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# IIT JEE lagrange's formula

Video transcript

If a and b are
vectors in space given by a is i minus 2j over
the square root of 5 and b is 2i plus j plus 3k
over the square root of 14, then the value of-- so we have
2a plus b dotted with-- that's a dot product there-- a
cross b cross a minus 2b is. And if you-- and I actually
tried this out a little bit before this video. If you actually just
tried to first take the cross product of a
and b, which you can do, and then try to take the
cross product of that with a minus 2b, it's going
to be unbelievably hairy, unbelievably
computationally intensive. And you're probably going
to make a careless mistake, especially when you take the
cross product with a minus 2b. It's a really, really,
really, really hairy problem. And in general, taking
cross products tend to be, but even more so when the
actual vectors look like this. So the trick of this
problem is to try to do it without actually
having to take cross products. And to do that, we're
going to use a little trick called the triple
product expansion or Lagrange's formula. And I actually just uploaded
a video proving the triple product expansion for those
of y'all who don't like-- and I'm one of those. I'm one of the y'all that
doesn't like just random formulas. I like to see
where it came from. But the triple product
expansion, it really allows us to rewrite a--
let me write it over here. It allows us to write
a cross b cross c. It allows us to write that. And I'm just speaking in general
terms of a, b, and c here, not these particular
a's and b's. It allows you to rewrite
that as the first vector. And when I talk about the first
vector and the first cross product, I'm talking
about the cross product you would have to take first. In this case, it would
be this one, where it's the cross
product on the right, I guess I should call it. Because there's
other ways that you could take the cross product
first where it's on the left. So you take this,
the first vector here, so it'll be b
scaled by a dot c-- let me scroll to the
left a little bit-- minus the second vector
in that second cross product minus c times a dot b. So when you have this kind of
triple product right over here, you're taking the cross
product of three vectors, you can simplify it. And it is a simplification
because the dot product is a much simpler operation to
compute than the cross product. So let's see how we
can apply it over here. It looks like we have a triple
product-looking thing here. We have the cross product of two
vectors crossed with a third. So it's a slightly different
thing than this over here. But what we can do is
we can swap it around so it's the same order. So this over here in green,
this is the exact same thing. If I just swap the order of
these two cross products, it'll be the negative. So this will be negative
a-- that's ugly parentheses. a minus 2b, so I'm putting
it first, cross a cross b. So all I did is I
rewrote this green part. When you swap the order
of the cross product, you put a negative out front. And actually, let me just
distribute this negative. So this is the same thing. If I multiply a
negative times this, it just swaps these two terms. So it's 2b minus
a cross a cross b. Now, the reason why I
like this form is it's exactly the form
that I have up here. It's exactly this form. And so I can use Lagrange's
formula or the triple product expansion. And so I'm just going to focus
on the green part right now. We can worry about this
thing out front that we're going to take the dot
product with later. Let's see if we can
simplify this green part. So this will be equal to
just using the triple product expansion. We're looking at our
second cross product. This is our second
cross product over here. So it's analogous to
that right over here. You take the first vector in
your second cross product. So that's the vector a. And I multiply that
times the dot product of this vector and that vector. So it's going to be 2b
minus a dotted with b, dotted with vector b. And then from that,
I'm just going to subtract the second vector. Let me do that in another color. I'm going to subtract b. So this is the same b over here. Maybe I should color code it. This a is that a there. And then this b is this b here. So b times the dot
product of-- so it's going to be other
than magenta again-- the dot product of this guy,
2b minus a dotted with a. So the dot product of
the other two vectors. Now let's see if we
can simplify this. So what we have in
parentheses here, so this over here in
green, the dot product, that's just the distributive
property applied. So this is going to be
2b dotted with b again. Or you might already know
that if I take a vector and I take the dot
product with itself, that's the same thing
as its magnitude. That's the same thing
as the magnitude of the vector squared. So let's just write that down. So 2b dot b is going to
be-- actually, let me just write it as 2b dot b. And then we have minus a dot b. So I just distributed. I didn't actually
use that notation. And all of that's going
to be times the vector a. And then we have
minus-- let me pick another color-- minus
all of this stuff. So we have 2b dot a. So let me write it this way. I'm going to just leave
the minus b out front. So let me write a minus, and
then I'll put the b over here. And so you have 2b
dot a-- that color. b dot a is the
same thing as a dot b, so I'll just write 2a dot
b, and then minus a dot a. So remember, we're just working
on this green part right here. But since we already
have all of these dot products of a with
themselves and a dot b and b with themselves,
maybe let's just see what happens if we
actually try to evaluate these, and maybe this will simplify
some of our algebra. But you wouldn't have to. You could just keep going
with the a's and the b's. But maybe these evaluate to
relatively straightforward numbers since we actually
know what our a's and b's are. So let's try that out. So what is a dot a
going to be equal to? Well, you're just
really squaring each of the components
of a and adding them up. So the first component,
the i component, is 1 over the square root of 5. So 1 over the square root of
5 squared is going to be 1/5. And then negative 2 over
the square root of 5 squared is going to be plus 4/5. So our intuition was correct. When you take the dot
product of a and a, it's a very simple number. It is 1. Let's do the same thing with b. If I take b dot
b, what do I get? So 2 over the square root
of 14 squared is 4/14. And then 1 over the square root
of 14 squared is plus 1/14. And then 3 over the
square root of 14 squared is 9/14, plus 9/14. And this, once
again, adds up to 1. This, once again, adds up to 1. So it looks like things
are simplifying a good bit. So let's just try a dot b. So that's going to be 1 over
the square root of 5 times 2 over the square root of 14. So it's going to be 2 over
the square root of 70. I just multiplied the x terms. And then we're going
to have negative 2 over the square root of 5 times
1 over the square root of 14. So that's negative 2 over
the square root of 70. And then plus 0. This has no k
component, so it's going to be 0 times this, whatever
the k component is here, but that's still going to be 0. So this just evaluates all to 0. So that simplifies
things a good bit. This green expression right
here simplifies to-- this is 2 times 1, which is
just 2, minus 0, right? a dot b we just
figured out is 0. So this is just 2
times a minus-- now what is this going
to be equal to? a dot b we just figured
out. a dot b is 0. And then a dot a we
figured out is 1. So it's minus negative, so
this becomes a positive b, so plus b. So all of this mess over
here simplified to 2a plus b, which is the
exact same thing we have over here-- 2a plus b. So let's actually-- that
we're dotting it with. So let me write it over here. So this whole thing
is this over here. And then we have to take the dot
product with 2a plus b again. So we're going to take the dot
product with 2a plus b again. It doesn't matter. a dot b
is the same thing as b dot a. And they're the same number,
so it matters even less. And here, the
distributive property applies to the
vector dot product. Let me make sure we remember. This is a dot product
that we're taking. This is a dot-- let me
put it in a darker color. This is a dot product. So this whole thing
was 2a plus b. I'm dotting it with
another 2a plus b. And now this, you just use the
distributive property here. So you have 2a dot 2a. That's going to be
4 times a dot a. And then 2a dotted
with b is going to be plus 2 times a dot b. And then b dot 2a is going to
be plus 2 b dot a or a dot b. I want to just switch
colors actually. b-- I'll do it in the
same color as the b. So b dot this is
going to be 2 a dot b. And then b dot b,
so plus b dot b. So what is this
going to evaluate to? a dot a is 1, so
it's going to be 4. a dot b is 0. And a dot b is 0. And then this is 1. So we have 4 plus 1. 4 plus 1 is equal to 5. So this whole mess, if
you're able to remember the triple product
expansion and kind of see the simplification
over here, it's actually a fairly
straightforward problem. This whole mess over here is 5. It evaluates to 5.