Having a bit of trouble with vectors

[Lilly]

Let u and v be two (unspecied) vectors in R^n, Use properties of the dot product to
prove that:
||u + v||^2 + ||u - v||^2 = 2||u||^2 + 2||v||^2

If someone could help me with some rules of vectors as to how to prove this, that would be a great help!

 

[Deleted member 4993]

Let u and v be two (unspecied) vectors in R^n, Use properties of the dot product to
prove that:
||u + v||^2 + ||u - v||^2 = 2||u||^2 + 2||v||^2

If someone could help me with some rules of vectors as to how to prove this, that would be a great help!

Use law of parallelogram for vector addition:


Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[Lilly]

oh ok, so it's pretty much the law of parallelogram for vector addition
so how am I able to prove the law of parallelogram?

 

[Deleted member 4993]

oh ok, so it's pretty much the law of parallelogram for vector addition
so how am I able to prove the law of parallelogram?

Have you not been taught the law of parallelogram for addition of two vectors?

 

[Lilly]

no, I haven't
which is kind of strange considering the question. I've tried searching it on the internet to try and apply it to this question but it's still confusing me.

 

[pka]

Let u and v be two (unspecied) vectors in R^n, Use properties of the dot product to
prove that: ||u + v||^2 + ||u - v||^2 = 2||u||^2 + 2||v||^2

You should know that \(\displaystyle \|V\|^2=V\cdot V\).
So \(\displaystyle \|V+U\|^2=(V+U)\cdot(V+U)=V\cdot V+2V\cdot U+U\cdot U~\).

Now you show some work.

 

[Lilly]

You should know that \(\displaystyle \|V\|^2=V\cdot V\).
So \(\displaystyle \|V+U\|^2=(V+U)\cdot(V+U)=V\cdot V+2V\cdot U+U\cdot U~\).

Now you show some work.

Ok, I see.
So, ||v+u||² = (v+u).(v+u) = v.v + 2(v.u) +u.u
and ||v-u||² = (v-u).(v-u) = v.v -2(v.u) + u.u

so, v.v + 2(u.v) + u.u + v.v - 2(v.u) + u.u
= v.v + u.u + v.v + u.u
= ||v||² + ||u||² + ||v||² + ||u||²
= 2||v||² + 2||u||²

 

[Deleted member 4993]

Good work, Lilly !!!