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# Vector Multiplication

## Practice

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### 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.

A × B = (AyBz − AzBy + (AzBx − AxBz + (AxBy − AyBx

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

B × C = (ByCz − BzCy + (BzCx − BxCz + (BxCy − ByCx

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

Are the two products equal or are they not?

(A × B) × C ≟ A × (B × C)

Let's make a direct comparison of the components.

 [(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, therefore the vector cross product is not an associative operation.

(A × B) × C ≠ A × (B × C)

Well now, that wasn't any fun, but fun be damned. This ain't no amusement park. It's a math proof.