1) If two vectors are collinear, i.e. angle between them is 0,their vector product is 0.
If two vectors are orthogonal, i.e. angle between them is pi/2, their scalar product is 0.
Thus, if one vector is i, the other one can be j, k, or any linear combination of j and k.
2) Suppose r = A X (B+C)-AXB-AXC----(i)
Now the scalar product of both sides with an arbitrary vector d,
d.r = d.[A X (B+C)-AXB-AXC]
=d.[A X (B+C)]-d.(AXB)-d.(AXC)---(ii)
(Since scalar product is distributive)
Now in a scalar triple product the positions of dot and cross can be interchanged without affecting its value.
Therefore from (ii),we get
So, Either d=0 or r=0 or d is perpendicular to r.But the vector d is arbitrary.
So d=0 or
3) If the body is at equilibrium, due to the 2nd Newton's Law,
A + B + C = 0.
By multiplying it (in sense of vector product) by A from theleft,
AxA + AxB + AxC = 0.
Some known rules:
1) Any vector product of collinear vectors is 0, so AxA =BxB = CxC = 0.
2) Changing the order of multipliers in the productchanges the sign: AxB = - BxA.
Thus, we have already AxB = - AxC = CxA. [i]
By multiplying the original equation in the same fashion byB, we get
BxA + BxC = 0;
AxB = - BxA = BxC. [ii]
Results [i] and [ii] provide enough evidence to say that A X B = B X C = CXA.
4) If a and b are vectors, and aXb is their vector product, the magnitude of it is
abs(aXb) = abs(a)*abs(b)*sin(a,b), where (a,b) is the angle between the vectors a and b.
If a.b is the scalar product of these two vectors, it is a.b= abs(a)*abs(b)*cos(a,b),
and the magnitude of the vector product is
abs(aXb) = (a.b) * sin(a,b) / cos(a,b) = (a.b)*tan(a,b).