MHB Problem of the Week #129 - September 15th, 2014

[Chris L T521]

Here's this week's problem!

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Problem: Prove for any $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}\in\mathbb{R}^3$,
\[(\mathbf{a}\times\mathbf{b}) \cdot (\mathbf{c}\times\mathbf{d}) = \begin{vmatrix}\mathbf{a}\cdot\mathbf{c} & \mathbf{b}\cdot\mathbf{c}\\ \mathbf{a}\cdot\mathbf{d} & \mathbf{b}\cdot\mathbf{d}\end{vmatrix}.\]

-----Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Euge, Kiwi, and Longines.

Here's Longines' solution, which uses a combination of the vector and scalar triple product: [sp]LHS = $(a \times b) \cdot (c \times d)$

RHS = $ \begin{vmatrix}
a \cdot b & b \cdot c\\
a \cdot d & b \cdot d
\end{vmatrix} $

Where $a,b,c,d \in R^3$
LHS :
Let $(c \times d) = \textbf{e}$

$ \implies (a \times b) \cdot \textbf{e}$Now using the properties of the scalar triple product, we have: $b \cdot (\textbf{e} \times a)$$= b \cdot ((c \times d) \times a)$$= b \cdot (( (a \cdot c)d - c(a \cdot d)) $ [Using the properties of the Vector Triple Product]

$= (a.c)(b.d)- (b.c)(a.d)$

$\implies \begin{vmatrix}
a \cdot b & b \cdot c\\
a \cdot d & b \cdot d
\end{vmatrix} $ = RHSWe have shown LHS = RHS for all $a,b,c,d \in R^3$[/sp]

Here's Euge's solution, which uses Einstein's notation: [sp]Using Einstein's notation, we write

$$(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d}) = \epsilon_{ijk} \epsilon^i_{\, \mu\nu}\, a^j b^k c^\mu d^\nu = (\delta_{j\nu} \delta_{k\mu} - \delta_{j\mu} \delta_{k\nu}) a^j b^k c^\mu d^\nu$$,

where the last equation follows from the $\epsilon-\delta$ identity. Since

$$ (\delta_{j\mu} \delta_{k\nu} - \delta_{j\nu} \delta_{k\mu}) a^j b^k c^\mu d^\nu = (\delta_{j\mu} a^j c^\mu) (\delta_{k\nu} b^k d^\nu) - (\delta_{j\nu} a^j d^\nu) (\delta_{k\mu} b^k c^\mu) = (\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d}) - (\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \cdot \mathbf{c})$$,

we deduce

$$ (\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d}) = (\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d}) - (\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \cdot \mathbf{c}) = \left|\begin{array}{cc}\mathbf{a} \cdot \mathbf{c} & \mathbf{b} \cdot \mathbf{c} \\ \mathbf{a} \cdot \mathbf{d} & \mathbf{b} \cdot \mathbf{d}\end{array}\right|.$$[/sp]