IIT JEE perpendicular planes (part 1) Video Lecture - Class 1

We define the equation
of the plane containing
the straight line-- They give us
the equation right over here--
And this plane has to be
perpendicular to the plane
containing the straight lines.
This thing, right over
here, x/3 is equal to y/4,
is equal to z/2.
And also it has the line x/4 is
equal to y/2, is equal to z/3.
Now just to visualize
this properly,
let me draw the planes.
The first plane in
question that we
need to find the equation
of-- Let me draw it.
So we need to find the
equation of this plane.
And I'm going to try to do
this whole problem using vector
notation, because
it's easier in my mind
to manipulate vectors in a way
that you can find perpendicular
vectors and
equations for planes.
There's other ways that
you could do this problem.
So we've defined the equation
of this plane right here.
We need to find the equation.
Now we know this plane
contains this line,
and this line, if we
look at right over here,
it contains the point 0, 0, 0.
If x and y and z are
equal to 0, then all sides
of these equations are 0.
So it contains
the point 0, 0, 0.
Or you could even view it as it
contains the position vector 0,
0, 0.
And it also contains
the point 2, 3, 4.
Because if x is 2, y is 3,
and z is 4, then all of these
are going to be equal to 1.
So let me draw that
right over here.
So it also contains the
point, or the position vector,
we could say 2, 3, 4.
So if we wanted to
just draw this line,
to point specify the line,
this line would look like that.
But if we wanted to
think about a vector
that this plane
contains, it would
be the vector that goes
from 0, 0, 0 to 2, 3, 4.
So it would be this
vector right here, which
really is the position
vector 2, 3, 4.
And I'm going to write
an ijk notation just
to make it clear that
this is a vector.
So this vector that lies
in the plane in question
is the vector 2i-- I'll write it
over here-- 2i plus 3j plus 4k.
I just wrote in unit
vector notation.
I'll say a, I'll
call this vector a.
Vector a is equal to that.
And I'll write a over here.
Now we need to figure out
the equation of this line.
And the best way to do
it, or at least the way
my brain is telling
me to do it, if I
can find another
vector in this plane,
then I can find a normal
vector to this plane.
And then using a normal vector
and some arbitrary other vector
on this plane, I can then figure
out the equation of the line.
Because the normal vector dotted
that arbitrary other vector
has to be equal to 0.
Now if none of that
makes any sense, maybe
as we go through the problem
it will make more sense.
So what I'm going to do is use
these other lines, or really
the other vectors,
that are implicitly
defined by these
lines to figure out
another vector in
this plane over here.
So let me draw what I'm talking
about, that might be easier.
So let me draw this other plane.
And I'll draw it in white.
And this other plane
also contains 0, 0, 0.
Or actually, both of these
lines over here contain 0, 0, 0.
So let me draw it over here.
So this other plane, I
think what I just said,
if I do it step by step, it
will make a lot more sense.
So the other plane, that's
when it comes vertically
above this plane, and it goes
vertically below the plane
as well.
And we see the
plane that contains
these two lines is
perpendicular to this.
They're perpendicular
to each other,
and we need to find the
equation of the yellow plane.
Now this blue line over here
contains the point 0, 0, 0
and the point 3, 4, 2.
Same logic that we've
got to get to this point.
If this is 3, this is 4,
this is 2, they all equal 1.
The equation holds.
So let me just draw
some arbitrary place.
So 3, 4, 2.
And I want to be
clear, I haven't even
defined the coordinate axis.
We don't know what angle
we're looking at this from.
This is just for
visualization purposes.
So this line right over
here, I could draw it,
would look something like this.
It would look
something like that.
It sits on that white plane.
We can do the same thing
for this magenta line.
The magenta line also
contains 0, 0, 0,
but it contains the
point 4, 2, and 3.
So let me just
draw it over here.
So that's the magenta line.
And you wouldn't even see
it if this was transparent.
You would see the line, it would
just keep going down like that.
Now, I'm going to
attempt to explain
how I'm going to tackle
this problem again.
I'm going to use vectors
defined, that are implicitly
defined, by this blue line
and this magenta line.
I'm going to take the cross
product of those two vectors
to find a vector that is normal
to this entire white plane.
And that vector has to sit
in this perpendicular plane.
And then I can use that vector
and vector a to figure out
a normal vector for this plane.
And then I'll be able to figure
out the equation of the plane.
Let me just do it, instead
of keep trying to--
So what's a vector?
What are two vectors that
sit in this white plane?
Well, the point 0,0, 0, or the
position vector is 0, 0, 0.
And the position vector
4,2,3 is in this plane.
So the vector, I'll
call that b, the vector
b is equal to 4i plus 2j plus
3k is also in this white plane.
And you really, I'm just
subtracting this position
vector from that position vector
to get this position vector,
which clearly sits in the plane.
This vector starts at the
origin goes to 4, 2, 3.
This plane contains
the origin is clearly
sitting in that plane.
And by that same logic I
could figure out a vector.
It's actually the
position vector
itself sits on this plane.
So this is the vector
I'll call that c
is equal to 3i plus 4j plus 2k.
The fact that everything
is at the origin kind of
simplifies it.
The position vectors
themselves sit on the plane.
If this wasn't
centered at the origin
we wouldn't be able to do that.
We would have to subtract
between these two vectors,
or between two other
points to actually get
vectors on the plane.
But with that out
of the way, I could
take the cross
product of b and c,
and that will give
us some vector that
is perpendicular
to both b and c.
Which means that
it's perpendicular,
or it's normal to this
entire white plane, which
means it must sit in
this yellow plane.
So I need to find,
I want to find,
a normal vector that's in the
yellow plane, just like that.
Let's call that vector d.
So the vector d is going
to be equal to this.
And so the cross product of b
and c, this is how I remember,
is the determinant
of you take ijk.
I'm going to explain
this in a lot more depth
in the linear algebra playlist.
But it's determinate of that,
The next row is vector b.
So it's 4, 2 and 3.
And then the next
row is the vector c.
So 3, 4, and 2.
And so this is going to be
equal to-- let's think about it.
So for i you cross out
that column, that row,
it's 2 times 2 minus 3 times 4.
So that's 4 minus 12.
So it's negative 8.
Negative 8 times
i minus-- remember
you have to swap signs.
You go plus, minus, plus when
you're taking the determinant--
so you ignore the j,
so it's 4 times 2.
4 times 2 is 8, minus 3
times 3, which is negative 1.
So this would be
a negative 1 here,
but I already have a
negative out there.
So it's going to be plus
1j and then finally, the k.
You cross out that row, that
column, 4 times 4 is 16.
16 minus 6 is 10.
So plus 10.
Let me make sure
I got that right.
So the first one, I
always have to make sure.
So i got 4 minus
12 is negative 8.
Then j, cross that out, 4 times
2 is 8, minus 9 is negative 1.
But the j has a negative,
so that's right.
And then k, 4 times 4 is 16.
16 minus 6 is 10.
So that looks right.
So plus 10k.
So this vector right here,
this is the vector d.
This is the vector d.
I just figured it out.
Now, if we take the cross
product of d and vector a,
the vector that
was-- you can kind of
view it as lying along
this line over here--
then we will get a vector
that is normal to this plane.
We could call that
vector n for normal.
That is normal to
the yellow plane.
And it would actually
sit in the white plane.
So let's figure that.
We'll see in a few
steps, probably
in the next video, why
that is actually useful.
So let's find what d cross-- so
if I take the cross product-- d
cross a is going to be
equal to our normal vector.
Which is equal to
the same thing, ijk.
We're going to take d,
which is negative 8, 1, 10.
And take the cross of that and
the vector a, which is 2, 3, 4,
and we will get--
and actually I'll
continue this in the next video.
Just so that I don't make
these videos too long.