Maths question

Let u=<u1,u2,u3> , v=<v1,v2,v3>
Prove that |u.v| <= |u||v|
How to prove? This is Vectors.
Answer #1
no idea
This post is considered spam.
Add an useful, elaborate and on-topic comment next time.
Kindly visit our rules:
www..org/rules
Medic
Answer #2
It is unclear if you are asking for vectors in R^3 (i.e. as in calculus 3) or if you are concerned with vectors in an inner product space. I will respond to both possibilities.
First, in the general inner product sense. Note: for any vector x we have ||x||^2 = x.x where “.” is the dot product. We will also make use of the following properties of dot products:
a.b = b.a
a . (b + c) = a.b + a.c
And for the proof:
||u + v||^2 + ||u – v||^2 = (u + v).(u+v) + (u – v).(u-v)
= u.(u+v) + v.(u + v) + u.(u – v) – v.(u – v)
= u.u + u.v + v.u + v.v + u.u – u.v -v.u + v.v
= 2u.u + 2v.v
= 2(u.u + v.v)
= 2(||u||^2 + ||v||^2)
This concludes the proof.
In particular if you are working in vectors in R^3 then let
u = (u1,u2,u3), v = (v1,v2,v3)
Then:
||u + v||^2 + || u – v ||^2 = ||(u1+v1, u2+v2,u3+v3)||^2 + ||(u1-v1, u2 – v2, u3-v3)||^2
= (u1 + v1)^2 + (u2 + v2)^2 + (u3+v3)^2 + (u1 – v1)^2
+ (u2 – v2)^2 + (u3 – v3)^2
After expanding:
= 2(u1)^2 + 2(u2)^2 + 2(u3)^2 + 2(v1)^2 + 2(v2)^2 + 2(v3)^2
= 2||u||^2 + 2||v||^2
= 2(||u||^2 + ||v||^2)
Geometrically this is the “parallelogram law”
It says that if we square the lengths of the two diagonals of a parallogram and add them together, then we should get the same number as we would by squaring the lengths of the four sides and adding them.