Vector

what can u tell about vectors?

Vectars??

The dot product is the product of the cosine of the angle between two vectors and their length/magnitude.
It can be written it as:

A . B = |A||B|cos theta

• Its clear that A.B = B.A (commutativity)
• also (kA).B = k(A.B), for any scalar k
• Its should also be noted that A.(B+C) = A.B + A.C (distributive)

Now define 3 perpendicular vectors i, j and k, we can now state


i.i = j.j = k.k = 1
i.j = j.k = k.i = 0

if we have our vectors A and B in component form:

A = i*x1 + j*y1 + k*z1
B = i*x2 + j*y2 + k*z2

We can then find a compenent form expression for the dot product.


A.B = (i*x1 + j*y1 + k*z1).(i*x2 + j*y2 + k*z2)

    = (i*x1 + j*y1 + k*z1).(i*x2)
    + (i*x1 + j*y1 + k*z1).(j*y2)
    + (i*x1 + j*y1 + k*z1).(k*z2)

    = (i*x1).(i*x2) + (j*y1).(i*x2) + (k*z1).(i*x2)
    + (i*x1).(j*y2) + (j*y1).(j*y2) + (k*z1).(j*y2)
    + (i*x1).(k*z2) + (j*y1).(k*z2) + (k*z1).(k*z2)

    = (i.i)*x1*x2 + (j.i)*y1*x2 + (k.i)*z1*x2
    + (i.j)*x1*y2 + (j.j)*y1*y2 + (k.j)*z1*y2
    + (i.k)*x1*z2 + (j.k)*y1*z2 + (k.k)*z1*z2

    = (1)*x1*x2 + (0)*y1*x2 + (0)*z1*x2
    + (0)*x1*y2 + (1)*y1*y2 + (0)*z1*y2
    + (0)*x1*z2 + (0)*y1*z2 + (1)*z1*z2

    = x1*x2 + y1*y2 + z1*z2

Now the cross product is defined as the product of the sine of the angle between two vectors, their magnitude and a unit vector perpendicular to those vectors.


AxB = |A||B|sin(theta)n
    = (i*x1 + j*y1 + k*z1)×(i*x2 + j*y2 + k*z2)

we can make the following observations about the cross of the vectors i, j, k


x| i  j  k
----------
i| 0  k -j
j|-k  0  i
k| j -i  0

it should be noted this table can be made in two ways, left or right handed. So there are really two definitions of cross product but its conventional to use this one.

• The cross product is distributive like the dot product.

again in component form:

A×B = (i*×1 + j*y1 + k*z1)×(i*×2 + j*y2 + k*z2)

    = (i*×1 + j*y1 + k*z1)×(i*×2)
    + (i*×1 + j*y1 + k*z1)×(j*y2)
    + (i*×1 + j*y1 + k*z1)×(k*z2)

    = (i*×1 + j*y1 + k*z1)×(i*×2)
    + (i*×1 + j*y1 + k*z1)×(j*y2)
    + (i*×1 + j*y1 + k*z1)×(k*z2)

    = (i*×1)×(i*×2) + (j*y1)×(i*×2) + (k*z1)×(i*×2)
    + (i*×1)×(j*y2) + (j*y1)×(j*y2) + (k*z1)×(j*y2)
    + (i*×1)×(j*z2) + (j*y1)×(k*z2) + (k*z1)×(k*z2)

    =     0*×1*×2 +    -k*y1*×2 +     j*z1*×2
    +     k*×1*y2 +     0*y1*y2 +    -i*z1*y2
    +    -j*×1*z2 +     i*y1*z2 +     0*z1*z2

    = i*(y1*z2-z1*y2) + j*(×1*y2-y1*×2) + k*(×1*y2-y1*×2)

now if you have ever done any matrix math, you may notice that this is equal to the determinant of the following 3x3 matrix

/x1 y1 z1\
|x2 y2 z2|
\ i  j  k/