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.