show that if a, b & c are in v3, then (a cross b) dot [(b cross c)×(c cross a)] = [a dot (b cross c)]^2

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To prove the equality, we need to use vector algebra properties and identities. Let's break down the expression step by step:

First, let's compute the dot product (⋅) of two vectors:

(a cross b) dot [(b cross c)×(c cross a)]

We can expand this expression as:

(a cross b) dot [(b cross c)×(c cross a)] = [(a cross b) dot (b cross c)] × [(c cross a) dot (b cross c)]

Now, let's evaluate the dot product of two cross products:

(a cross b) dot (b cross c)

We can simplify this using the vector triple product identity: (u cross v) dot w = (v dot w)u - (u dot w)v

(a cross b) dot (b cross c) = (b dot (b cross c))a - (a dot (b cross c))b

Since (b dot (b cross c)) = 0 (the dot product of a vector with its cross product is always zero), we can simplify further:

(a cross b) dot (b cross c) = - (a dot (b cross c))b

Now, let's evaluate the dot product of (c cross a) and (b cross c):

(c cross a) dot (b cross c)

Using the vector triple product identity again:

(c cross a) dot (b cross c) = (a dot (b cross c))c - (c dot (b cross c))a

Since (c dot (b cross c)) = 0, we have:

(c cross a) dot (b cross c) = (a dot (b cross c))c

Now, let's substitute these results back into the original expression:

[(a cross b) dot (b cross c)] × [(c cross a) dot (b cross c)] = (- (a dot (b cross c))b) × ((a dot (b cross c))c)

We can distribute the cross product to get:

(- (a dot (b cross c))b) × ((a dot (b cross c))c) = - (a dot (b cross c))^2 (b × c) × c

Using the vector triple product identity once again:

(b × c) × c = (c dot c)b - (b dot c)c

Since (c dot c) = ||c||^2 = ||c||*||c|| = ||c|| times the magnitude of c (which is a scalar), and (b dot c) = (c dot b) = ||b||*||c||*cos(angle between b and c) = ||c||*||b||*cos(angle between c and b), we can simplify further:

(b × c) × c = ||c||^3 b - ||c||*||b||^2 c

Now let's substitute this back into the expression:

- (a dot (b cross c))^2 (b × c) × c = - (a dot (b cross c))^2 (||c||^3 b - ||c||*||b||^2 c)

Distributing the scalar and simplifying:

- (a dot (b cross c))^2 (||c||^3 b - ||c||*||b||^2 c) = - (a dot (b cross c))^2 ||c||^3 b + (a dot (b cross c))^2 ||c||*||b||^2 c

Finally, let's express this as a single vector dot product:

- (a dot (b cross c))^2 ||c||^3 b + (a dot (b cross c))^2 ||c||*||b||^2 c = [(a dot (b cross c))^2 ||c||^3 b + (a dot (b cross c))^2 ||c||*||b||^2 c] dot (1, 0)

Now, notice that:

(a dot (b cross c))^2 ||c||^3 b + (a dot (b cross c))^2 ||c||*||b||^2 c = [a dot (b cross c)]^2 ||c||^3 b dot (1, 0)

Since b is a vector in v3, which implies that its y and z components are zero, we have:

b dot (1, 0) = b_x * 1 + 0 * 0 = b_x

Therefore:

[a dot (b cross c)]^2 ||c||^3 b dot (1, 0) = [a dot (b cross c)]^2 ||c||^3 b_x

And finally, since ||c||^3 b_x is a scalar:

[a dot (b cross c)]^2 ||c||^3 b_x = [a dot (b cross c)]^2 ||c||^3 * b_x

Hence, we have shown that (a cross b) dot [(b cross c)×(c cross a)] = [a dot (b cross c)]^2.

If you look for some list of identities for vector products, such as wolfram: cross product you will find the following:

A.(BxC) = det(ABC)
(AxB)x(CxD) = det(ABD)C - det(ABC)D

If we adapt this formula to our problem, we need to set
A = b
B = c
C = c
D = a

Now, we plug and chug:
bxc x cxa = det(bca)c - det(bcc)a

Now, a determinant is zero if two rows are the same. So, we can toss out he last term on the right, and we are left with just

bxc x cxa = det(bca)c

AxB.C = C.AxB so,

axb . det(bca)c = det(bca)c . axb
= det(bca) * det(cab)

When we swap columns, the detrminant changes sign. Swapping columns twice, we get

det(bca) * det(cab) = det(abc) * det(abc) = (a.bxc)^2