SOLUTION: What is a value of a where vektors are (2,0,a), (1,a,1) and (2,1,a) make basis for R^3?

Question 161324: What is a value of a where vektors are (2,0,a), (1,a,1) and (2,1,a) make basis for R^3?
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
Let's call the vectors u,v, and w.
u=(2,0,a)
v=(1,a,1)
w=(2,1,a)
For the set to be a basis, their dot products must be zero (orthogonality condition).
u*v=0
v*w=0
w*u=0
.
.
.
1.u*v=2*1+0*a+a*1=0
1.
.
.
.
2.v*w=1*2+a*1+1+a=0
2.
.
.
.
3.w*u=2*2+1*0+a*a=0
3.
.
.
.
From 1,

This value for a also needs to satisfy eq. 2 and eq. 3,
Checking with 2,
2+2a=0
2+2(-2)=-2
the value a does not solve 2 and you can check that it doesn't solve 3 either.
If the vectors formed a basis in R3, they must be orthogonal.
From 1,2, and 3, you cannot find an "a" that solves all three equations.
There is no solution.
Additionally, you could have also concluded this since eq. 3 has no real solution.
RELATED QUESTIONS
Let S={v1, v2, v3}where v1^T=[1 0] v2^T=[0 1] v3^T=[-1 1] (answered by venugopalramana)

Given the following matrices over Z5: b1 b4 b7 b2 b5 b8 = B b3 b6 (answered by CPhill)

Given the following matrices: b1 b4 b7 b2 b5 b8 = B b3 b6 b9 (answered by CPhill)

For a fixed rate, a fixed principal amount, and a fixed compounding cycle, the return is... (answered by Nate)

Given the representation matrix of a linear transformation over the basis B, how to find... (answered by CPhill)

Hi,my name is Natalia. I solved two problems, but I'm not sure that I did it right. I... (answered by venugopalramana)

(a) If 3(a^2 + b^2 + c^2) = (a+b+c)^2, then the relation between a,b,c is ? (b) If P/a (answered by JBnovelwriter,ikleyn)

Consider the accompanying matrix. Use the test for linear independence to find a basis... (answered by venugopalramana)

Given n ≥ 1,(Mn×n(R), +, -) is a real vector space, where Mn×n(R) is the set of n ×... (answered by ikleyn)