Vector Multiplication

Glenn Elert

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

= (A_yB_z − A_zB_y) î + (A_zB_x − A_xB_z) ĵ + (A_xB_y − A_yB_x) k̂

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

B × C

= (B_yC_z − B_zC_y) î + (B_zC_x − B_xC_z) ĵ + (B_xC_y − B_yC_x) k̂

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

(A × B) × C

= (A_yB_z − A_zB_y) î × C_x î
+ (A_zB_x − A_xB_z) î × C_y ĵ
+ (A_xB_y − A_yB_x) î × C_z k̂
+ (A_yB_z − A_zB_y) ĵ × C_x î
+ (A_zB_x − A_xB_z) ĵ × C_y ĵ
+ (A_xB_y − A_yB_x) ĵ × C_z k̂
+ (A_yB_z − A_zB_y) k̂ × C_x î
+ (A_zB_x − A_xB_z) k̂ × C_y ĵ
+ (A_xB_y − A_yB_x) k̂ × C_z k̂

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

(A × B) × C

= (A_zB_xC_y − A_xB_zC_y) k̂
− (A_xB_yC_z − A_yB_xC_z) ĵ
− (A_yB_zC_x − A_zB_yC_x) k̂
+ (A_xB_yC_z − A_yB_xC_z) î
+ (A_yB_zC_x − A_zB_yC_x) ĵ
− (A_zB_xC_y − A_xB_zC_y) î

Then simplify.

(A × B) × C

= [(A_xB_yC_z + A_xB_zC_y) − (A_yB_xC_z + A_zB_xC_y)] î
+ [(A_yB_xC_z + A_yB_zC_x) − (A_xB_yC_z + A_zB_yC_x)] ĵ
+ [(A_zB_xC_y + A_zB_yC_x) − (A_xB_zC_y + A_yB_zC_x)] k̂

Repeat by crossing A into the second equation.

A × (B × C)

= A_x î × (B_yC_z − B_zC_y) î
+ A_x î × (B_zC_x − B_xC_z) ĵ
+ A_x î × (B_xC_y − B_yC_x) k̂
+ A_y ĵ × (B_yC_z − B_zC_y) î
+ A_y ĵ × (B_zC_x − B_xC_z) ĵ
+ A_y ĵ × (B_xC_y − B_yC_x) k̂
+ A_z k̂ × (B_yC_z − B_zC_y) î
+ A_z k̂ × (B_zC_x − B_xC_z) ĵ
+ A_z k̂ × (B_xC_y − B_yC_x) k̂

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

A × (B × C)

= (A_xB_zC_x − A_xB_xC_z) k̂
− (A_xB_xC_y − A_xB_yC_x) ĵ
− (A_yB_yC_z − A_yB_zC_y) k̂
+ (A_yB_xC_y − A_yB_yC_x) î
+ (A_zB_yC_z − A_zB_zC_y) ĵ
− (A_zB_zC_x − A_zB_xC_z) î

Then simplify.

A × (B × C)

= [(A_yB_xC_y + A_zB_xC_z) − (A_yB_yC_x + A_zB_zC_x)] î
+ [(A_xB_yC_x + A_zB_yC_z) − (A_xB_xC_y + A_zB_zC_y)] ĵ
+ [(A_xB_zC_x + A_yB_zC_y) − (A_xB_xC_z + A_yB_yC_z)] 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.

[(A_xB_yC_z + A_xB_zC_y) − (A_yB_xC_z + A_zB_xC_y)] î	≟	[(A_yB_xC_y + A_zB_xC_z) − (A_yB_yC_x + A_zB_zC_x)] î

[(A_yB_xC_z + A_yB_zC_x) − (A_xB_yC_z + A_zB_yC_x)] ĵ	≟	[(A_xB_yC_x + A_zB_yC_z) − (A_xB_xC_y + A_zB_zC_y)] ĵ

[(A_zB_xC_y + A_zB_yC_x) − (A_xB_zC_y + A_yB_zC_x)] k̂	≟	[(A_xB_zC_x + A_yB_zC_y) − (A_xB_xC_z + A_yB_yC_z)] 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.