document.write( "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? \n" ); document.write( "
Algebra.Com's Answer #118841 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
Let's call the vectors u,v, and w.
\n" ); document.write( "u=(2,0,a)
\n" ); document.write( "v=(1,a,1)
\n" ); document.write( "w=(2,1,a)
\n" ); document.write( "For the set to be a basis, their dot products must be zero (orthogonality condition).
\n" ); document.write( "u*v=0
\n" ); document.write( "v*w=0
\n" ); document.write( "w*u=0
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\n" ); document.write( "1.u*v=2*1+0*a+a*1=0
\n" ); document.write( "1.\"2%2Ba=0\"
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\n" ); document.write( "2.v*w=1*2+a*1+1+a=0
\n" ); document.write( "2.\"2%2B2a=0\"
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\n" ); document.write( "3.w*u=2*2+1*0+a*a=0
\n" ); document.write( "3.\"4%2Ba%5E2=0\"
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\n" ); document.write( "From 1,
\n" ); document.write( "\"2%2Ba=0\"
\n" ); document.write( "\"a=-2\"
\n" ); document.write( "This value for a also needs to satisfy eq. 2 and eq. 3,
\n" ); document.write( "Checking with 2,
\n" ); document.write( "2+2a=0
\n" ); document.write( "2+2(-2)=-2
\n" ); document.write( "the value a does not solve 2 and you can check that it doesn't solve 3 either.
\n" ); document.write( "If the vectors formed a basis in R3, they must be orthogonal.
\n" ); document.write( "From 1,2, and 3, you cannot find an \"a\" that solves all three equations.
\n" ); document.write( "There is no solution.
\n" ); document.write( "Additionally, you could have also concluded this since eq. 3 has no real solution.
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