# Scalar triple product - why equivalent to determinant?

I'm looking at the scalar triple product and I'm wondering: is there any demonstration (possibly a simple one) that

$$\mathbf{a} \cdot \left(\mathbf{b} \times \mathbf{c} \right)= \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix}$$

These two things seem totally unrelated to me.

• why don't you compute both sides ? – Dominic Michaelis Feb 25 '13 at 21:27
• What do you mean with "both sides" ? – John Smith Feb 25 '13 at 21:29
• left and right side of the equation mark – Berci Feb 25 '13 at 21:36

The easiest demonstration is just to cmpute both sides of the equation:

$$b\times c=(b_2c_3-b_3c_2,b_3c_1-b_1c_3,b_1c_2-b_2c_1)$$ $$a\cdot(b\times c)=a_1b_2c_3-a_1b_3c_2+a_2b_3c_1-a_2b_1c_3+a_3b_1c_2-a_3b_2c_1$$

That's the determinant. Note that each term of the sum is one possible permutation multiplied by its parity, which happens to be the definition of determinant.

• Everyone has given an excellent answer but I like this most of all for its simplicity. Thank you I didn't think of it in the first place! – John Smith Feb 25 '13 at 21:37
• @JohnSmith Glad I helped! – MyUserIsThis Feb 25 '13 at 22:45
• @JohnSmith I really hope you're not going to prove something like $$\nabla\times(\mathbf u\times\mathbf u)=(\nabla\cdot\mathbf v)\mathbf u-(\nabla\cdot\mathbf u)\mathbf v+(\mathbf v\cdot\nabla)\mathbf u-(\mathbf u\cdot\nabla)\mathbf v$$ by direct calculating, and instead will use handy index notation :) – Kaster Feb 25 '13 at 23:28

Here's a slightly different perspective.

We use the mnemonic device...

$$\langle a_1,a_2,a_3 \rangle \times \langle b_1,b_2,b_3 \rangle = \begin{vmatrix} {\bf i} & {\bf j} & {\bf k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$$

...to remember the formula for the cross product.

If you think about the dot product in the following (somewhat odd way), the triple scalar product identity becomes a triviality: Dot product-ing replaces standard unit vectors with corresponding vector components:

$$(a_1{\bf i} + a_2{\bf j}+a_3{\bf k}) \bullet (b_1{\bf i} + b_2{\bf j}+b_3{\bf k}) = b_1\fbox{a_1}+b_2\fbox{a_2}+b_3\fbox{a_3}$$

So ${\bf i,j,k}$ have been replaced by $a_1,a_2,a_3$ respectively. Thus

$$\langle c_1,c_2,c_3 \rangle \bullet (\langle a_1,a_2,a_3 \rangle \times \langle b_1,b_2,b_3 \rangle) = \begin{vmatrix} \fbox{c_1} & \fbox{c_2} & \fbox{c_3} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$$

I am pretty sure you don't want this answer, but I couldn't resist.

The term on the left is an alternating $3$-linear form in $(a,b,c)$.

So it is equal to a constant times the determinant on the right..

Compute the constant with the canonical basis. It's $1$.

So the lhs is equal to the rhs.

Note: see here if you want to read about this characterization of the determinant and see here if you want a proof of the fact that the space of $n$-linear alternating forms is one-dimensional.

• Thats the way i would take too :) +1 – Dominic Michaelis Feb 25 '13 at 21:32
• I feel that this answer should reference a proof of the fact that "The vector space $W$ of all alternating multilinear $n$-forms on an $n$-dimensional vector space $V$ has dimension one." – us2012 Feb 25 '13 at 22:50
• @us2012 You're right. Here you go. – Julien Feb 26 '13 at 2:06
• @julien Great, that makes this answer complete! +1. (Although I feel a bit sad about destroying your perfect 7500 reputation ;) – us2012 Feb 26 '13 at 2:08

The truth is you've been lied to, or at least that the usual notation makes the innate connection way more difficult to see than necessary.

To see how the two are the same, let me tell you about the wedge product. The wedge product of vectors is like the cross product, in that it is anticommutative--$a \wedge b = -b \wedge a$--but it does not produce a vector. Instead, we directly interpret its result as a planar object--in fact, as the planar object that would be perpendicular to the vector from the cross product.

We formalize that relationship as follows: we say that $a \times b= -i a \wedge b$, where the $i = \hat x \wedge \hat y \wedge \hat z$ is the unit pseudoscalar, representing a volume. The pseudoscalar itself is an object of interest, as it converts wedges to dot products and vice versa when you move it through expressions. In fact,

$$a \cdot (b \times c) = a \cdot (i [b \wedge c]) = i (a \wedge b \wedge c)$$

How does all this wedge stuff connect to linear algebra? Quite simply, actually. Most linear operators will "distribute" over wedges in an intuitive way. That is, for a linear operator $\underline T$,

$$\underline T(a \wedge b) = \underline T(a) \wedge \underline T(b)$$

This is unlike the cross product, which doesn't follow such a simple law. The advantage of this is that, uniquely,

$$\underline T(a \wedge b \wedge c) = \alpha a \wedge b \wedge c$$

for some scalar $\alpha$, for every $a, b, c$. We call $\alpha$ the determinant! It is the special number by which every volume object is dilated or shrunk by the linear transformation, and here it becomes geometrically clear that that is the case: $a \wedge b \wedge c$ is literally multiplied by $\alpha = \det T$ as a result of the transformation.

Now, build up a linear transformation as so. Let $l, m, n$ be vectors, so that a transformation looks like

$$\underline T(a) = (a \cdot e_1) l + (a \cdot e_2) m + (a \cdot e_3) n$$

Then $l,m,n$ are the columns of the matrix representation of $\underline T$, and $\underline T(i) = l \wedge m \wedge n$. This completes the connection between the determinant and the scalar triple product.

Let's use index (tensor) notation $$\mathbf a \cdot (\mathbf b \times \mathbf c) = a_i (\mathbf b \times \mathbf c)_i = a_i \varepsilon_{ijk} b_jc_k = \varepsilon_{ijk} a_i b_j c_k = det \left | \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ b_1 & b_2 & b_3 \end{array}\right|$$

• can you explain a bit more in your answer how the various steps are obtained at ? For e.g. how the covariant index of a and the cross product is equal to the original scalar triple product ? – gansub Jan 3 '17 at 6:48

Note that when you compute the determinant in the traditional fashion, you start with the first row, so you're going to have

$$a_1 \cdot \textrm{something} - a_2 \cdot \textrm{something} + a_3\cdot \textrm{something}$$

That looks an awful lot like

$$a \cdot \textrm{something} = (a_1,a_2,a_3)\cdot (\textrm{something}, -\textrm{something}, \textrm{something})$$

When you look at the cross product $b\times c$, you find that the somethings are basically the minors of that matrix.

1. Check it for the standard basis vectors ${\bf i}, {\bf j}, {\bf k}$ in every possible order (these are all simple).
2. Show that if some triple of vectors $a,b,c$ satisfy the equation, then the triples $\lambda a,b,c$; $\ a,\lambda b,c$, $\ a,b,\lambda c$ also satisfy, and if $a',b,c$ also satisfies then so does $a+a',\, b,c$, and similarly for the sum of $b$'s and $c$'s.
3. Conclude that every triple satisfies.

The cross product can be defined as the unique vector $a \times b$ that satisfies $\langle x, a \times b \rangle = \det \begin{bmatrix} a & b & x \end{bmatrix}$ for all $x$ (or $\det \begin{bmatrix} a^T \\ b^T \\ x^T \end{bmatrix}$, if you prefer the equivalent, transposed version, or $\det \begin{bmatrix} x^T \\ a^T \\ b^T \end{bmatrix}$ if you prefer the transposed and rotated version).

If $a \times b = c_x {\bf i} + c_y {\bf j} + c_z {\bf k}$, you can see that $c_x = \langle (1,0,0)^T, a \times b \rangle = \det \begin{bmatrix} (1,0,0)^T \\ a^T \\ b^T \end{bmatrix}$, and similarly for $c_y,c_z$. Hence the formal notation $a \times b = \det \begin{bmatrix} {\bf i} & {\bf j} & {\bf k} \\ & a^T \\ & b^T \end{bmatrix}$.