a) prove the identity:
Ax(BXC)=B(A*C)-C(A*B)
A,B,C vectors
x- cross product
*-dot product
b) use part a) to prove that cross product is not associative.
google vector triple product and you will find lots of proofs. One that is relatively straightforward is at
http://www.msci.memphis.edu/faculty/dwiggins/files/triplecross.htm
You already know that AxB = -BxA, so that will help in the part (b).
a) To prove the identity Ax(BxC) = B(A*C) - C(A*B), we'll need to utilize the properties of both the cross product (x) and dot product (*). Let's break down the steps:
Step 1: Expand Ax(BxC)
The cross product of B and C can be written as BxC = |B||C| sin(theta) n, where |B| and |C| are the magnitudes of vectors B and C, theta is the angle between them, and n is the unit vector perpendicular to both B and C.
Now we have Ax(BxC) = A x (|B||C| sin(theta) n). Using the distributive property of the cross product, we can write this as Ax(BxC) = |B||C| (A x sin(theta) n).
Step 2: Distribute A to sin(theta) n
We know that A x n = |A||n| sin(phi) m, where |A| is the magnitude of vector A, phi is the angle between vectors A and n, and m is the unit vector perpendicular to both A and n.
So, A x sin(theta) n = |A||sin(theta)| (n x n) = 0, as the cross product of a vector with itself is zero.
Therefore, Ax(BxC) = 0.
Step 3: Expand B(A*C) and C(A*B)
B(A*C) = B * (|A||C| cos(phi) m), where cos(phi) is the cosine of the angle between vectors A and C.
Now, using the distributive property of the dot product, we can write this as B(A*C) = |A||B||C| cos(phi) (B * m).
Similarly, C(A*B) = |A||B||C| cos(theta) (C * m).
Step 4: Subtract C(A*B) from B(A*C)
We have B(A*C) - C(A*B) = |A||B||C| cos(phi) (B * m) - |A||B||C| cos(theta) (C * m).
Since B * m and C * m are dot products of perpendicular unit vectors, they are both equal to zero.
Finally, B(A*C) - C(A*B) simplifies to zero.
Therefore, Ax(BxC) = B(A*C) - C(A*B).
b) To prove that the cross product is not associative, we will use the identity from part a) and consider three vectors A, B, and C.
Suppose we have vectors A = i, B = j, and C = k, where i, j, and k are the unit vectors along the x, y, and z axes, respectively.
Using the identity from part a), we have A × (B × C) = B(A · C) - C(A · B).
A × (B × C) = A × (B × C) = i × (j × k) = i × (-k) = -j.
On the other hand, (A × B) × C = (i × j) × k = k × k = 0.
Since -j is not equal to 0, we can conclude that A × (B × C) is not equal to (A × B) × C.
Therefore, the cross product is not associative.