Question

Find an orthonormal basis of the plane ${\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{x}}_{\mathbf{3}}\mathbf{=}\mathbf{0}$.

Short answer

The solution is the vectors ${\overrightarrow{\mathrm{u}}}_1=\left(\frac{\sqrt{2}}{2},0,-\frac{\sqrt{2}}{2\sqrt{2}}\right)$and ${\overrightarrow{\mathrm{u}}}_3=\left(-\frac{\sqrt{6}}{6},\frac{2\sqrt{6}}{6},\frac{-\sqrt{6}}{6}\right)$form an orthonormal basis for the given plane

Step-by-step solution

`Explanation of the solution

Consider the vectors that satisfy the equation of the given plane as follows.

$$\begin{gathered} \left\{({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3):{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3=0\right\} \\ =\left\{({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3):{\mathrm{x}}_3=-{\mathrm{x}}_1-{\mathrm{x}}_2\right\} \\ =\left\{({\mathrm{x}}_1,{\mathrm{x}}_2,-{\mathrm{x}}_1,-{\mathrm{x}}_2):{\mathrm{x}}_1,{\mathrm{x}}_2\in \mathrm{R}\right\} \\ =\left\{({\mathrm{x}}_1,0,-{\mathrm{x}}_1,)+(0,{\mathrm{x}}_2,-{\mathrm{x}}_2):{\mathrm{x}}_1,{\mathrm{x}}_2\in \mathrm{R}\right\} \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} =\left\{{\mathrm{x}}_1(1,0,-1)+{\mathrm{x}}_2(0,1,-1):{\mathrm{x}}_1,{\mathrm{x}}_2\in \mathrm{R}\right\} \\ =\mathrm{span}\left\{(1,0,-1),(0,1,-1)\right\} \end{gathered}$$

Thus, vectors ${\overrightarrow{\mathrm{V}}}_1=(1,0,-1)$and ${\overrightarrow{\mathrm{V}}}_2=(0,1,-1)$span the set of all vectors that belong to the given plane.

Additionally, since they have different non-zero entries they are linearly independent.

Therefore, they form a basis for the set of all vectors that belongs to this plane.

Now, let’s find an orthonormal basis$\left\{\overrightarrow{u_1},\overrightarrow{u_2}\right\}$of this plane using the Gram-Schmidt algorithm as follows.

$$\begin{gathered} {\overrightarrow{u}}_1=\frac{1}{||{\overrightarrow{v}}_1||}\overrightarrow{v_2} \\ =\frac{1}{\sqrt{1^2+0^2+{\left(-1\right)}^2}}\left(1,0,-1\right) \\ =\frac{1}{\sqrt{2}}\left(1,0,-1\right) \\ {\overrightarrow{u}}_1=\left(\frac{\sqrt{2}}{2},0,-\frac{\sqrt{2}}{2\sqrt{2}}\right) \end{gathered}$$

Now, simplify for${\overrightarrow{v}}_2^{\perp }$as follows.

$$\begin{gathered} {\overrightarrow{v}}_2^{\perp }={\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_2\right){\overrightarrow{u}}_2 \\ =\left(0,1,-1\right)-\left(\frac{\sqrt{2}}{2}.0+0.1+\left(-\frac{\sqrt{2}}{2}\right)\right)\left(\frac{\sqrt{2}}{2},0,-\frac{\sqrt{2}}{2}\right) \\ =\left(0,1,-1\right)-\frac{\sqrt{2}}{2}\left(\frac{\sqrt{2}}{2},0,-\frac{\sqrt{2}}{2}\right) \\ =\left(-\frac{1}{2},1,\frac{-1}{2}\right) \end{gathered}$$

Now, simplify for as follows.

$$\begin{gathered} {\overrightarrow{v}}_3=\frac{1}{||\overrightarrow{v_2^{\perp }}||}{v}_2^{\perp } \\ =\frac{1}{\sqrt{{\left(-{\displaystyle \frac{1}{2}}\right)}^2+1^2+{\left(-{\displaystyle \frac{1}{2}}\right)}^2}}\left(-\frac{1}{2},1,\frac{-1}{2}\right) \\ =\frac{1}{\sqrt{\left({\displaystyle \frac{6}{4}}\right)}}\left(-\frac{1}{2},1,\frac{1}{2}\right) \\ =\frac{2}{\sqrt{6}}\left(-\frac{1}{2},1,\frac{1}{2}\right) \\ {\overrightarrow{u}}_3=\left(-\frac{\sqrt{6}}{6},\frac{2\sqrt{6}}{6},\frac{-\sqrt{6}}{6}\right) \end{gathered}$$

Therefore, the vectors ${\overrightarrow{u}}_1=\left(\frac{\sqrt{2}}{2},0,-\frac{\sqrt{2}}{2\sqrt{2}}\right)$and ${\overrightarrow{u}}_3=\left(-\frac{\sqrt{6}}{6},\frac{2\sqrt{6}}{6},\frac{-\sqrt{6}}{6}\right)$form an orthonormal basis for the given plane.