Vector Multiplication

Practice

practice problem 1

Write something.

solution

Answer it.

practice problem 2

Write something.

solution

Answer it.

practice problem 3

Derive the law of cosines from the dot product.

solution

Begin by defining an arbitrary vector C as the difference between two other vectors A and B, then take the dot product of C with itself.

Let …
     
  C  = A − B
 
Then …    
 
  C · C  = (A − B) · (A − B)
  C2  = (A · A) − (A · B) − (B · A) + (B · B)
  C2  = A2 + B2 − 2AB cosθ
 

practice problem 4

Test the cross product for associativity by determining if this equation is true.
 
(A × B) × C ≟ A × (B × C)
 

solution

Behold! A big damn pile of symbols.

We've already shown that …

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

Change the symbols around, swapping A with B and B with C.

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

Now for the tedious part. Take the first equation and cross it into C.

(A × B) × C  =  (AyBz − AzBy) î  ×  Cx î  +  (AzBx − AxBz) î  ×  Cy ĵ  +  (AxBy − AyBx) î  ×  Cz k̂
   +  (AyBz − AzBy) ĵ  ×  Cx î  +  (AzBx − AxBz) ĵ  ×  Cy ĵ  +  (AxBy − AyBx) ĵ  ×  Cz k̂
   +  (AyBz − AzBy) k̂  ×  Cx î  +  (AzBx − AxBz) k̂  ×  Cy ĵ  +  (AxBy − AyBx) k̂  ×  Cz k̂

Eliminate the zero terms. Watch the signs on the other terms.

(A × B) × C  =  (AzBxCy − AxBzCy)  k̂  −  (AxByCz − AyBxCz)  ĵ
   −  (AyBzCx − AzByCx)  k̂  +  (AxByCz − AyBxCz)  î
   +  (AyBzCx − AzByCx)  ĵ  −  (AzBxCy − AxBzCy)  î

Then simplify.

(A × B) × C  =  (AxByCz + AxBzCy)  −  (AyBxCz + AzBxCy)  î
   +  (AyBxCz + AyBzCx)  −  (AxByCz + AzByCx)  ĵ
   +  (AzBxCy + AzByCx)  −  (AxBzCy + AyBzCx)  k̂

Repeat by crossing A into the second equation.

A × (B × C)  =  Ax î  ×  (ByCz − BzCy) î  +  Ax î (BzCx − BxCz) ĵ  ×   +  Ax î  ×  (BxCy − ByCx) k̂
   +  Ay ĵ  ×  (ByCz − BzCy) î  +  Ay ĵ (BzCx − BxCz) ĵ  ×   +  Ay ĵ  ×  (BxCy − ByCx) k̂
   +  Az k̂  ×  (ByCz − BzCy) î  +  Az k̂ (BzCx − BxCz) ĵ  ×   +  Az k̂  ×  (BxCy − ByCx) k̂

Eliminate the zero terms. Watch the signs on the other terms.

A × (B × C)  =  (AxBzCx − AxBxCz) k̂  −  (AxBxCy − AxByCx) ĵ
   −  (AyByCz − AyBzCy) k̂  +  (AyBxCy − AyByCx) î
   +  (AzByCz − AzBzCy) ĵ  −  (AzBzCx − AzBxCz) î

Then simplify.

A × (B × C)  =  (AyBxCy + AzBxCz)  −  (AyByCx + AzBzCx)  î
   +  (AxByCx + AzByCz)  −  (AxBxCy + AzBzCy)  ĵ
   +  (AxBzCx + AyBzCy)  −  (AxBxCz + AyByCz)  k̂

Well now, that wasn't any fun, but fun be damned. This ain't no amusement park. It's a math proof. The real question is, are the two products equal or are they not equal? Let's make a direct comparison of the components of both products.

(A × B) × C  ≟  A × (B × C)
(AxByCz + AxBzCy)  −  (AyBxCz + AzBxCy)  î  ≟  (AyBxCy + AzBxCz)  −  (AyByCx + AzBzCx)  î
(AyBxCz + AyBzCx)  −  (AxByCz + AzByCx)  ĵ  ≟  (AxByCx + AzByCz)  −  (AxBxCy + AzBzCy)  ĵ
(AzBxCy + AzByCx)  −  (AxBzCy + AyBzCx)  k̂  ≟  (AxBzCx + AyBzCy)  −  (AxBxCz + AyByCz)  k̂

I don't see one triplet of subscripts in the same order as any other triplet of subscripts. Could these two products equal? Maybe. I don't have the time to sort through all the possibilities. Are they equal in general? Most definitely not. Therefore, the cross product of two vectors is not an associative operation.

(A × B) × C ≠ A × (B × C)

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