Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer
    
Very nice argument! That's what I was hoping for. Thanks! –  Little Brown One Jul 28 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.

share|improve this answer
    
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. –  Little Brown One Jul 28 at 11:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.