# Mathematics exercise?

It says:

In R^3 is considered the inner product <x,y>=x₁y₁ + (x₁ + x₂).(y₁ + y₂) + x₃.y₃

where W = {u ∈ R^3 : (2,-3,-1) ⊥ u}:

a)Give an orthogonal basis B ={v₁, v₂, v₃} of R^3 such that W = span{v₁, v₂}.

b)Find, if possible, α ∈ R ||u - 2v|| = √(5) where u=(1,1,0) and v=(0,α,1).

Thanks!!

Relevance
• Anonymous
2 years ago

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• Anonymous
2 years ago

I’m guessing, but I think this is what is happening...

In the same way that the usual dot product a•b = 0 when a and b are orthogonal, I’m guessing orthogonality here means the inner product is zero: <a,b> = 0 means a and b are orthogonal (a⊥b).

{u ∈ R^3: (2,-3,-1) ⊥ u} is the set of vectors {u} which are orthogonal to (2, -3, 1). This is analogous to the simple 3D situation where (2,-3,-1) is normal to a plane and {u} are vectors in/parallel to the plane (all of which must be orthogonal to (2, -3, 1)).

Also, I assume we are meant to define the norm of a vector ||x|| using:

||x||² = <x,x>

____________________

a) B ={v₁, v₂, v₃}

Let v₃ = (2, -3, 1)

Then we need <v₃, v₁> = 0 and <v₃, v₂> = 0 and v₁ and v₂ to be linearly independent

<v₃,v₁> = 0

Let v₁ = (a.b,c), then using the definition <x,y>=x₁y₁ + (x₁ + x₂).(y₁ + y₂) + x₃.y₃ gives:

< v₃,v₁> = 2a + (2-3)(a+b) + 1c

. . . . . . .= 2a -a – b + c

. . . . . . .= c – a- b

We want < v₃, v₁> = 0 and can pick any values for a, b and c which make this true. So pick some simple values e.g.

a = 0, b=1, c =1

v₁ = (0, 1, 1)

Repeat the same process for v₂ but choose different values for a, b and c.

a = 1. b=0, c=1

v₂ = (1, 0, 1)

By inspection it is obvious that v₁ and v₂ are linearly independent.

v₁ = (0, 1,1)

v₂ = (1, 0, 1)

v₃ = (2, -3, 1)

_________________

b)

u - 2v = (1, 1, 0) – 2(0, α, 1)

. . . . . = (1, 1-2α, -2)

||u – 2v||² = <(1, 1-2α, -2), (1, 1-2α, -2)>

= 1*1 + (1+1-2α)(1+1-2α) + (-2)(-2)

= 1 + 4(1-α)² + 4

= 5 + 4(1-α)²

Since ||u - 2v|| = √(5)

5 + 4(1-α)² = 5

4(1-α)² = 0

α = 1