Vector Multiplication

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Discussion

scalar-vector multiplication

Multiplication of a vector by a scalar changes the magnitude of the vector, but leaves its direction unchanged. The scalar changes the size of the vector. The scalar "scales" the vector. For example, the polar form vector …

r = r ˆr + θ ˆθ

multiplied by the scalar a is …

a r = ar ˆr + θ ˆθ

Multiplication of a vector by a scalar is distributive.

a(A + B) = a A + a B

Consequently, the rectangular form vector …

r = x î + y ĵ

multiplied by the scalar a is …

ar = ax î + ay ĵ

dot product

Geometrically, the dot product of two vectors is the magnitude of one times the projection of the other along the first.


[magnify]

The symbol used to represent this operation is a small dot at middle height (·), which is where the name "dot product" comes from. Since this product has magnitude only, it is also known as the scalar product.

A·B = AB cos θ

The dot product is distributive …

A·(B + C) = A·B + A·C

and commutative …

A·B = B·A

Since the projection of a vector on to itself leaves its magnitude unchanged, the dot product of any vector with itself is the square of that vector's magnitude.

A·A = AA cos 0° = A2

Applying this corollary to the unit vectors means that the dot product of any unit vector with itself is one. In addition, since a vector has no projection perpendicular to itself, the dot product of any unit vector with any other is zero.

î · î = ĵ · ĵ = ˆk · ˆk = (1)(1)(cos 0°) = 1

î · ĵ = ĵ · ˆk = ˆk · î = (1)(1)(cos 90°) = 0

Using this knowledge we can derive a formula for the dot product of any two vectors in rectangular form. The resulting product looks like it's going to be a terrible mess, but consists mostly of terms equal to zero.

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 = AxBx + AyBy + AzBz

The dot product of two vectors is thus the sum of the products of their parallel components. From this we can derive the Pythagorean Theorem in three dimensions.

A·A = AA cos 0° = AxAx + AyAy + AzAz

A2 = Ax2 + Ay2 + Az2

cross product

Geometrically, the cross product of two vectors is the area of the parallelogram between them.


[magnify]

The symbol used to represent this operation is a large diagonal cross (×), which is where the name "cross product" comes from. Since this product has magnitude and direction, it is also known as the vector product.

A × B = AB sin θ ˆn

The vector ˆn ("n hat") is a unit vector perpendicular to the plane formed by the two vectors. The direction of ˆn is determined by the right hand rule, which will be discussed shortly.

The cross product is distributive …

 A × (B + C) = (A × B) + (A × C)

but not commutative …

A × B = −B × A

Reversing the order of cross multiplication reverses the direction of the product.

Since two identical vectors produce a degenerate parallelogram with no area, the cross product of any vector with itself is zero …

A × A = 0

Applying this corollary to the unit vectors means that the cross product of any unit vector with itself is zero.

î × î = ĵ × ĵ = ˆk × ˆk = (1)(1)(sin 0°) = 0

It should be apparent that the cross product of any unit vector with any other will have a magnitude of one. (The sine of 90° is one, after all.) The direction is not intuitively obvious, however. The right hand rule for cross multiplication relates the direction of the two vectors with the direction of their product. Since cross multiplication is not commutative, the order of operations is important.

  1. Hold your right hand flat with your thumb perpendicular to your fingers. Do not bend your thumb at anytime.
  2. Point your fingers in the direction of the first vector.
  3. Orient your palm so that when you fold your fingers they point in the direction of the second vector.
  4. Your thumb is now pointing in the direction of the cross product.

A right-handed coordinate system, which is the usual coordinate system used in physics and mathematics, is one in which any cyclic product of the three coordinate axes is positive and any anticyclic product is negative. Imagine a clock with the three letters x-y-z on it instead of the usual twelve numbers. Any product of these three letters that runs around the clock in the same direction as the sequence x-y-z is cyclic and positive. Any product that runs in the opposite direction is anticyclic and negative.

Cross products of cyclic unit vectors are positive. Cross products of anticyclic unit vectors are negative.
magnify

Using this knowledge we can derive a formula for the cross product of any two vectors in rectangular form. The resulting product looks like it's going to be a terrible mess, and it is!

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

There is a simpler way to write this. For those of you familiar with matrices, the cross product of two vectors is the determinant of the matrix whose first row is the unit vectors, second row is the first vector, and third row is the second vector. Symbolically …

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

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

Summary

Problems

practice

  1. Write something.
    • Answer it.
  2. Write something.
    • Answer it.
  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
     
  4. Test the cross product for associativity by determining if this equation is true.
     
    (A × B) × C ≟ A × (B × C)
     

    Behold! A big damn pile of symbols.

    We've already shown that …

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

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

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

    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 ˆk
       +  (AyBz − AzBy) ĵ  ×  Cx î  +  (AzBx − AxBz) ĵ  ×  Cy ĵ  +  (AxBy − AyBx) ĵ  ×  Cz ˆk
       +  (AyBz − AzBy) ˆk  ×  Cx î  +  (AzBx − AxBz) ˆk  ×  Cy ĵ  +  (AxBy − AyBx) ˆk  ×  Cz ˆk

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

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

    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) ˆk
       +  Ay ĵ  ×  (ByCz − BzCy) î  +  Ay ĵ (BzCx − BxCz) ĵ  ×   +  Ay ĵ  ×  (BxCy − ByCx) ˆk
       +  Az ˆk  ×  (ByCz − BzCy) î  +  Az ˆk (BzCx − BxCz) ĵ  ×   +  Az ˆk  ×  (BxCy − ByCx) ˆk

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

    A × (B × C)  =  (AxBzCx − AxBxCz) ˆk  −  (AxBxCy − AxByCx) ĵ
       −  (AyByCz − AyBzCy) ˆk  +  (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)  ˆk

    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)  ˆk  ≟  (AxBzCx + AyBzCy)  −  (AxBxCz + AyByCz)  ˆk

    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)

numerical

  1. Two related questions.
    1. What is the angle between two vectors if their magnitudes are 3 and 4 and their cross product is 5?
    2. What unit does this cross product have if each vector was expressed in meters?
  2. Two related questions.
    1. What is the angle between two sides of a triangle if the lengths of the sides are 3 m and 4 m and the area of the triangle is 5 m2?
    2. What is the length of the remaining side?

algebraic

  1. Test the scalar-vector product for distributivity by determining if this equation is true.
     
    (a + b) C ≟ (a C) + (b C)
     
  2. Test the scalar-vector product for associativity by determining if this equation is true.
     
    a (b C) ≟ (ab) C
     
  3. Test the dot product for distributivity over addition by determining if this equation is true.
     
    A · (B + C) ≟ (A · B) + (A · C)
     
  4. Test the cross product for distributivity over addition by determining if this equation is true.
     
    A × (B + C) ≟ (A × B) + (A × C)
     

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