### 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 R^{3} 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 **w**_{1} so that
{Vector([2,-1,1]), **w**_{1}}
is a basis for some plane P_{1} [Hint: take something off of the
line L]

**Part 2**:
Find a vector **w**_{2}
not on the same line through the origin as **w**_{1} from
Part 1
so that {Vector([2,-1,1]), **w**_{2}} is also a basis for the
*same*
plane P_{1} 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
P_{1} [Hint - linear combinations are in the same geometric space
so think about what linear combinations

*a* Vector([2,-1,1]) + *b* **w**_{1}

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 P_{2} 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.