Question

Solve `vec u - x vec v = 3 vec u - x^2 vec v` for `vec u` parallel to `vec v` ?

Answer 1

See below.

Explanation:

If `vec u` and `vec v` are parallel then

`vec v = lambda vec u` for some `lambda in RR` and then

`vec u -x lambda vec u = 3 vec u -x^2 lambda vec u` or

`(1-x lambda-3+x^2 lambda) vec u = vec 0`

but `vec u ne vec 0` then

`1-x lambda-3+x^2 lambda=0`

and solving for `x`

`x = 1/2(1 pm 1/(sqrt[lambda/(8 + lambda)]))`