Answer to Question #276634 in Linear Algebra for Nikhil

Question #276634

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

1
Expert's answer
2021-12-07T13:34:32-0500

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

So for x1=(2,0,3)x_1=(2,0,3) we have x2=(0,3,0)x_2=(0,-3,0) with property x1x2=20+0(3)+30=0x_1\cdot x_2=2\cdot0+0\cdot(-3)+3\cdot 0=0 .

We serach third vector in the form x3=(a,b,c)x_3=(a,b,c) . We must provide that x1x3=2a+0b+3c=2a+3c=0x_1\cdot x_3=2\cdot a+0\cdot b+3\cdot c=2\cdot a+3\cdot c=0 and x2x3=0a+(3)b+0c=(3)b=0x_2\cdot x_3=0\cdot a+(-3)\cdot b+0\cdot c=(-3)\cdot b=0

Thus we have the linear system:

{2a+3c=03b=0\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:

x1=x1x1=(2,0,3)22+02+32=(2,0,3)13=(213,0,313)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)

x2=x2x2=(0,3,0)02+(3)2+02=(0,3,0)3=(0,1,0)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)

x3=x3x3=(3,0,2)(3)2+02+22=(3,0,2)13=(313,0,213)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)



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