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# IIT JEE position vectors

Video transcript

Let P, Q, R, and S be the points
on the plane with position vectors negative 2i minus j,
4i, 3i plus 3j, and negative 3i plus 2j, respectively. The quadrilateral
PQRS must be a-- So let's just graph each
of these positions vectors, or graph the points
that they're specifying. So let me draw my coordinate
axes right over here. So this is my vertical. I can draw a
straighter line that. There you go. That's about as
good as I can do. Let me draw the horizontal. That's a better job, I think. And then let me mark it off. Let's see. We go as high as 4. We go up to 4i. So 1, 2, 3, 4. We goes as low was negative 2i. So 1, 2. Just making sure I
make it a little bit even, because we're going
to have to figure out the shape of whatever we draw. Then we're going to go up 3. 1, 2, 3. And we actually go
down 1, as well. So I think that'll give us-- and
we could keep marking them off if we wanted to, but
we don't need to, based on the positions
that have been specified. So let's draw these vectors. So let's draw vector P first. Negative 2i minus j. So negative 2i, i is
along the horizontal, or as you can imagine,
it's the x-coordinate. So negative 2i minus j. So it is this position
vector right over here. If we put it in
standard form, starting with its base at the origin. It's that position vector. But it's specifying
that point over there, which is the coordinate
negative 2, negative 1. Let me just write it down. Negative 2, negative 1. Now let's do 4i. So 4i is literally-- that's
just 4 in the x direction. 1, 2, 3, 4. There's no j component,
no vertical component. So it specifies that
position right over there, or it specifies the point 4, 0. Then we have 3i plus 3j. So 1, 2, 3. 1, 2, 3. That is that point right there. The position vector
specifies that point, or specifies the point 3, 3. And then finally,
we have negative 3i. Let me do this in another color. Negative 3i. So 1, 2, 3, plus 2j, 1, 2. So right over there. So that is that last
position vector, and it's specifying the
point negative 3, 2. And they want to know,
what's the quadrilateral made by these four positions? So let's draw it. Let's connect the dots. So this is P, go to Q. It gets
us that line right over there. Then from Q to R, we're
going to go like that. And then from R to S, we're
going to go like that. Actually, it's going to
go a little bit more-- it's going to go like that. Sorry, that's from R
to S. And then S to P is going to be like that. So just eyeballing it,
it's clearly not a square. It's clearly not a rectangle. It's coming off at
different angles. It's clearly not a rhombus. A rhombus would have
all of the same sides. So a square would
be all 90 degrees, would be all the same sides and
all the angles are 90 degrees. That would be a square. It's clearly not a square. A rectangle would be all
the angles are 90 degrees, but two sides are
going to be the same, and the other two sides
are going to be the same. These do not look
like 90 degree angles. And you could try if you want. I mean, just eyeballing
it, it's pretty clear these aren't
90 degree angles. But if you wanted to, you
could find the slope of this and find the slope of that
line, and if they're not the negative inverse
of each other, then they're not
perpendicular, and they're not coming at 90 degree angles. And so the last one
is are they a rhombus. A rhombus is a parallelogram
while the sides are the same. So a rhombus would
look like that. Clearly not a rhombus. These sides are
clearly much longer than those sides over there. Now just from
deductive reasoning, it looks like the answer
is A. it's a parallelogram. But we can verify it. In order for this to
be a parallelogram, these two sides
have to be parallel, and these two sides
have to be parallel. Or another way to
think of it, they would have to have
the same slope. So we could verify it. If we were taking this on
exam under time pressure, we would just cross these
out, and it looks pretty good. We would just go with A.
But let's just verify it. So what's the slope down here? So it's change in
y over change in x. So 0 minus negative 1,
over 4 minus negative 2, or this is equal to 1
over 4 minus negative 2, is positive 6. This is 1/6. Or another way to think
about it, we ran 6 and we only rised 1. This is 1/6. Now what happens over here? Well, it's the same thing. We're going from
negative 3 to three. So we're running 6. So we say 3 minus
negative 3 is equal to 6. And how much did we rise? We have 3 minus 2,
which is equal to 1. So this is also a slope of 1/6. So these two lines
are clearly parallel. And now let's do it
for these other two. What's the slope over here? And actually, you could
just do rise over run. So that's exactly what
we were doing before, but just to give you the
taste for everything. So we ran, to go
from here to here, we have to move to the right 1. So we run 1. And how much do we rise? Well, we had to go down 3. So it's negative 3. So the slope here is negative 3. And what about here? To go from this point to this
point, we went to the right 1. So we ran 1. And we had to go
down-- we started at 2, and we ended up at negative 1. We had to go down 3. So the slope here
is also negative 3. So these lines
are also parallel. So we're definitely dealing
with a parallelogram, which is neither a rhombus
nor a rectangle.