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Katie writes,
Prove that (u cross v) dot ( (v cross w) cross (w cross u) ) = (u dot (v
cross w) )2
Any ideas as to where to start even would be much appreciated!
My first thought on seeing this is "what a mess!" But as the cobwebs cleared, I noticed the left side is a triple product, so maybe it's the volume of a parallelepiped. The three vectors being multiplied together are the three cross products of u, v, and w, or the "directed areas" of the three faces that meet at one corner.
Proof that (u�v)�((v�w)�(w�u)) = (u�(v�w))2
Let's start by looking at the cross product (v�w)�(w�u). v�w is perpendicular to w, and so is w�u perpendicular to w, so it follows that when you cross these two vectors, both perpendicular to w, the resulting cross product is a scalar multiple of w.
This thinking is confirmed, when, by Lagrange's Formula, a�(b�c) = (c�a)b - (b�a)c, it follows that
(v�w)�(w�u) = (u�(v�w))w - (w�(v�w))u
Because w and v�w are orthogonal w�(v�w)=0, so the formula simplifies to
(v�w)�(w�u) = (u�(v�w))w
which I put in bold face, because it's a fairly useful fact in its own right. (Another way to get this result is from the general formula for the cross product of two cross products, in which the second determinant is zero.)
If we let "a" represent the scalar triple product, a = u�(v�w), we see from the bold-face formula, above, that
(v�w)�(w�u) = aw,
and so, because a = u�(v�w) = w�(u�v),
(u�v)�((v�w)�(w�u) )
= (u�v)�aw
= aw�(u�v)
= a2
= (u�(v�w))2
If a, b, c, and d are vectors, then the cross product of a�b and c�d is given by
(a�b)�(c�d) = det(a b d) c - det(a b c) d
where det is the determinant of the square matrix formed by placing the three vectors, one above the other.
Triple Product -- a·(b�c) is a scalar representing the "signed volume" of a parallelepiped
Triangle Area Using Vectors, part 1 and part 2
Triangle Area using Determinant
Geometry and Trigonometry, and in particular, the Points and Lines section.
The webmaster and author of this Math Help site is Graeme McRae.