Answer to Question #275035 in Linear Algebra for Nikhil

Question

Find an orthonormal basis of R^3 of which (1√10,0,-3/√10) is one element

Answer 1

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

$$||\vec v_1 ||=\sqrt{(1/\sqrt{10})^2+(0)^2+(-3/\sqrt{10})^2}=1$$

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

$$(1/\sqrt{10})(x)+(0)(y)+(-3/\sqrt{10})(z)=0$$

$$x-3z=0$$

For example, the vector $\vec u=\langle 3, 1, 1 \rangle$ 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 $\vec v_1$

$$\vec w=\vec u\times\vec v_1=\begin{vmatrix} \vec i & \vec j & \vec k \\ 3 & 1 & 1 \\ 1/\sqrt{10} & 0 & -3/\sqrt{10} \end{vmatrix}$$

$$=\vec i\begin{vmatrix} 1 & 1 \\ 0 & -3/\sqrt{10} \end{vmatrix}-\vec j\begin{vmatrix} 3 & 1 \\ 1/\sqrt{10} & -3/\sqrt{10} \end{vmatrix}$$

$$+\vec k\begin{vmatrix} 3 & 1 \\ 1/\sqrt{10} & 0 \end{vmatrix}$$

$$=(-3/\sqrt{10})\vec i+(10/\sqrt{10})\vec j+(-1/\sqrt{10})\vec k$$

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

$$||\vec u||=\sqrt{(3)^2+(1)^2+(1)^2}=\sqrt{11}$$

$$||\vec w||=\sqrt{(-3/\sqrt{10})^2+(10/\sqrt{10})^2+(-1/\sqrt{10})^2}$$

$$=\sqrt{11}$$

Therefore, the set

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

$$=\bigg\{\begin{bmatrix} 1/\sqrt{10} \\ 0 \\ -3/\sqrt{10} \end{bmatrix},\begin{bmatrix} 3/\sqrt{11} \\ 1/\sqrt{11}\\ 1/\sqrt{11} \end{bmatrix},\begin{bmatrix} -3/\sqrt{110} \\ 10/\sqrt{110}\\ -1/\sqrt{110} \end{bmatrix}\bigg\}$$

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