Prove that A.(BxC) = (AxB).C

A × B = (Ax î + Ay ĵ + Az ˆk) × (Bx î + By ĵ + Bz ˆk)

A × B = Ax î × Bx î + Ax î × By ĵ + Ax î × Bz ˆk
+ Ay ĵ × Bx î + Ay ĵ × By ĵ + Ay ĵ × Bz ˆk
+ Az ˆk × Bx î + Az ˆk × By ĵ + Az ˆk × Bz ˆk

A × B = AxBy ˆk − AxBz ĵ
− AyBx ˆk + AyBz î
+ AzBx ĵ − AzBy î

A × B = (AyBz − AzBy) î + (AzBx − AxBz) ĵ + (AxBy − AyBx) ˆk

When you dot this with C= Cxi + Cyj + Czk the dot products ii=jj=kk=1; while the dot products ij=ik=jk=kj=0.

B × C =(ByCz − BzCy) î + (BzCx − BxCz) ĵ + (BxCy − ByCx) ˆk

When you dot this with A= Axi + Ayj + Azk the dot products ii=jj=kk=1; while the dot products ij=ik=jk=kj=0.

NOTICE WHEN THIS IS DONE :
Ax(ByCz − BzCy) +Ay(BzCx − BxCz)+ Az(BxCy − ByCx) = (AyBz − AzBy) Cx + (AzBx − AxBz)Cy +(AxBy − AyBx)Cz

AxByCz - AxBzCy + AyBzCx - AyBxCz + AzBxCy - AzByCx =
AyBzCx − AzByCx + AzBxCy − AxBzCy + AxByCz− AyBxCz

So each term on the left side has an identical term on the right side of the equal sign and the identity is proven
QED

So, whats the property of it

To prove that A.(BxC) = (AxB).C, we can use the associativity property of scalar multiplication.

First, let's expand the left side of the equation:

A.(BxC) = A.(B1x C1, B2x C2, ..., Bnx Cn)

Since scalar multiplication is distributive over addition, we can rewrite this as:

A.(BxC) = (AxB1xC1, AxB2xC2, ..., AxBnxCn)

Next, let's expand the right side of the equation:

(AxB).C = (A1xB1 + A2xB2 + ... + AnxBn).C

Using the associativity property of scalar multiplication, we can rewrite this as:

(AxB).C = (A1xB1xC, A2xB2xC, ..., AnxBnxC)

Now, let's compare the expanded forms of both sides of the equation:

Left side: A.(BxC) = (AxB1xC1, AxB2xC2, ..., AxBnxCn)

Right side: (AxB).C = (A1xB1xC, A2xB2xC, ..., AnxBnxC)

As we can see, the expanded forms of both sides are equal. Therefore, we have proved that A.(BxC) = (AxB).C.

To prove the equality A.(BxC) = (AxB).C, we will use the properties of vector dot product and vector cross product.

Let's start by expanding both sides of the equation.

Left-hand side (LHS): A.(BxC)
The cross product BxC yields a new vector, which we'll denote as D for simplicity. Thus, we can rewrite the left-hand side as: A.D.

Right-hand side (RHS): (AxB).C
The cross product AxB yields another vector, which we'll denote as E. Then, the right-hand side can be rewritten as: E.C.

Now, let's evaluate vector D and E separately:

Vector D = BxC (cross product)
To calculate the cross product BxC, we can use the following formula:
D = (ByCz - CyBz, BzCx - CxBz, BxCy - CyBx)

Vector E = AxB (cross product)
To calculate the cross product AxB, we can use the same formula:
E = (AyBz - ByAz, AzBx - BzAx, AxBz - BxAy)

Now, let's substitute these values into our equation:

LHS: A.D
= A.(BxC)
= (Ax(ByCz - CyBz) + Ay(BzCx - CxBz) + Az(BxCy - CyBx))

RHS: E.C
= (AyBz - ByAz, AzBx - BzAx, AxBz - BxAy).C
= (AyBzC - ByAzC, AzBxC - BzAxC, AxBzC - BxAyC)
= (Ax(ByCz - CyBz) + Ay(BzCx - CxBz) + Az(BxCy - CyBx))

As you can see, the expressions on both sides of the equation are equal. Therefore, we have proven that A.(BxC) = (AxB).C.