Revisiting Vectors & Solid Geometry (Part 2)

Products of Two Vectors:

Dot Product:

In the diagram below we  show the geometry relevant to the scalar product of two vectors, say A ·B.

                                                                                                             















The scalar product of two vectors  A  and B  is  also called the dot product because of the dot symbol used to denote it.   Thus:

 A ·B    =  ‖A‖ ‖B‖   cos Θ, hence also:

cos(Θ) = (A ·B)/ [A][B]

If the two vectors are at a right angle to each other then Θ =  90 degrees and:

 cos(Θ) = (A ·B)/ [A][B]   =  0  

Since vector multiplication is commutative, we also have:

A ·B    =   B ·A    

Vector Cross Product:

The vector cross product A X B    is illustrated below:

Here, we let the angle subtended between A and B be Θ  with:  0   <  Θ  <  π.  Then unless A and B are parallel, they now determine a plane.  Let n be a unit vector perpendicular to the plane and pointing in the direction a right-handed thread screw would advance when its head is rotated from A to B through the angle Θ.   The vector product or cross product is then:

A X B    =   n  ‖A‖ ‖B‖   sin  Θ

It is clear that if A and B are parallel (Θ = 0) so:

A X B    =     0

Also, if A and B are reversed in the depiction above, then it follows the unit vector n is replaced by – n and hence:

B X A      =   - A X B   

 Let us say the vectors are respectively  represented by:

A  =  a  _x i   +  a  _y  j   +   a  _z    k     

B  =   b  _x  i   +  b  _y   j   +   b  _z  k     

Then the cross product can be conveniently written as a 3rd order determinant:

A X  B  =   [  i         j         k ]

                    [  a  _x     a  _y     a  _z ]

                    [  b  _x     b  _y    b  _z]

Triple Scalar Product:

We refer to the product: (A X B) ·  C   as the triple scalar product, given that it is derived from vector dimensions in a solid geometry setting.  On inspection of the diagram below, of a paralleilepiped, we see the vector N = A X B is normal to the base (determined by the vectors A, B) so equals to the area of that base. Thus:

(A X B) ·  C        =   ‖N‖ ‖C‖   cos  Θ

 

Thus:  N = A X B= area of the base

And:  ‖C‖   cos  Θ   =  + altitude of the box

Given A is defined:  A  =  a ₁ i   + a ₂   j   +   a ₃  k    

And:

B  =  b ₁  i   + b ₂    j   +   b ₃  k    

And C similarly, with:

(A X B) ·  C    =   A  · B X  C   

The triple scalar product can be expressed:

(A X B) ·  C          =   [a  ₁     a  ₂       a  ₃  ]

                                      [ b  ₁     b  ₂      b  ₃]

                                      [  c  ₁     c  ₂       c  ₃]

Triple Vector Product:

Represent the product of three vectors A, B and C with two equations that are companions of each other:

(A X B) X  C    =   (A ·  C  ) B   -  (B ·  C  ) A            

And:

A X  (B  X  C)   =  (A ·  C  ) B   -    (A ·  B  ) C   

 Example Problem:     Show the two identities are equal by computing:

 A X  (B  X  C)  two different ways,    if:

A = i – j + 2k

B = 2i + j + k

C = i + 2j - k

Solution:

We first use:

(A X B) X  C    =   (A ·  C  ) B   -  (B ·  C  ) A            

Whence:

A ·  C    =  -3  And  B ·  C  =  3

And:

(A ·  C  ) B   -  (B ·  C  ) A      =  -3B  - 3 A

(A ·  C  ) B   -  (B ·  C  ) A      =  -3B  - 3 A

=    -3 (2i + j + k)  -  3(i – j + 2k ) = -6i – 3i – 3k – 6k

=    -  9i    - 9 k

Using  the method of determinants:

A X  B  =   [  i         j         k ]

                    [  a  _x     a  _y      a  _z]

                    [  b  _x     b  _y     b  _z]

So:

A X  B  =   [  i         j         k ]

                     [  1        -1         2]

                     [  2        1         1  ]

=   – 3i + 3j +  3 k

But:

(A X B) X  C      =   [  i         j         k ]

                                    [  -3        3        3 ]

                                    [ 1          2         -1]

=    -  9i    - 9 k

Problems:

1)     For a vector A = 3i – 2j – k find the magnitude of the vector product: A X A.

2)  A cube has edges of length 2a, with center at 0 and its sides are parallel to the coordinate planes of an xyz coordinate system.  What are the position vectors of the corners?

3) Find A X B if A = 2i - 2j -k, and B = i + j + k.

4) Find a vector perpendicular to both of the vectors A = i + j + k  and   B = i + j

5) Let   A = i + j,   B = 2i - 3j -k   and: C = 4j - 3k

Show that: (A X B) X  C    =  A X (B X  C) 

Using the determinant method.