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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.