# PlanetPhysics/Vector Triple Product
## Vector Triple Product

Combining three vectors into a product is called a triple product. The vector triple product is the vector product of two vectors of which one is itself a vector product. Such as
$\mathbf {A} \times \left(\mathbf {B} \times \mathbf {C} \right)$ $\left(\mathbf {A} \times \mathbf {B} \right)\times \mathbf {C}$
The vector $\mathbf {A} \times \left(\mathbf {B} \times \mathbf {C} \right)$ is perpendicular to $\mathbf {A}$ and to $\left(\mathbf {B} \times \mathbf {C} \right)$. But $\left(\mathbf {B} \times \mathbf {C} \right)$ is perpendicular to the plane of $\mathbf {B}$ and $\mathbf {C.}$ Hence $\mathbf {A} \times \left(\mathbf {B} \times \mathbf {C} \right)$, being perpendicular to $\left(\mathbf {B} \times \mathbf {C} \right)$ must lie in the plane of $\mathbf {B}$ and $\mathbf {C}$ and thus take the form
$\mathbf {A} \times \left(\mathbf {B} \times \mathbf {C} \right)=x\mathbf {B} +y\mathbf {C}$
where $x$ an $y$ are two scalars. In like manner also the vector $\left(\mathbf {A} \times \mathbf {B} \right)\times \mathbf {C}$, being perpendicular to $\left(\mathbf {A} \times \mathbf {B} \right)$ must lie in the plane of $\mathbf {A}$ and $\mathbf {B}$. Hence it will be of the form
$\left(\mathbf {A} \times \mathbf {B} \right)\times \mathbf {C} =m\mathbf {A} +n\mathbf {B}$
where $m$ and $n$ are two scalars. From this it is evident that in general
$\left(\mathbf {A} \times \mathbf {B} \right)\times \mathbf {C} \,\,is\,not\,equal\,to\,\,\mathbf {A} \times \left(\mathbf {B} \times \mathbf {C} \right)$
The parentheses therefore cannot be removed or interchanged. It is essential to know which cross product is formed first and which second. This product is termed the vector triple product in contrast to the scalar triple product.

### Geometric Interpretation

The vector triple product may be used to express that component of a vector ${\mathbf {B} }$ which is perpendicular to a given vector ${\mathbf {A} }$. This geometric use of the product is valuable not only in itself but for the light it sheds upon the properties of the product. Let ${\mathbf {A} }$ (below figure) be a given vector and ${\mathbf {B} }$ another vector whose components parallel and perpendicular to ${\mathbf {A} }$ are to be found. Let the components pf ${\mathbf {B} }$ parallel and perpendicular to ${\mathbf {A} }$ be ${\mathbf {B} '}$ and ${\mathbf {B} ''}$ respectively. Draw ${\mathbf {A} }$ and ${\mathbf {B} }$ from a common origin. The product ${\mathbf {A} }\times {\mathbf {B} }$ is perpendicular to the plane of ${\mathbf {A} }$ and ${\mathbf {B} }$. The product ${\mathbf {A} }\times \left({\mathbf {A} }\times {\mathbf {B} }\right)$ lies in the plane of ${\mathbf {A} }$ and ${\mathbf {B} }$. It is furthermore perpendicular to ${\mathbf {A} }$. Hence it is collinear with ${\mathbf {B} ''}$. An examination of the figure will show that the direction of ${\mathbf {A} }\times \left({\mathbf {A} }\times {\mathbf {B} }\right)$ is opposite to that of ${\mathbf {B} ''}$.
\begin{figure} \includegraphics[scale=.5]{Fig1.eps} \vspace{20 pt} \end{figure}
Hence
${\mathbf {A} }\times \left({\mathbf {A} }\times {\mathbf {B} }\right)=-c{\mathbf {B} ''}$
where $c$ is some scalar constant.
Now
${\mathbf {A} }\times \left({\mathbf {A} }\times {\mathbf {B} }\right)=-A^{2}B\sin({\mathbf {A} },{\mathbf {B} }){\mathbf {\hat {b}} ''}$
where ${\mathbf {\hat {b}} ''}$ is the unit vector in the direction of ${\mathbf {B} ''}$.
But
$-c{\mathbf {B} ''}=-cB\sin({\mathbf {A} },{\mathbf {B} }){\mathbf {\hat {b}} ''}$
Hence
$c=A^{2}={\mathbf {A} }\cdot {\mathbf {A} }$
Therefore
   ${\mathbf {B} ''}=-{\frac {{\mathbf {A} }\times \left({\mathbf {A} }\times {\mathbf {B} }\right)}{{\mathbf {A} }\cdot {\mathbf {A} }}}$
The component of ${\mathbf {B} }$ perpendicular to ${\mathbf {A} }$ has been expressed in terms of the vector triple product of ${\mathbf {A} }$, ${\mathbf {A} }$, and ${\mathbf {B} }$. The component ${\mathbf {B} '}$ parallel to ${\mathbf {A} }$ is found using the dot product as a projection and leads to
   ${\mathbf {B} '}={\frac {{\mathbf {A} }\cdot {\mathbf {B} }}{{\mathbf {A} }\cdot {\mathbf {A} }}}{\mathbf {A} }$
Hence
   ${\mathbf {B} }={\mathbf {B} '}+{\mathbf {B} ''}={\frac {{\mathbf {A} }\cdot {\mathbf {B} }}{{\mathbf {A} }\cdot {\mathbf {A} }}}{\mathbf {A} }-{\frac {{\mathbf {A} }\times \left({\mathbf {A} }\times {\mathbf {B} }\right)}{{\mathbf {A} }\cdot {\mathbf {A} }}}$
Next, consider the product when two of the vectors are the same. By equation (3)
   ${\mathbf {A} }\cdot {\mathbf {A} }{\mathbf {B} }={\mathbf {A} }\cdot {\mathbf {B} }{\mathbf {A} }-{\mathbf {A} }\times \left({\mathbf {A} }\times {\mathbf {B} }\right)$
Or
   ${\mathbf {A} }\times \left({\mathbf {A} }\times {\mathbf {B} }\right)={\mathbf {A} }\cdot {\mathbf {B} }{\mathbf {A} }-{\mathbf {A} }\cdot {\mathbf {A} }{\mathbf {B} }$
This proves the formula in case two vectors are the same.

### Property I .

The vector triple product $\mathbf {A} \times \left(\mathbf {B} \times \mathbf {C} \right)$ may be expressed as the sum of two terms as
$\mathbf {A} \times \left(\mathbf {B} \times \mathbf {C} \right)=\mathbf {B} \left(\mathbf {A} \cdot \mathbf {C} \right)-\mathbf {C} \left(\mathbf {A} \cdot \mathbf {B} \right)$
To prove this, express $\mathbf {A}$ in terms of the three non-coplanar vectors $\mathbf {B}$, $\mathbf {C}$, and $\left(\mathbf {B} \times \mathbf {C} \right)$.
   $\mathbf {A} =b\mathbf {B} +c\mathbf {C} +a\left(\mathbf {B} \times \mathbf {C} \right)$
where $a$, $b$, $c$ are scalar constants. Then
   $\mathbf {A} \times \left(\mathbf {B} \times \mathbf {C} \right)=b\mathbf {B} \times \left(\mathbf {B} \times \mathbf {C} \right)+c\mathbf {C} \times \left(\mathbf {B} \times \mathbf {C} \right)+a\left(\mathbf {B} \times \mathbf {C} \right)\times \left(\mathbf {B} \times \mathbf {C} \right)$
The vector product of any vector by itself is zero. Hence
$\left(\mathbf {B} \times \mathbf {C} \right)\times \left(\mathbf {B} \times \mathbf {C} \right)=0$
so
   $\mathbf {A} \times \left(\mathbf {B} \times \mathbf {C} \right)=b\mathbf {B} \times \left(\mathbf {B} \times \mathbf {C} \right)+c\mathbf {C} \times \left(\mathbf {B} \times \mathbf {C} \right)$
Using the relationship developed under the geometric interpretation
$\mathbf {A} \times \left(\mathbf {A} \times \mathbf {B} \right)=\mathbf {A} \cdot \mathbf {B} \,\mathbf {B} -\mathbf {A} \cdot \mathbf {B} \,\mathbf {C}$
with different vectors yields
$\mathbf {B} \times \left(\mathbf {B} \times \mathbf {C} \right)=\mathbf {B} \cdot \mathbf {C} \,\mathbf {B} -\mathbf {B} \cdot \mathbf {B} \,\mathbf {C}$
$\mathbf {C} \times \left(\mathbf {B} \times \mathbf {C} \right)=-\mathbf {C} \times \left(\mathbf {C} \times \mathbf {B} \right)=-\mathbf {B} \cdot \mathbf {C} \,\mathbf {C} -\mathbf {C} \cdot \mathbf {C} \,\mathbf {B}$
Hence
   $\mathbf {A} \times \left(\mathbf {B} \times \mathbf {C} \right)=\left[\left(b\mathbf {B} \cdot \mathbf {C} +c\mathbf {C} \cdot \mathbf {C} \right)\mathbf {B} -\left(b\mathbf {B} \cdot \mathbf {B} +c\mathbf {C} \cdot \mathbf {B} \right)\mathbf {C} \right]$
But from (1)
$\mathbf {A} \cdot \mathbf {B} =b\mathbf {B} \cdot \mathbf {B} +c\mathbf {C} \cdot \mathbf {B} +a\left(\mathbf {B} \times \mathbf {C} \right)\cdot \mathbf {B}$
$\mathbf {A} \cdot \mathbf {C} =b\mathbf {B} \cdot \mathbf {C} +c\mathbf {C} \cdot \mathbf {C} +a\left(\mathbf {B} \times \mathbf {C} \right)\cdot \mathbf {C}$
using a property from the scalar triple product
$\left(\mathbf {B} \times \mathbf {C} \right)\cdot \mathbf {B} =0$
$\left(\mathbf {B} \times \mathbf {C} \right)\cdot \mathbf {C} =0$
we get
$\mathbf {A} \cdot \mathbf {B} =b\mathbf {B} \cdot \mathbf {B} +c\mathbf {C} \cdot \mathbf {B}$
$\mathbf {A} \cdot \mathbf {C} =b\mathbf {B} \cdot \mathbf {C} +c\mathbf {C} \cdot \mathbf {C}$
Substituting these values into (4)
   $\mathbf {A} \times \left(\mathbf {B} \times \mathbf {C} \right)=\left(\mathbf {A} \cdot \mathbf {C} \right)\mathbf {B} -\left(\mathbf {A} \cdot \mathbf {B} \right)\mathbf {C}$
Finally, noting that the dot product gives a scalar, we can use the Vector Identity $a\mathbf {B} =\mathbf {B} a$
to get
   $\mathbf {A} \times \left(\mathbf {B} \times \mathbf {C} \right)=\mathbf {B} \left(\mathbf {A} \cdot \mathbf {C} \right)-\mathbf {C} \left(\mathbf {A} \cdot \mathbf {B} \right)$
The relation is therefore proved for any three vectors $\mathbf {A}$, $\mathbf {B}$ and $\mathbf {C}$ and is often referred to as the {\mathbf BACK CAB} rule.

### Property II.

From the three letters ${\mathbf {A} }$, ${\mathbf {B} }$, ${\mathbf {C} }$ by different arrangements, four allied products in each of which ${\mathbf {B} }$ and ${\mathbf {C} }$ are included in parentheses may be formed. These are
${\mathbf {A} }\times \left({\mathbf {B} }\times {\mathbf {C} }\right)$ ${\mathbf {A} }\times \left({\mathbf {C} }\times {\mathbf {B} }\right)$ $\left({\mathbf {C} }\times {\mathbf {B} }\right)\times {\mathbf {A} }$ $\left({\mathbf {B} }\times {\mathbf {C} }\right)\times {\mathbf {A} }$
As a vector product changes its sign whenever the order of two factors is interchanged, the above products evidently satisfy the equations
${\mathbf {A} }\times \left({\mathbf {B} }\times {\mathbf {C} }\right)=-{\mathbf {A} }\times \left({\mathbf {C} }\times {\mathbf {B} }\right)=\left({\mathbf {C} }\times {\mathbf {B} }\right)\times {\mathbf {A} }=-\left({\mathbf {B} }\times {\mathbf {C} }\right)\times {\mathbf {A} }$
The expansion for a vector triple product in which the parenthesis comes first may therefore be obtained directly from that already found when the parenthesis comes last.
$\left({\mathbf {A} }\times {\mathbf {B} }\right)\times {\mathbf {C} }=-{\mathbf {C} }\times \left({\mathbf {A} }\times {\mathbf {B} }\right)=-{\mathbf {C} }\cdot {\mathbf {B} }{\mathbf {A} }+{\mathbf {C} }\cdot {\mathbf {A} }{\mathbf {B} }$
The formulas then become
${\mathbf {A} }\times \left({\mathbf {B} }\times {\mathbf {C} }\right)={\mathbf {B} }{\mathbf {A} }\cdot {\mathbf {C} }-{\mathbf {C} }{\mathbf {A} }\cdot {\mathbf {B} }$ $\left({\mathbf {A} }\times {\mathbf {B} }\right)\times {\mathbf {C} }={\mathbf {B} }{\mathbf {A} }\cdot {\mathbf {C} }-{\mathbf {A} }{\mathbf {C} }\cdot {\mathbf {B} }$
These reduction formulas are of such constant occurrence and great importance that they should be committed to memory.