Vectors and scalars

Question

all the seven questions

1. The resultant of scalar product and vector product of two given vectors is zero. if one vector is i. What is the other vector?
2. If A,B,C are any three vectors, show that Ax(B+C) = AxB +AxC

3. A body is in equillibrium under the action of three vectors A,B and C simultaneously. Show that AxB = BxC = CxA

4. Express the magnitude of axb in terms of scalar products

5. Find a.b and axb if a = i+2j+k and |b| = 3acting along c = i+j+k
6. Prove that the vector area of triangle whose vertices are a,b,c is 1/2(bxc + cxa + axb)

Comments

do you need answers of all of the above question?

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yes
or otherwise solve as much questions you can

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ok giving you the answers

Answer 1

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

d.r=(dXA).(B+C)-(dXA).B-(dXA).C
=(dXA).B+(dXA).C-(dXA).B-(dXA).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)-AXB-AXC=0
i.e AX(B+C)=AXB+AXC

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).

Answer 2

Answer 5.  a= i + 2j + k 
|b| = magnitude of b = 3 
Firstly; the magnitude of a vector α, is given by: 
|α| = √[(α1)² + (α2)²], 
so for example, the magnitude of |a| is: 
|a| = √[(1)² + (2)² + (1)²] = √[1 + 4 +1] = √6 = 2.45 
also the dot product is given by: 
a.b = a1b1 + a2b2 + a3b3 
now onto the answer: 
vector c has a direction of (i+j+k)/√3 
and since vector b is along c vector b = 3(i+j+k)/√3 = √3(i+j+k) 
a.b = [i + 2j + k] . [ √3(i+j+k) ] 
applying the above formula: 
(1 x [√3 x 1]) + (2 x [√3 x 1]) + (1 x [√3 x 1]) = √3 + [2 x √3 ] + √3 = 6.928 or = 4√3 
now the cross product utilises a matrix (which is hard to do here, but i will provide the formula); it states: 
axb = <a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1> 
axb = 
| i j k | 
| 1 2 1 | 
| √3 √3 √3 | 
= i | (2 x √3 ) - (1 x √3 ) | - j | (1 x √3) - (√3 x 1)| + k | (1 x √3) - ( 2 x √3) | 
= √3 i - 0j + (-)√3 k 
= √3 i - √3 k 
= √3 (i-k) 
summary: 
a.b = 4√3 
axb = √3 (i-k)