What is therotical proof for Vector Product of two vectors?...Why is it perpendicular to the plane?

Answer 1

There is no proof per se of the vector product. It's a definition; not a theorem. It's defined because doing so and using it simplifies many equations in physics.

Answer 2

Pollux is right.

When I spotted this question I was eager to point out that it's a definition not a theorem.

It main usefulness, and original motivation, comes from physics, If you've ever twisted a wrench then you've generated a vector quatity, torque, by using the cross product of force and distance. If you've used a cheater bar on a wrench then you've increased the distance vector to increase the torque vector while using the same force vector.

A right handed coordinate system is one in which X cross Y = Z,
where X,Y,Z are vectors. It's also the right hand rule.

So, vector cross products are used big time in physics.

Answer 3

Even though its a definition, you can prove that it's always perpendicular to the plane in any particular dimention by taking the vector product of two generic vectors (e.g. x ) then showing that the resulting vector is perpendicular to every vector in the span (i.e. the plane) of the two vectors you multipled. This would be a lot of messy algebra, but it would prove that it is always perpendicular.

Answer 4

I expect that your question will be answered by knowing the historical derivation of vector algebra.

Why did the inventors of vector algebra make the definitions that we use today?

The mathmetician Hamilton figured out how to make an algebra to describe rotations in three dimensional space.

He defined quantities i, j, k that had the folllowing propertities.

i * i = j * j = k * k = -1

i * j = k ; j * k = i ; k * i = j

j * i = -k ; k * j = -i ; i * k = - j

This algebra, developed by Hamilton, and called quaterions,
has evolved into vector algebra and the original quaterion multiplication has been split into dot product and cross product.

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

i x i = j x j = k x k = 0

i x j = k ; j x k = i ; k x i = j

j x i = -k ; k x j = -i ; i x j = - j

Answer 5

this is basically a definition used via mathematicians and physicists. a x b = moda. modb. sintheta. n(hat) is the axiom somebody invented some years in the past it quite is sensible for some problems. extra problems upward thrust up while one realises that a x b isn't possibly a real vector yet a pseudo-vector yet this would not subject many human beings!

Answer 6

Yes, it is definition; Let
vector A=Ax*i + 0*j +0*k and
vector B= Bx*i +By*j +0*k = |B|*cos(phi)*i +|B|*sin(phi)*j,
where: –180 then according to definition cross-product C = [A*B] =

|``i`````j`````k|
|Ax````0````0| = Ax*By*k = Ax*|B|*sin(phi)*k,
|Bx```By```0|

hence dot products (C*A)=0 and (C*B)=0;
thus C is normal to both A and B; and |C| = |A|*|B|*sin(phi), where 0<=phi<180;

Answer 7

it is the outer product you mean

for the result of v = a x b
v is perpendicular to a and b

you can see this by the orthogonal projection of v onto the space spanned by axb
the projection will have only one point, thus _|_

Answer 8

The Vector Product of two vectors is:

Z = DET(i,j,k; x1,x2,x3; y1,y2,y3) =
(x2·y3-y2·x3, y1·x3-x1·y3, x1·y2-y1·x2)

z is perpendicular to x?

= x1(x2·y3-y2·x3)
+x2(y1·x3-x1·y3)
+x3(x1·y2-y1·x2)
= 0

Implies z is perp. to x.

z is perpendicular to y?

= y1(x2·y3-y2·x3)
+y2(y1·x3-x1·y3)
+y3(x1·y2-y1·x2)
= 0

Implies z is perp. to y and x and,
therefore, to the plane.