### 4.5 Practice Problem

Look at Vector([2,-1,1]) and let L=Span of (2,-1,1) = {t (2,-1,1) where t is real}. Notice L is a line through the origin in R3 and we can graph the vector in Maple (without the arrowhead) as:
with(LinearAlgebra): with(plots):
a:=spacecurve({[2*t,-1*t,1*t,t=0..1]},color=red):
display(a);

Part 1: Find a vector w1 so that {Vector([2,-1,1]), w1} is a basis for some plane P1 [Hint: take something off of the line L]
Part 2: Find a vector w2 not on the same line through the origin as w1 from Part 1 so that {Vector([2,-1,1]), w2} is also a basis for the same plane P1 in Part 1 [Hint: if you are stuck, then jump to Part 4].
Part 3: In Maple, use spacecurve commands and display (as above) to show that all three vectors lie in the same plane but no 2 are on the same line [use different colors like black, blue, green..., and one display command like display(a,b,c);, and rotate to see this]
Part 4: Describe all the vectors w for which {Vector([2,-1,1]), w} is a basis for the same plane P1 [Hint - linear combinations are in the same geometric space so think about what linear combinations
a Vector([2,-1,1]) + b w1
you can use that will give a basis ie what a's and b's you can use to not give you Vector([2,-1,1])].
Part 5: Find a vector u so that {Vector([2,-1,1]), u} is a basis for a different plane P2 through the origin.
Part 6: Add u to your graph from Part 3 to show it lies outside the plane.

Print your Maple work and graphs.