Answer:
W⊥= (a, b, c)
a = v - (1/2)w
b = v
c = w
Basis for W⊥ = <(1,1,0),(-1/2,0,1)>
Step-by-step explanation:
The orthogonal complement of W is the set of vectors (a,b,c) that satisfy:
(x,y,z)·(a,b,c) = 0
ax + by + cz = 0
So, taking into account that x=12t, y=-12t and z=6t, the equation above is equal to:
12ta + -12tb + 6tc = 0
12a - 12b + 6c = 0
Then, if we made b=v and c=w and solve for a, we get:
12a - 12v + 6w = 0
a = (12v - 6w)/12
a = v - (1/2)w
Therefore, the orthogonal complement W⊥ of W has the form:
W⊥ = (a,b,c) = (v - (1/2)w, v, w)
On the other hand, we can write W⊥ as:
W⊥ = (v - (1/2)w, v, w)
W⊥ = (v,v,0) + ((-1/2)w,0,w)
W⊥ = v(1,1,0) + w(-1/2,0,1)
That means that we can write any vector of W⊥ as a linear combinations of the vectors (1,1,0) and (-1/2,0,1), so they are a basis for W⊥
5.8