Find two non parallel vectors perpendicular to the line

(x,y,z)= (1,-1,2)+t ( 1,-2,1)

Two vectors are perpendicular if their dot product is zero.

If A = (a1, a2, a3)

And B = (b1, b2, b3).

Then the dot product of A and B is A•B = a1.b1 + a2.b2 + a3.b3.

A vector C = (cx, cy) has a perpendicular vector Cp = (cy, -cx).

N.B. their dot product, C•Cp = 0.

Start with the vector V = (1+ t, -1 – 2t, 2 + t), our original vector.

Using the properties of perpendicular vectors, choose a vector A = (-1 – 2t, -1 – t, 0).

Where V = (vx, vy, vz) and A = (vy, -vx, 0).

Then the dot product A•V = (-1 – 2t, -1 – t, 0) •(1+ t, -1 – 2t, 2 + t )

A•V = -1 – 2t – t – 2t^2 + 1 + t + 2t + 2t^2 + 0(2 + t)

A•V = 0 + 0 + 0 = 0.

__A____•____V = 0, hence they are perpendicular.__

__A = (-1 – 2t, -1 – t, 0)__, and both A and V depend upon the same value for t.

Let B = (α, β, γ).

Then B•V = (α, β, γ) •(1 + t, -1 – 2t, 2 + t) = α(1 + t) + β(-1 -2t) + γ(2 + t).

B•V = (α - β + 2γ) + t(α - 2β + γ)

Then,

α - β + 2γ = 0

α - 2β + γ = 0

Setting γ = 1,

We get, __α = -3, β = -1, γ = 1__

We can choose another perpendicular vector to be B = (-3, -1, 1).

Then the dot product B•V = (-3, -1, 1) •(1 + t, -1 – 2t, 2 + t)

B•V = -3 – 3t + 1 + 2t + 2 + t = 0

B•V = 0, hence they are perpendicular.

Hence __B = (-3, -1, 1)__, with no dependency upon the parameter t.

Since A and B are not scalar multiples of each other then they are non-parallel.

__The two non-parallel vectors are: A = (-1 – 2t, -1 - t, 0), B = (-3, -1, 1)__