squared plus zed squared.
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Kohl's beta which will be
the angle that Opie makes with
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the Y axis, and so it will be
why over the square root of X
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squared plus Y squared plus zed
squared and cause gamma.
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Gamma is the angle that Opie
makes with the Z Axis, and so it
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will be zed over the square root
of X squared plus Y squared
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close zed square.
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So these are our direction
cosines. These are expressions
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for being able to calculate
them, but there is something
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that we can notice about them.
What happens if we square them
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and add them? So what do we
get if we take 'cause squared
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Alpha plus cause squared beta
plus cause squared gamma?
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So let's just calculate
this expression.
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Kohl's squared Alpha is going
to be X squared over
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X squared plus Y squared
plus said squared.
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Call squared beta is going to
be Y squared over X squared
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plus Y squared plus zed squared.
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And cost squared gamma is
going to be zed squared
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over X squared plus Y
squared plus zed squared.
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Now we're looking at adding all
of these three expressions
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together. Cost squared Alpha
plus cost squared beta plus cost
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squared gamma. Well, they've all
got exactly the same denominator
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X squared plus Y squared plus
said squared, so we can just add
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together X squared plus Y
squared plus 10 squared in the
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numerator. So that's X squared
plus Y squared zed squared all
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over X squared plus Y squared
plus said squared. Of course,
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that's just one.
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So the squares of the direction
cosines added together give us
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one. What possible use could
that be to us? Well, one of
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the things it does mean is
that we have the vector,
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let's say cause.
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Alpha I plus
cause beta J
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plus cause Gamma
K.
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That vector is a unit vector.
It's a unit vector because if
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we calculate its magnitude
that's cost squared Alpha plus
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cost squared beta plus cost
squared gamma is equal to 1.
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Take the square root. That's
one. So this is a unit vector.
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Further, this is X over X
squared plus Y squared plus Z
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squared Y over X squared plus
Y squared plus said squared.
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And zed over X squared plus Y
squared plus said Square and so
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it's in the same direction as
our original OP. Our original
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position vector opi bar.
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And that means that this is a
unit vector in the direction of
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OP bar and that may prove to be
quite useful later on when we
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want to look at unit vectors in
particular directions. For now,
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let's just have a look at doing
a little bit of calculation.
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Let's say we've got a point.
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That has
coordinates 102.
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Under point that has
coordinates 2 - 1.
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4.
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The question that we might ask
is if we form the vector AB.
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What's the magnitude of a bee?
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And what are its direction
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cosines? We just have a
look at this. Let's
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remember that, oh, a bar.
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Is.
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I.
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No JS.
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And two K's.
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That OB bar.
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Will be. Two I.
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Minus one
J plus 4K.
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We want to know what's the
magnitude of the vector AB bar.
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Just draw quick picture just to
remind ourselves of how to get
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there. There's A and its
position vector with respect to.
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Oh there's B with its position
vector with respect to. If we're
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wanting a baby that's from there
to there and so we can see that
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by going from A to B, we can go
round AO plus OB.
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And so therefore, that is OB bar
minus Oh, a bar. So that's what
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we need to do here. A bar must
be OB bar minus oh, a bar.
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And all we do to do the
subtraction is what you would
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do naturally, which is to
subtract the respective bits
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so it's two I take away I.
That's just an eye bar.
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Minus J takeaway no
JS, so that's minus J
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Bar and 4K takeaway
2K. That's plus 2K.
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So now we have our vector AB
bar. We can calculate its
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magnitude AB modulus of a bar
that's just a be the length from
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A to B, and that's the square
root of 1 squared plus minus one
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squared +2 squared altogether.
That's 1 + 1 + 4 square
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root of 6, and the direction
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cosines. Our cause Alpha.
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That's The X coordinate
over the modulus, so that's
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one over Route 6.
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Kohl's beta that's minus
one over Route 6. the Y
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coordinate over the
modulus and cause gamma.
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The zed coordinate
over the modulus.
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Now this is a fairly standard
calculation. The sort of
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calculation that it will be
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expected. You'll be able to do
and simply be able to work your
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way through it very quickly.
Very, very easily, so you have
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to be able to practice some of
these. You have to be able to
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work with it very rapidly, very,
very easily, but always keep
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this diagram in mind.
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That to get from A to B to form
the vector AB bar, you go a
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obarr plus Obiba and so.
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It's the result, so to form a B
it's Obi bar, take away OA bar.
Title:
www.mathcentre.ac.uk/.../Cartesian%20components%20of%20vectors.mp4
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