# Scalar triple product in terms of scalar products only

I am trying to express the scalar triple product $\bf a \cdot (b \times c)$ only in terms of scalar products $\bf a \cdot b$, $\bf b \cdot c$, $\bf c \cdot a$ and the lengths of vectors $a$, $b$ and $c$. I start by writing $$\mathbf{a} = B\, \mathbf{b} + C\, \mathbf{c} + D\, \mathbf{b \times c}$$ and then performing the dot products with $\bf a, b, c, b \times c$, I arrive at the following equations $$a^2 = B \, (\mathbf{a \cdot b}) + C \, (\mathbf{c \cdot a}) + D \, (\mathbf{a \cdot (b\times c)})$$ $$\mathbf{a \cdot b} = B\, b^2 + C\, (\mathbf{b \cdot c})$$ $$\mathbf{c \cdot a} = B\, (\mathbf{b \cdot c})+ C\, c^2$$ $$\mathbf{a \cdot (b\times c)} = D\, \mathbf{(b\times c)}^2$$ Unknowns $B$ and $C$ can be easily eliminated, but when eliminating $D$, I necessarily have to square $\bf a \cdot (b \times c)$ to get the final solution $$[\mathbf{a \cdot (b \times c)}]^2 = a^2b^2c^2 - a^2(\mathbf{b \cdot c}) - b^2(\mathbf{c \cdot a}) - c^2(\mathbf{a \cdot b}) + 2(\mathbf{a \cdot b})(\mathbf{b \cdot c})(\mathbf{c \cdot a})$$

Is there a way of getting a similar expression for $\mathbf{a \cdot (b \times c)}$ directly? I would like to see an expression which clearly demonstrates the cyclic property of the scalar triple product. The square in my expression ruins it...

-

Using only the scalar product, vector addition, scalar multiplication, and operations on the reals, you can determine the magnitude but not the sign of the triple product.

Consider any reflecting linear transformation $T$, such as $(x,y,z) \mapsto (x,y,-z)$. Then we have:

$(T{\bf a}) \cdot (T{\bf b}) = {\bf a} \cdot {\bf b}$: the scalar product is preserved, but

$(T{\bf a}) \cdot ((T{\bf b}) \times (T{\bf c})) = - {\bf a} \cdot ({\bf b} \times {\bf c})$: the triple product is negated.

-
Very nice argument! That's what I was hoping for. Thanks! – LBO Jul 28 '14 at 11:37

A "similar" expression is the determinant representation $$\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=\operatorname{det} S,\qquad S=\left( \begin{array}{ccc} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{array}\right).$$ The expression that you have obtained can be easily deduced from this one: $$\Bigl(\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})\Bigr)^2=\operatorname{det} SS^T= \operatorname{det}\left(\begin{array}{ccc} \mathbf{a}\cdot\mathbf{a} & \mathbf{a}\cdot\mathbf{b} & \mathbf{a}\cdot\mathbf{c} \\ \mathbf{b}\cdot\mathbf{a} & \mathbf{b}\cdot\mathbf{b} & \mathbf{b}\cdot\mathbf{c} \\ \mathbf{c}\cdot\mathbf{a} & \mathbf{c}\cdot\mathbf{b} & \mathbf{c}\cdot\mathbf{c} \end{array}\right).$$ However the non-squared triple product cannot be polynomially expressed in terms of scalar products only.

-
I am aware of the determinant representation, but it uses components, i.e., a specific basis vectors, which I wanted to avoid. However, your calculation is much more elegant than mine. – LBO Jul 28 '14 at 11:36
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}