**Answer:**

**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**)-AX**B**-**A**X**C**----(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.**(**A**X**B**)-**d**.(**A**X**C**)---(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

**d.r**=(**d**X**A**).(**B**+**C**)-(**d**X**A)**.**B**-(**d**X**A**).**C**

=(**d**X**A**).**B**+(**d**X**A)**.**C**-(**d**X**A**).**B**-(**d**X**A**).**C**=0

So, Either d=0 or r=0 or d is perpendicular to r.But the vector d is arbitrary.

So d=0 or

AX**(B+C)-A**X**B-A**X**C=0**

i.e** A**X**(B+C)=A**X**B+A**X**C**

**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,

we get

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).