The Vector Triple Product and Identities

Transcript

0:00Introduction

Hello everyone! Here is a small question. For numbers, two times three is six, and six times four is twenty four. Or three times four is twelve, and two times twelve is again twenty four. So the brackets do not matter, right? Now, does that still hold for the cross product of vectors? Today we find out, and build the vector triple product formula. It comes up often in JEE vector problems. So, let's begin.

0:29Do the brackets matter for the cross product?

Let us test it with i hat, j hat, and j hat again; the little caps mark unit vectors along the axes, and the cross sign is the cross product. First, i hat cross j hat, inside the brackets, gives k hat. So, what is k hat cross j hat? It is minus i hat, yes, exactly! Going from k to j is against the cyclic order, so we get a minus sign. Okay? Now move the brackets. Any vector crossed with itself gives the zero vector, the zero with an arrow on top. So j hat cross j hat is the zero vector, and so is i hat cross it. Right? Minus i hat is not equal to the zero vector; that is the equals sign with a slash. So the brackets really matter: the cross product is not associative. Clear?

1:22The vector triple product

Now take three vectors a, b and c; the small arrows on top mark vectors. Inside the bracket, b cross c is perpendicular to both b and c, so it stands straight out of their plane. With me? Now cross a with it, from the left. Let me ask you this: which two vectors is the new vector perpendicular to? It is perpendicular to a, and to b cross c, yes! Every cross product is perpendicular to both of its vectors. Okay? But b cross c stands straight out of the plane of b and c. So anything perpendicular to it lies flat in that plane, and is built out of b and c. Got it? And here is exactly how. The vector a cross, b cross c, equals a dot c, times b, minus a dot b, times c. The dot is the dot product, so these are just numbers. Right? To remember it, say outer dot far, times near, minus outer dot near, times far. The outer vector is a, outside the bracket; inside, b is near and c is far. So a dot c, times b, minus a dot b, times c. Clear?

2:34Move the brackets, change the answer

Now here is a trap. Put the bracket on the left: a cross b first, then cross c. Can we copy the same pattern straight away? No, careful here, because the single vector c now sits on the right. Swapping the order of a cross product changes its sign, so this is minus c cross, a cross b. With me? Now c is outer, a is near and b is far. Use the rule, keep the minus sign, and we get a dot c, times b, minus b dot c, times a, so this one is built from a and b. Watch the brackets. Okay? There is one special case, when the outer vector repeats. The vector a cross, a cross b, equals a dot b, times a, minus mod a squared, times b. The vertical bars are the modulus, and a dot a is mod a squared. Got it?

3:28Two cross products together

Now take four vectors, a, b, c and d. The dot product of a cross b with c cross d equals a dot c, times b dot d, minus a dot d, times b dot c. So a pairs with c and b with d, minus the crossed pairs. Right? And their cross product? Treat a cross b as one outer vector, and use our formula. We get a b d in square brackets, times c, minus a b c in square brackets, times d. The square brackets are the scalar triple product, just a number. With me? Treat c cross d as the one vector instead, and we get it in terms of b and a, okay? It is a c d in square brackets, times b, minus b c d in square brackets, times a, which is the same vector again. Got it? Now here is one last beauty. Take a cross, b cross c, shift the letters round in a cycle, and add all three. What do you get? We get the zero vector! Expand each with outer dot far, and the six terms cancel in pairs. This is the Jacobi identity. Clear?

4:40Example 1: a cross (b cross c) by the formula, then directly

Now, example one, the easier one. The vector a is i hat plus j hat plus k hat, b is i hat minus j hat, and c is j hat plus k hat. Find a cross, b cross c, by the formula, and then directly. Pause the video and try it first. Done? Okay, let's check. The outer vector is a, b is near and c is far. So a dot c is zero plus one plus one, which is two, and a dot b is one minus one plus zero, which is zero. Right? So the answer is two times b, minus zero times c. That is two i hat minus two j hat. Okay? Now the long way, to be sure. The cross product b cross c is minus one, minus one, one; the numbers in round brackets are the components. Will a cross this give the same answer? Let us see. One, one, one, cross minus one, minus one, one, gives two, minus two, zero. Yes, that is two i hat minus two j hat again, a multiple of b, in the plane of b and c. Clear?

5:55Example 2: angles that unit vector a makes with b and c

Now a harder one, at JEE level. Here, a, b and c are unit vectors, and a cross, b cross c, equals one over root two, times b plus c. Also, b and c are not parallel. Find the angles that a makes with b and with c. Pause the video and try it first. Done? Alright, let's solve it. Open the left side with outer dot far: a dot c, times b, minus a dot b, times c. The right side is one over root two times b, plus one over root two times c. Okay? Since b and c are not parallel, neither is a multiple of the other. So the two sides match only when the coefficients of b and of c match. That is why the question says non parallel. Got it? Matching the coefficients of b, a dot c is one over root two. And matching the coefficients of c, a dot b is also one over root two, right? Oh no, careful here, there is a minus sign in front of a dot b. Minus a dot b is one over root two, and the double arrow means implies, so a dot b equals minus one over root two. With me? Since all three are unit vectors, each dot product is just the cosine of the angle between them. Clear? The cosine with b is minus one over root two, so that angle is three pi over four. With c, the cosine is one over root two, so that angle is pi over four. Here pi is the Greek letter pi, and both angles are in radians. Miss that minus sign, and the angle with b goes wrong. Okay?

7:39Quick revision: formula, brackets matter, match coefficients

So, quick revision. First, a cross, b cross c, equals a dot c, times b, minus a dot b, times c, and it lies in the plane of b and c. Second, brackets matter: a cross b, cross c, is a dot c, times b, minus b dot c, times a, so the answer changes. Third, compare the coefficients of b and c only when they are non parallel. Now go and try the practice problems on this page. All the best!