We need to find vectors:
e 1 ′ = ( x 1 , y 1 , z 1 ) e 2 ′ = ( x 2 , y 2 , z 2 ) e 3 ′ = ( x 3 , y 3 , z 3 ) e'_1 = (x_1,y_1,z_1) \\
e'_2 = (x_2,y_2,z_2) \\
e'_3 = (x_3,y_3,z_3) \\ e 1 ′ = ( x 1 , y 1 , z 1 ) e 2 ′ = ( x 2 , y 2 , z 2 ) e 3 ′ = ( x 3 , y 3 , z 3 ) The condition equals to three systems of three equations. Each system will give us one vector for the dual base.
First vector:
{ x 1 + z 1 = 1 x 1 + y 1 = 0 y 1 + z 1 = 0 \begin{cases}
x_1+z_1=1 \\
x_1+y_1=0 \\
y_1+z_1=0 \\
\end{cases} ⎩ ⎨ ⎧ x 1 + z 1 = 1 x 1 + y 1 = 0 y 1 + z 1 = 0
x 1 = 1 2 ; y 1 = − 1 2 ; z 1 = 1 2 x_1=\frac{1}{2}; y_1=-\frac{1}{2}; z_1=\frac{1}{2} x 1 = 2 1 ; y 1 = − 2 1 ; z 1 = 2 1
e 1 ′ = ( 1 2 , − 1 2 , 1 2 ) e'_1 = (\frac{1}{2},-\frac{1}{2},\frac{1}{2}) \\ e 1 ′ = ( 2 1 , − 2 1 , 2 1 ) Second vector:
{ x 2 + z 2 = 0 x 2 + y 2 = 1 y 2 + z 2 = 0 \begin{cases}
x_2+z_2=0 \\
x_2+y_2=1 \\
y_2+z_2=0 \\
\end{cases} ⎩ ⎨ ⎧ x 2 + z 2 = 0 x 2 + y 2 = 1 y 2 + z 2 = 0
x 2 = 1 2 ; y 2 = 1 2 ; z 2 = − 1 2 x_2=\frac{1}{2}; y_2=\frac{1}{2}; z_2=-\frac{1}{2} x 2 = 2 1 ; y 2 = 2 1 ; z 2 = − 2 1
e 2 ′ = ( 1 2 , 1 2 , − 1 2 ) e'_2 = (\frac{1}{2},\frac{1}{2},-\frac{1}{2}) \\ e 2 ′ = ( 2 1 , 2 1 , − 2 1 ) Third vector:
{ x 3 + z 3 = 0 x 3 + y 3 = 0 y 3 + z 3 = 1 \begin{cases}
x_3+z_3=0 \\
x_3+y_3=0 \\
y_3+z_3=1 \\
\end{cases} ⎩ ⎨ ⎧ x 3 + z 3 = 0 x 3 + y 3 = 0 y 3 + z 3 = 1
x 3 = − 1 2 ; y 3 = 1 2 ; z 3 = 1 2 x_3=-\frac{1}{2}; y_3=\frac{1}{2}; z_3=\frac{1}{2} x 3 = − 2 1 ; y 3 = 2 1 ; z 3 = 2 1
e 3 ′ = ( − 1 2 , 1 2 , 1 2 ) e'_3 = (-\frac{1}{2},\frac{1}{2},\frac{1}{2}) \\ e 3 ′ = ( − 2 1 , 2 1 , 2 1 )
Answer: e 1 ′ = ( 1 2 , − 1 2 , 1 2 ) e'_1 = (\frac{1}{2},-\frac{1}{2},\frac{1}{2}) \\ e 1 ′ = ( 2 1 , − 2 1 , 2 1 ) ; e 2 ′ = ( 1 2 , 1 2 , − 1 2 ) e'_2 = (\frac{1}{2},\frac{1}{2},-\frac{1}{2}) \\ e 2 ′ = ( 2 1 , 2 1 , − 2 1 ) ; e 3 ′ = ( − 1 2 , 1 2 , 1 2 ) e'_3 = (-\frac{1}{2},\frac{1}{2},\frac{1}{2}) \\ e 3 ′ = ( − 2 1 , 2 1 , 2 1 ) .
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