consider two vectors a=
, b= x,y,p and q are real numbers representing the length of the components of the unit vectors in the x and y direction (here x and y refer to the cordinate axis not the x,y real numbers mentioned earlier) usually labelled i and j (This is actually a special case because i and j are orthogonal ie at right angles to one another. In the general case they need not be.)
what this means is that a = < x, y > = xi + yj and b=
= pi + qj
remember x,y,p and q are scalar real numbers without direction, they contribute the magnitudes whilst i and j impart direction with just unit magnitude.
what is i.i ? By definition IiI.IiI cos0 = 1 x 1 x1 = 1 = IjI.IjI cos0
what about i.j and j.i well because they are at right angles to each other and cos 90 = 0 they are both zero
so what about a.b?
(xi + yj).(pi + qj) = xp(i.i) + xq(i.j) + yp(j.i) + yq(j.j)
= xp(1) + xq(0) + yp(0) + yq(1) = xp + yq which is a real number since
x,y,p and q are all real numbers.
This shouldn't come as any surprise as you accepted
a.b=lal.lbl.cos theta. but this is also a real number
IaI is just a number representing the magnitude of a
IbI is just a number representing the magnitude of b
and cos theta is of course a number
therefore for a = = xi + yj and b = = pi + qj
a.b=lal.lbl.cos theta = xp + yq
now because IaI = sqrt(x^2 + y^2) and IbI = sqrt(p^2 + q^2) by Pythagoras theorem
lal.lbl.cos theta = xp + yq
cos theta = (xp + yq)/lal.lbl =
(xp + yq)/(sqrt(x^2 + y^2)*sqrt(p^2 + q^2)
which gives a direct means of calculating cos theta and hence theta the angle between the two vectors a and b