Triple Product

Learning goals: one last thing to introduce on the way to a coordinate formula for cross products.

3 Staying in R for a moment, lets say that we have three vectors a = (a1, a2, a3), b = (b1, b2, b3) and c = (c1, c2, c3). By dragging each to start at the other head in every combination, we would get a box with angled sides. This is called a parallelepiped.

Definition: the triple product of a, b, and c (in that order!) is the volume of this

a1 a2 a3

box. It is denoted by b1 b2 b3 . (There isn’t a more compact notation.) c c c 1 2 3

Now it’s easy to find the volume of such a shape—it is simply base times height! Let’s temporarily let v = a × b, and look at the figure edge-on along the plane of the vectors a and b. Note that v will be perpendicular to this plane. �

From this perspective, it is easy to calculate the height of the θ parallelepiped—it is ||c||cos(θ) from simple trig. So since ||v|| is the � area of the base, the volume of the parallelepiped will be ||v||||c||cos(θ). From the picture we can see that this is v⋅c. In other words, the triple product is equal to the quantity (a × b) ⋅ c.

But what if c is on the opposite side of the plane from a × b? Then the angle will be larger then 90° and we would have a negative volume. Is that OK? Sure! We will say that we are working with signed volume, and volume is positive if the order of the vectors is right-handed, while it is negative if the vectors are given in a left-handed order.

This is no different than signed area where if we go around a region counterclockwise we get positive area, while clockwise gives us a negative area.

Note that we can rotate the order of the vectors from a, b, c to c, a, b or b, c, a and the right- or left-handedness will not change, nor will the volume of course. So we have just shown that the triple products (a × b) ⋅ c = (b × c) ⋅ a = (c × a) ⋅ b. We can, of course, also use commutativity of the dot product. For instance the middle terms would also be equal to a ⋅ (b × c). So as long as we don’t switch the order of the letters, we can switch the order of the dot and cross.

Now we don’t have a formula for the cross product in terms of coordinates, so how can we figure out the triple product? It turns out this is easier to figure out than the cross product, so we’ll do it first and then use it to find the cross product’s formula.

a1 a2 a3

Rules for computing b1 b2 b3 : c c c 1 2 3 1) Swapping two rows changes the sign of the triple product. This is because in one of those variations of dots and crosses, the two rows that you swap are being crossed together. When you swap the rows, you have to use anticommutativity of the cross product to see that the sign of the result now changes. Geometrically, we are putting the vectors in a different order, and the three will switch from left- to right-handed arrangements or vice versa.

a1 a2 a3 a1 a2 a3

2) You can factor a number out of a row: b1 b2 b3 = k b1 b2 b3 and the same goes kc kc kc c c c 1 2 3 1 2 3 for the other two rows. Geometrically, all we are doing is stretching one dimension of the box, so we should stretch the volume by the same factor (which could be negative, which would switch the handedness of the system!). Algebraically, we can arrange the computation of dots and crosses so that we are dotting by the row being multiplied, and we know that constants factor out of dot products.

3) You may add a multiple of one row to another and this does not change the value of the triple product. Here is what is happening geometrically when we add a multiple of a to b:

Clearly, using this as the base won’t change the area of the base, and the height of the parallelepiped is unaffected, so the triple product remains the same. You can’t change the sign because since you are sliding b in a direction parallel to a it can never end up on the other side of a which is what you would need to reverse the handedness.

From the algebraic point of view, let’s say that we add a multiple of a to b. Then one order of terms is that the triple product equals (c × a) ⋅ b. But when we’ve added a multiple of a to b we get (c × a) ⋅ (b + ka) = (c × a) ⋅ b + (c × a) ⋅ ka by distributitivity of dot product. But the second piece of this sum is zero because c × a is perpendicular to any multiple of a.

Now we use these operations to simplify the matrix until we can easily recognize the triple product. The process is easier to explain with an example.

2 1 −1 Example: compute the triple product 3 4 2 . First, we’ll use rule two to factor a two out 1 2 1

2 1 −1 1 1/ 2 −1/ 2 of the first row: 3 4 2 = 2 3 4 2 . Now we’ll add the multiple –3 times the 1 2 1 1 2 1

1 1/ 2 −1/ 2 1 1/ 2 −1/ 2 first row to the second row to obtain 2 3 4 2 = 2 0 5 / 2 7 / 2 . We then 1 2 1 1 2 1 eliminate the bottom-left entry by adding –1 times the first row to the last to obtain 1 1/ 2 −1/ 2 2 0 5 / 2 7 / 2 . We factor ½ out of each row, and then add (–3/5) of the second row to the 0 3/ 2 3/ 2

1 1/ 2 −1/ 2 1 1/ 2 −1/ 2 1 1/ 2 −1/ 2 1 1 1 third: 2 0 5 / 2 7 / 2 = 2⋅ ⋅ 0 5 7 = 0 5 7 . Factor out the 2 2 2 0 3/ 2 3/ 2 0 3 3 0 0 −6 / 5

1 1/ 2 −1/ 2 1 1/ 2 −1/ 2 1 1 −6 entries in each row: 0 5 7 = ⋅5⋅ 0 1 7 / 5 . Now the elimination can 2 2 5 0 0 −6 / 5 0 0 1

1 0 0 continue upward, until we are left with −3 0 1 0 . The nifty thing is, we recognize right 0 0 1 away that this is the triple product of the vectors that form the unit box, so this triple product is one! The final result is that the original triple product is –3.

Note that we are free to do the steps in any order, so if you see something that will make the computations easier, feel free. For instance, we could have swapped the first and last rows to start, avoiding the ½’s that appear all over the place.

Another thing is that once you have eliminated everything downward, you can stop—you already know what the final numbers will be.

Finally, if you end up completely eliminating one of the rows, the final answer must be zero, because you have shifted things around to notice that you have a flat parallelepiped, whose volume is zero, of course.

Now you might have noticed that the notation we are using looks like the determinant of a matrix. That’s because it is the determinant of a matrix. You may have been taught a technique for finding determinants that involves reading across a row or down a column, crossing out stuff, taking the determinant of what’s left, etc. That is actually the hard way of doing it. The technique we have just discussed is the correct way to find determinants. And it works for any size matrix. In two dimensions it finds the area of a parallelogram whose sides are the two vectors given by the rows of the matrix. In more dimension, you get the higher-dimensional analog of volume.

There are tricks for finding the determinants of small (2 × 2 and 3 × 3 matrices) but be warned not to try anything similar on larger matrices.

2 5 For 2 × 2, you multiply down each diagonal and then subtract: to find we multiply 6 9 down the diagonal 2⋅9 = 18. Then we multiply down the backwards diagonal 5⋅6 = 30. Subtract the backward from the forward to get 18 – 30 = –12. We check: 2 5 2 5 1 5 / 2 → → 2(−6) → −12 . 6 9 0 −6 0 1

For a 3 × 3 we have to duplicate the first two columns, and then multiply down each of the three forward diagonals, and subtract the products of the three backward diagonals. Trying this with 2 1 −1 2 1 our previous example matrix, we get 3 4 2 3 4 . The forward diagonals multiply to 8, 1 2 1 1 2

2, and –6 (adding to 4) and the backward diagonals to –4, 8, and 3 (total: 7). Well, 4 – 7 = –3.