The length of the vector v ⃗ 1 \vec v_1 v 1 is 1 1 1 as
∣ ∣ v ⃗ 1 ∣ ∣ = ( 0 ) 2 + ( 3 / 13 ) 2 + ( 2 / 13 ) 2 = 1 ||\vec v_1 ||=\sqrt{(0)^2+(3/\sqrt{13})^2+(2/\sqrt{13})^2}=1 ∣∣ v 1 ∣∣ = ( 0 ) 2 + ( 3/ 13 ) 2 + ( 2/ 13 ) 2 = 1 We want to find two vectors v ⃗ 2 , v ⃗ 3 \vec v_2,\vec v_3 v 2 , v 3 such that { v ⃗ 1 , v ⃗ 2 , v ⃗ 3 } \{\vec v_1,\vec v_2,\vec v_3\} { v 1 , v 2 , v 3 } is an orthonormal basis for R 3 . \R^3. R 3 .
Let u ⃗ = ⟨ x , y , z ⟩ \vec u=\langle x, y, z \rangle u = ⟨ x , y , z ⟩ be a vector that is perpendicular to v ⃗ 1 . \vec v_1. v 1 .
Then we have u ⃗ ⋅ v ⃗ 1 = 0 , \vec u\cdot \vec v_1=0, u ⋅ v 1 = 0 , and hence we have the relation
0 ( x ) + ( 3 / 13 ) ( y ) + ( 2 / 13 ) ( z ) = 0 0(x)+(3/\sqrt{13})(y)+(2/\sqrt{13})(z)=0 0 ( x ) + ( 3/ 13 ) ( y ) + ( 2/ 13 ) ( z ) = 0
3 y + 2 z = 0 3y+2z=0 3 y + 2 z = 0 For example, the vector u ⃗ = ⟨ 1 , 2 , − 3 ⟩ \vec u=\langle 1, 2, -3 \rangle u = ⟨ 1 , 2 , − 3 ⟩ satisfies the relation, anf hence u ⃗ ⋅ v ⃗ 1 = 0. \vec u\cdot \vec v_1=0. u ⋅ v 1 = 0.
Let us define the third vector w ⃗ \vec w w to be the cross product of u ⃗ \vec u u and v ⃗ 1 \vec v_1 v 1
w ⃗ = u ⃗ × v ⃗ 1 = ∣ i ⃗ j ⃗ k ⃗ 1 2 − 3 0 3 / 13 2 / 13 ∣ \vec w=\vec u\times\vec v_1=\begin{vmatrix}
\vec i & \vec j & \vec k \\
1 & 2 & -3 \\
0 & 3/\sqrt{13} & 2/\sqrt{13}
\end{vmatrix} w = u × v 1 = ∣ ∣ i 1 0 j 2 3/ 13 k − 3 2/ 13 ∣ ∣
= i ⃗ ∣ 2 − 3 3 / 13 2 / 13 ∣ − j ⃗ ∣ 1 − 3 0 2 / 13 ∣ =\vec i\begin{vmatrix}
2 & -3 \\
3/\sqrt{13} & 2/\sqrt{13}
\end{vmatrix}-\vec j\begin{vmatrix}
1 & -3 \\
0 & 2/\sqrt{13}
\end{vmatrix} = i ∣ ∣ 2 3/ 13 − 3 2/ 13 ∣ ∣ − j ∣ ∣ 1 0 − 3 2/ 13 ∣ ∣
+ k ⃗ ∣ 1 2 0 3 / 13 ∣ +\vec k\begin{vmatrix}
1 & 2 \\
0 & 3/\sqrt{13}
\end{vmatrix} + k ∣ ∣ 1 0 2 3/ 13 ∣ ∣
= ( 13 / 13 ) i ⃗ − ( 2 / 13 ) j ⃗ + ( 3 / 13 ) k ⃗ =(13/\sqrt{13})\vec i-(2/\sqrt{13})\vec j+(3/\sqrt{13})\vec k = ( 13/ 13 ) i − ( 2/ 13 ) j + ( 3/ 13 ) k By the property of the cross product, the vector w ⃗ \vec w w is perpendicular to both u ⃗ , v ⃗ 1 . \vec u, \vec v_1. u , v 1 .
∣ ∣ u ⃗ ∣ ∣ = ( 1 ) 2 + ( 2 ) 2 + ( − 3 ) 2 = 14 ||\vec u||=\sqrt{(1)^2+(2)^2+(-3)^2}=\sqrt{14} ∣∣ u ∣∣ = ( 1 ) 2 + ( 2 ) 2 + ( − 3 ) 2 = 14
∣ ∣ w ⃗ ∣ ∣ = ( 13 / 13 ) 2 + ( − 2 / 13 ) 2 + ( 3 / 13 ) 2 ||\vec w||=\sqrt{(13/\sqrt{13})^2+(-2/\sqrt{13})^2+(3/\sqrt{13})^2} ∣∣ w ∣∣ = ( 13/ 13 ) 2 + ( − 2/ 13 ) 2 + ( 3/ 13 ) 2
= 14 =\sqrt{14} = 14 Therefore, the set
{ v ⃗ 1 , v ⃗ 2 , v ⃗ 3 } \{\vec v_1,\vec v_2,\vec v_3\} { v 1 , v 2 , v 3 }
= { [ 0 3 / 13 2 / 13 ] , [ 1 / 14 2 / 14 − 3 / 14 ] , [ 13 / 182 3 / 182 2 / 182 ] } =\bigg\{\begin{bmatrix}
0 \\
3/\sqrt{13}\\
2/\sqrt{13}
\end{bmatrix},\begin{bmatrix}
1/\sqrt{14} \\
2/\sqrt{14}\\
-3/\sqrt{14}
\end{bmatrix},\begin{bmatrix}
13/\sqrt{182} \\
3/\sqrt{182}\\
2/\sqrt{182}
\end{bmatrix}\bigg\} = { ⎣ ⎡ 0 3/ 13 2/ 13 ⎦ ⎤ , ⎣ ⎡ 1/ 14 2/ 14 − 3/ 14 ⎦ ⎤ , ⎣ ⎡ 13/ 182 3/ 182 2/ 182 ⎦ ⎤ }
is an orthonormal basis for R 3 . \R^3. R 3 .
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