Answer to Question #276634 in Linear Algebra for Nikhil

Question

Complete { (2, 0, 3)} to form an orthonormal basis of R^3

Answer 1

Orthogonal vector to one given vector (a,b,c) we take as (0,-c,b).

So for $x_1=(2,0,3)$ we have $x_2=(0,-3,0)$ with property $x_1\cdot x_2=2\cdot0+0\cdot(-3)+3\cdot 0=0$ .

We serach third vector in the form $x_3=(a,b,c)$ . We must provide that $x_1\cdot x_3=2\cdot a+0\cdot b+3\cdot c=2\cdot a+3\cdot c=0$ and $x_2\cdot x_3=0\cdot a+(-3)\cdot b+0\cdot c=(-3)\cdot b=0$

Thus we have the linear system:

$\begin{cases} 2a+3c=0 \\ -3b=0 \end{cases}$

This system has 3 unknowns: a,b,c, we may take one of them with arbitrary non zero value, let we assign c=2, then from the system we get a=-3,b=0 therefore x₃=(-3,0,2).

After orthogonalization we provide normalization:

$x'_1=\frac{x_1}{||x_1||}=\frac{(2,0,3)}{\sqrt{2^2+0^2+3^2}}=\frac{(2,0,3)}{\sqrt{13}}=\left( \frac{2}{\sqrt{13}},0,\frac{3}{\sqrt{13}} \right)$

$x'_2=\frac{x_2}{||x_2||}=\frac{(0,-3,0)}{\sqrt{0^2+(-3)^2+0^2}}=\frac{(0,-3,0)}{3}=\left( 0,-1,0 \right)$

$x'_3=\frac{x_3}{||x_3||}=\frac{(-3,0,2)}{\sqrt{(-3)^2+0^2+2^2}}=\frac{(-3,0,2)}{\sqrt{13}}=\left( \frac{-3}{\sqrt{13}},0,\frac{2}{\sqrt{13}} \right)$