Answer to Question #274142 in Linear Algebra for Nikhil

Question

Find an orthonormal basis of R^3, of which (0,3√13, 2√13) is one element

Answer 1

The length of the vector $\vec v_1$ is $1$ as

$$||\vec v_1 ||=\sqrt{(0)^2+(3/\sqrt{13})^2+(2/\sqrt{13})^2}=1$$

We want to find two vectors $\vec v_2,\vec v_3$ such that $\{\vec v_1,\vec v_2,\vec v_3\}$is an orthonormal basis for $\R^3.$

Let $\vec u=\langle x, y, z \rangle$ be a vector that is perpendicular to $\vec v_1.$

Then we have $\vec u\cdot \vec v_1=0,$ and hence we have the relation

$$0(x)+(3/\sqrt{13})(y)+(2/\sqrt{13})(z)=0$$

$$3y+2z=0$$

For example, the vector $\vec u=\langle 1, 2, -3 \rangle$ satisfies the relation, anf hence $\vec u\cdot \vec v_1=0.$

Let us define the third vector $\vec w$ to be the cross product of $\vec u$ and $\vec v_1$

$$\vec w=\vec u\times\vec v_1=\begin{vmatrix} \vec i & \vec j & \vec k \\ 1 & 2 & -3 \\ 0 & 3/\sqrt{13} & 2/\sqrt{13} \end{vmatrix}$$

$$=\vec i\begin{vmatrix} 2 & -3 \\ 3/\sqrt{13} & 2/\sqrt{13} \end{vmatrix}-\vec j\begin{vmatrix} 1 & -3 \\ 0 & 2/\sqrt{13} \end{vmatrix}$$

$$+\vec k\begin{vmatrix} 1 & 2 \\ 0 & 3/\sqrt{13} \end{vmatrix}$$

$$=(13/\sqrt{13})\vec i-(2/\sqrt{13})\vec j+(3/\sqrt{13})\vec k$$

By the property of the cross product, the vector $\vec w$ is perpendicular to both $\vec u, \vec v_1.$

$$||\vec u||=\sqrt{(1)^2+(2)^2+(-3)^2}=\sqrt{14}$$

$$||\vec w||=\sqrt{(13/\sqrt{13})^2+(-2/\sqrt{13})^2+(3/\sqrt{13})^2}$$

$$=\sqrt{14}$$

Therefore, the set

$$\{\vec v_1,\vec v_2,\vec v_3\}$$

$$=\bigg\{\begin{bmatrix} 0 \\ 3/\sqrt{13}\\ 2/\sqrt{13} \end{bmatrix},\begin{bmatrix} 1/\sqrt{14} \\ 2/\sqrt{14}\\ -3/\sqrt{14} \end{bmatrix},\begin{bmatrix} 13/\sqrt{182} \\ 3/\sqrt{182}\\ 2/\sqrt{182} \end{bmatrix}\bigg\}$$

is an orthonormal basis for $\R^3.$