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
Advertising, public
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