Resolution of Vectors into Components Video Lecture - Physics Class 11 - NEET

Voiceover:We have two vectors here.
Vector A, it has a magnitude of three
so the length of this blue arrow is three.
Its direction, it forms a 33
degree angle with the positive,
I guess you could say the positive x axis.
I haven't drawn that here
and vector B has a magnitude of two,
the length of this arrow is two
and it forms a 135 degree
angle with a positive x axis.
What I want to think about in this video
and maybe the next one depends
on how we do with the time
is what is the magnitude
and direction of the sum
of these two vectors.
What is the magnitude and
direction of the vector A plus B?
I encourage you to pause this video
and try to work this through on your own
before I work through it.
Well to think about this,
I'm going to decompose
each of these vectors
into their horizontal and
their vertical components.
For example vector A could be viewed
as the sum of this
horizontal pointing vector
plus this vertical pointing vector.
Another way of thinking
about this horizontal vector,
this right over here,
this is going to be a scaled up version
of the unit vector I
and so this is going
to be something times I
and this vector right over
here is going to be something
times the unit vector in
the vertical direction.
The vertical direction might
look something like that
or let's see if this is
three then the unit vector,
yeah, actually would
look something like that.
That's what the J unit vector looks like
and this is what the I unit
vector would look like.
This is a scaled up version of it.
It's something times I and
this is something times J.
The same exact argument right over here.
This is going to be something
times the I unit vector,
its horizontal component
and its vertical component
is going to be something
times the J unit vector.
What we really need to do
is figure out the magnitudes
of these horizontal
and vertical components
and then we know how much to scale up I
and how much to scale up J.
Let's think through this a little bit.
This one might jump out at you immediately
is this is a 30, 60, 90 triangle.
This is a 30, 60, 90 triangle
then this side right over here
is going to be half the
length of the hypotenuse.
This is going to be half of three,
so it's three over two and
this one right over here,
let me do that in same color.
This is going to be three,
that's not the same,
that's a different color.
This is going to be three over two
and then this length right over here.
This is going to be square root of three
times the shorter side.
This is three times the
square root of three over two.
I just took this and multiplied
it by the square root of three
and once again that just came
out of 30, 60, 90 triangles.
Now another way that you could tackle this
is to use your, what would you know
about your trig functions?
You say, "Okay I know
this angle right here"
"is a 30 degree angle,
this is the opposite side."
You could say, "Well,
the opposite over three"
"is going to be equal to
the sine of 30 degrees,"
let me right this down.
Soh cah toa.
Sine of 30 degrees is going to
be equal to the opposite side
over the length of the hypotenuse.
Over three or we could
say that the opposite side
is equal to three times
the sine of 30 degrees.
If you put this in your calculator,
sine of 30 degrees is one half
and so you're going to
get three halves here.
Similarly for this side,
this side is adjacent
to the 30 degree angle.
You could say cosine, cosine
is adjacent over hypotenuse.
You could say that cosine of 30 degrees
is equal to adjacent over the hypotenuse
or multiplying both sides by three.
The adjacent is going to be equal to three
times the cosine of 30 degrees
and the cosine of 30 degrees
if you type in your calculator,
you're actually going to
get some type of a decimal
but it's square root of three over two.
Three square root of three over two
and you can verify this if you like.
We see here, sine of 30 is indeed one half
and cosine of 30
well you get this kind of crazy decimal
but notice that is the same thing
as square root of three divided by two.
The exact same value right over here
but we were able to figure that out
just using what we know
about 30, 60, 90 triangles.
Now this triangle, you might
say, "Well, what's the angle?"
What angles do we know about it?
Well this angle right over here
is this is supplementary
to this 135 degree angle.
This is a 45 degree angle.
This is 90 so this is 45.
This is a 45, 45, 90 degree triangle.
Now you could say that
using the trig identities
and what we did right
over here that this length
is going to be the hypotenuse
times the cosine of 45 degrees
and you could say that
this length right over here
is going to be the hypotenuse
times the sine of 45 degrees.
Using the exact same logic here
and as you get more practice
with finding the components,
you'll realize, okay,
you take the hypotenuse
times the cosine of this angle.
You're going to get the adjacent side.
If you do the hypotenuse
times the sine of this angle,
you're going to get the opposite side
but we could use this sine of 45 degrees.
Once again if you put in your calculator,
you get a crazy decimal
but we can figure that out,
we know 45, 45, 90 triangles.
Sine of 45 degrees is
square root of two over two.
Cosine of 45 degrees is also
square root of two over two.
I know my writing is
getting a little bit messy
and actually I should have
done all of these in green,
square root of two over two.
What's the length of that green vector?
What's two times square
root of two over two,
that's going to be square root of two.
What's the length of this orange vector?
It's two times square root of two over two
so it's going to be square root of two.
Now we know the magnitudes
of the component vector
is a horizontal and vertical
components of each of these
so now we can write these out as the sum
of these horizontal and vertical vectors.
Vector A, we can write
as square root of three
or three square roots of
three over two times I.
That's this vector right over here,
we scaled up the I unit vector
by three squares root of three over two
plus three halves times J.
This right over here is
this vector right over here.
Scaled up version of the J unit vector
and if you add this orange
vector to this green vector,
you get vector A.
Similarly, vector B is equal to the length
of the horizontal component
and we got to be very careful.
It's length is square root of two
but it's going in the leftward direction.
We're going to put a
negative on it times I.
If we just squared it, I is
doing something like this,
square root of two times
I would look like that.
Negative square root of two
would point it to the left.
This is negative square
root of two times I
and then we're going to have
plus square root of two times J.
Now that we've broken them
up in their components,
we're ready to figure out what,
at least broken up into its components
what A plus B is.
A plus B is equal to, well
it's going to be the sum
of all of these things.
Let me just write that down.
It's going to be that, copy and paste.
That plus this, let me
just copy and paste it
and you're going to get that
but of course we can simplify this.
We can add the I unit
vectors to each other.
If I have three times
the square root of three
over two Is and then I have another
negative square root of two I,
I can add that together.
This is going to be equal to three
times the square root of three over two
minus square root of two times I
and then I can add this to this
and I'm going to get plus three halves
plus square root of two times J.
It looks a little bit complicated
but we could type it in to our calculator
and get approximations of
each of these two values
and we essentially have
a at least a broken down
into its components
representation of A plus B.
In the next video, we're
now going to take this
and figure out the actual
magnitude and direction
of A plus B.