Answer to Question #223069 in Linear Algebra for tanu

Question

Find two vectors of norm 1 that are orthogonal to the vectors u = (2,1, −4,0), v = (−1, −1,2,2),

and w = (3,2,5,4).

Answer 1

Let vector $x=(a, b, c, d)$ is orthogonal to the vectors $u=(2,1,-4,0), v=(-1,-1,2,2),$

and $w=(3,2,5,4).$ Then

$$\begin{matrix} 2a+b-4c=0 \\ -a-b+2c+2d=0 \\ 3a+2b+5c+4d=0 \\ \end{matrix}$$

Augmented matrix

$$A=\begin{pmatrix} 2 & 1 & -4 & 0 & & 0 \\ -1 & -1 & 2 & 2 & & 0 \\ 3 & 2 & 5 & 4 & & 0 \\ \end{pmatrix}$$

$R_1=R_1/2$

$$\begin{pmatrix} 1 & 1/2 & -2 & 0 & & 0 \\ -1 & -1 & 2 & 2 & & 0 \\ 3 & 2 & 5 & 4 & & 0 \\ \end{pmatrix}$$

$R_2=R_2+R_1$

$$\begin{pmatrix} 1 & 1/2 & -2 & 0 & & 0 \\ 0 & -1/2 & 0 & 2 & & 0 \\ 3 & 2 & 5 & 4 & & 0 \\ \end{pmatrix}$$

$R_3=R_3-3R_1$

$$\begin{pmatrix} 1 & 1/2 & -2 & 0 & & 0 \\ 0 & -1/2 & 0 & 2 & & 0 \\ 0 & 1/2 & 11 & 4 & & 0 \\ \end{pmatrix}$$

$R_2=-2R_2$

$$\begin{pmatrix} 1 & 1/2 & -2 & 0 & & 0 \\ 0 & 1 & 0 & -4& & 0 \\ 0 & 1/2 & 11 & 4 & & 0 \\ \end{pmatrix}$$

$R_1=R_1-R_2/2$

$$\begin{pmatrix} 1 & 0 & -2 & 2 & & 0 \\ 0 & 1 & 0 & -4& & 0 \\ 0 & 1/2 & 11 & 4 & & 0 \\ \end{pmatrix}$$

$R_3=R_3-R_2/2$

$$\begin{pmatrix} 1 & 0 & -2 & 2 & & 0 \\ 0 & 1 & 0 & -4& & 0 \\ 0 & 0 & 11 & 6 & & 0 \\ \end{pmatrix}$$

$R_3=R_3/11$

$$\begin{pmatrix} 1 & 0 & -2 & 2 & & 0 \\ 0 & 1 & 0 & -4& & 0 \\ 0 & 0 & 1 & 6/11 & & 0 \\ \end{pmatrix}$$

$R_1=R_1+2R_3$

$$\begin{pmatrix} 1 & 0 & 0 & 34/11 & & 0 \\ 0 & 1 & 0 & -4& & 0 \\ 0 & 0 & 1 & 6/11 & & 0 \\ \end{pmatrix}$$

$$x=(-\dfrac{34}{11}t, 4t, -\dfrac{6}{11}t, t), t\in \R$$

$$|x|=\dfrac{|t|}{11}\sqrt{(-34)^2+(44)^2+(-6)^2+(11)^2}=\dfrac{57}{11}|t|$$

$$t=\frac{11}{57}:\,x_1=(-\dfrac{34}{57}, \dfrac{44}{57}, -\dfrac{6}{57}, \dfrac{11}{57})$$

$$t=-\frac{11}{57}:\,x_2=(\dfrac{34}{57}, -\dfrac{44}{57}, \dfrac{6}{57}, -\dfrac{11}{57})$$