Answer to Question #213086 in Linear Algebra for Dhruv rawat

Question

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

Answer 1

Let v = <2,0,3>

We need to find three vectors in R³ such that they form an orthogonal basis.

Let the vectors be u₁, u₂ , u₃ .

Let v = $\dfrac{1}{3}$ <$\dfrac{2}{3}, 0, \dfrac{3}{3}$ >

v = $\dfrac{1}{3}$ u₁

Hence, we compute that u₁ = <$\dfrac{2}{3}, 0, \dfrac{3}{3}$ >

Our first goal is to find the vectors u₂ and u₃  such that {u₁ ,u₂ , u₃} is an orthogonal basis for R³ .

Let there be a vector x = <x, y, z>

such that x . u₁ = 0.

Hence form the above equation we have

$\dfrac{2}{3} x + z = 0$

Now let vector u₂ = <-3, 0, 2> satisfies the above relation.

So, we conclude that u₂ = <-3, 0, 2>

We can find u₃ = u₁ X u₂

u₃ = <$\dfrac{2}{3}, 0, \dfrac{3}{3}$ > X < - 3, 0, 2>

u₃ = $\begin{vmatrix} i & j &k\\ 2/3 & 0 & 1\\ -3 & 0 &2 \end{vmatrix}$

u₃ = <0, $-\dfrac{13}{3}$ , 0>

Thus we have found the vectors u₁, u₂ and u₃ .

To normalize them we divide the vectors by their length. Let v₁ , v₂ and v₃ be the corresponding unit vectors.

v₁ = $\dfrac{3}{\sqrt{13}} <\dfrac{2}{3}, 0, 1>$

v₂ = $\dfrac{1}{{\sqrt{13}}} <-3, 0, 2>$

v₃ = $\dfrac{3}{13} <0, -\dfrac{13}{3}, 0>$

{v₁, v₂, v₃} = {$\dfrac{3}{\sqrt{13}} <\dfrac{2}{3}, 0, 1>$ , $\dfrac{1}{{\sqrt{13}}} <-3, 0, 2>$ , $\dfrac{3}{13} <0, -\dfrac{13}{3}, 0>$ } form an orthonormal

basis on R³ containing the vector <2, 0, 3> .