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Two vectors are perpendicular to each other if their dot product is zero,

Let us suppose that $\stackrel{\mathbf{鈫�}}{\mathbf{x}}$and $\stackrel{\mathbf{鈫�}}{\mathbf{y}}$are two perpendicular then these two vectors are said to be perpendicular, if
$\stackrel{\mathbf{鈫�}}{\mathbf{x}}\mathbf{.}\stackrel{\mathbf{鈫�}}{\mathbf{y}}\mathbf{=}\mathbf{0}$

The three vectors are given by,

_{$\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]$, $\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$,$\left[\begin{array}{c}1\\ 9\\ 9\\ 7\end{array}\right]$}

Let us assume that the,

$\stackrel{鈫�}{\mathrm{v}}=\left[\begin{array}{c}{\mathrm{v}}_1\\ {\mathrm{v}}_2\\ {\mathrm{v}}_3\\ {\mathrm{v}}_4\end{array}\right]$

The dot product of$\stackrel{鈫�}{\mathrm{v}}$ vector with the given three vectors should be equal to zero,

$\begin{array}{rcl}\left(1\right){\mathrm{v}}_1+\left(1\right){\mathrm{v}}_2+\left(1\right){\mathrm{v}}_3+\left(1\right){\mathrm{v}}_4& =& 0\\ {\mathrm{v}}_1+{\mathrm{v}}_2+{\mathrm{v}}_3+{\mathrm{v}}_4& =& 0\end{array}$

The dot product of$\stackrel{鈫�}{\mathrm{v}}$ vector with the second given vector,

role="math" localid="1660716565341" $\begin{array}{rcl}\left(1\right){\mathrm{v}}_1+\left(2\right){\mathrm{v}}_2+\left(3\right){\mathrm{v}}_3+\left(4\right){\mathrm{v}}_4& =& 0\\ 1{\mathrm{v}}_1+2{\mathrm{v}}_2+3{\mathrm{v}}_3+4{\mathrm{v}}_4& =& 0\end{array}$

The dot product of$\stackrel{鈫�}{\mathrm{v}}$ vector with the third given vector,

$\begin{array}{rcl}\left(1\right){\mathrm{v}}_1+\left(9\right){\mathrm{v}}_2+\left(9\right){\mathrm{v}}_3+\left(7\right){\mathrm{v}}_4& =& 0\\ 1{\mathrm{v}}_1+9{\mathrm{v}}_2+9{\mathrm{v}}_3+7{\mathrm{v}}_4& =& 0\end{array}$

Make$3脳5$ matrix from the above three calculated equation, so that columns are the components of v and the last column is the value of dot product which is equal to 0.

$\left[\begin{array}{ccccc}1& 1& 1& 1& 0\\ 1& 2& 3& 4& 0\\ 1& 9& 9& 7& 0\end{array}\right]$

Now reduce the above matrix,

Subtract row 1 from row 2 and row 3,

$\left[\begin{array}{ccccc}1& 1& 1& 1& 0\\ 0& 1& 2& 3& 0\\ 1& 8& 8& 6& 0\end{array}\right]$

Use operation data-custom-editor="chemistry" ${\mathrm{R}}_1鈫�{\mathrm{R}}_1-{\mathrm{R}}_2$and${\mathrm{R}}_3鈫�{\mathrm{R}}_3-8{\mathrm{R}}_2$

$\left[\begin{array}{cccc}1& 0& -1& -2\\ 0& 1& 2& 3\\ 0& 0& -8& -18\end{array}\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

Use operation, data-custom-editor="chemistry" ${\mathrm{R}}_3鈫�-\frac{{\mathrm{R}}_3}{8}$

$\left[\begin{array}{ccccc}1& 0& -1& -2& 0\\ 0& 1& 2& 3& 0\\ 0& 0& 1& \frac{9}{4}& 0\end{array}\right]$

Use operation, data-custom-editor="chemistry" ${\mathrm{R}}_1鈫�{\mathrm{R}}_1+{\mathrm{R}}_3$anddata-custom-editor="chemistry" ${\mathrm{R}}_2鈫�{\mathrm{R}}_2-2{\mathrm{R}}_3$

$\left[\begin{array}{ccccc}1& 0& 0& \frac{1}{4}& 0\\ 0& 1& 2& \frac{-3}{2}& 0\\ 0& 0& 1& \frac{9}{4}& 0\end{array}\right]$

Now from here we can write each component of the vectordata-custom-editor="chemistry" $\stackrel{鈫�}{\mathrm{v}}$ in terms ofdata-custom-editor="chemistry" ${\mathrm{v}}_4$ as follows,

$\begin{array}{rcl}{\mathrm{v}}_1+\frac{1}{4}{\mathrm{v}}_4& =& 0\\ {\mathrm{v}}_1& =& -\frac{1}{4}{\mathrm{v}}_4\end{array}$

Similarly will have,

$$\begin{gathered} {\mathrm{v}}_2=\frac{3}{2}{\mathrm{v}}_4 \\ {\mathrm{v}}_3=\frac{9}{4}{\mathrm{v}}_4 \end{gathered}$$

Let us assume that for some arbitrarydata-custom-editor="chemistry" ${\mathrm{v}}_4=4\mathrm{t}$ (for real number t), we have the solution as,

data-custom-editor="chemistry" $\stackrel{鈫�}{\mathrm{v}}=\left[\begin{array}{c}-\mathrm{t}\\ 6\mathrm{t}\\ -9\mathrm{t}\\ 4\mathrm{t}\end{array}\right]$.