Question

The dot product of two vectors,

$\overrightarrow{\mathbf{x}}\mathbf{=}\left[\begin{array}{c}{x}_1\\ x_2\\ .\\ .\\ x_n\end{array}\right]\mathbf{\&}\mathbf{}\overrightarrow{\mathbf{y}}\mathbf{=}\left[\begin{array}{c}{y}_1\\ y_2\\ .\\ .\\ y_n\end{array}\right]$in data-custom-editor="chemistry" ${\mathbf{R}}^{\mathbf{n}}$is defined by $\overrightarrow{\mathbf{x}}\mathbf{.}\overrightarrow{\mathbf{y}}\mathbf{=}{\mathbf{x}}_{\mathbf{1}}{\mathbf{y}}_{\mathbf{1}}\mathbf{+}{\mathbf{x}}_{\mathbf{2}}{\mathbf{y}}_{\mathbf{2}}\mathbf{+}{\mathbf{x}}_{\mathbf{3}}{\mathbf{y}}_{\mathbf{3}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathbf{x}}_{\mathbf{n}}{\mathbf{y}}_{\mathbf{n}}$

Note that the dot product of two vectors is a scalar.

We say that the vectors data-custom-editor="chemistry" $\overrightarrow{\mathbf{x}}$and data-custom-editor="chemistry" $\overrightarrow{\mathbf{y}}$are perpendicular if

data-custom-editor="chemistry" $\overrightarrow{\mathbf{x}}\mathbf{.}\overrightarrow{\mathbf{y}}\mathbf{=}\mathbf{0}$

Find all vectors indata-custom-editor="chemistry" ${\mathbf{R}}^{\mathbf{3}}$ perpendicular to

$\left[\begin{array}{c}1\\ 3\\ -1\end{array}\right]$.

Short answer

Thus, all the vectors that are perpendicular to$\left[\begin{array}{c}1\\ 3\\ -1\end{array}\right]$ are in the plane is$\left\{\mathrm{x},\mathrm{y},\mathrm{z}|\mathrm{x}+3\mathrm{y}-\mathrm{z}=0\right\}$.

Step-by-step solution

Concept Introduction

Two vectors are perpendicular to each other if their dot product is zero, Let us suppose that $\overrightarrow{\mathbf{x}}$and $\overrightarrow{\mathbf{y}}$are two perpendicular then these two vectors are said to be perpendicular, if

$\overrightarrow{\mathbf{x}}\mathbf{.}\overrightarrow{\mathbf{y}}\mathbf{=}\mathbf{0}$

Given data

The two vectors are given by,

$\overrightarrow{\mathrm{x}}=\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ .\\ .\\ {\mathrm{x}}_{\mathrm{n}}\end{array}\right]\&\overrightarrow{\mathrm{y}}=\left[\begin{array}{c}{\mathrm{y}}_1\\ {\mathrm{y}}_2\\ .\\ .\\ {\mathrm{y}}_{\mathrm{n}}\end{array}\right]$

And it is defined by,

data-custom-editor="chemistry" $\overrightarrow{\mathrm{x}}.\overrightarrow{\mathrm{y}}={\mathrm{x}}_1{\mathrm{y}}_1+{\mathrm{x}}_2{\mathrm{y}}_2+...+{\mathrm{x}}_{\mathrm{n}}{\mathrm{y}}_{\mathrm{n}}$

Find all the vectors that are perpendicular to the given vectors.

Let us assume that the $\overrightarrow{\mathrm{v}}$vector is $\left[\begin{array}{c}1\\ 3\\ -1\end{array}\right]$.

Let us assume that the vector perpendicular to the $\overrightarrow{\mathrm{w}}=\left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]$is $\overrightarrow{\mathrm{v}}$then,

$\overrightarrow{\mathrm{v}}.\overrightarrow{\mathrm{w}}=0$

Substitute the values of $\overrightarrow{\mathrm{v}}$and $\overrightarrow{\mathrm{w}}$into the above equation,

$\begin{array}{rcl}\left[\begin{array}{c}1\\ 3\\ -1\end{array}\right].\left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]& =& 0\\ & =& 1.\mathrm{x}+3.\mathrm{y}+\left(-1\right)\mathrm{z}\\ & =& \mathrm{x}+3\mathrm{y}-\mathrm{z}\\ & \Rightarrow & \overrightarrow{\mathrm{v}}.\overrightarrow{\mathrm{w}}=\mathrm{x}+3\mathrm{y}-\mathrm{z}\end{array}$

As the above equaton is of the form of $\mathrm{ax}+\mathrm{by}+\mathrm{cz}=0$, which is the equation of the plane.

thus, all the vectors that are perpendicular to$\left[\begin{array}{c}1\\ 3\\ -1\end{array}\right]$ are in the plane is $\left\{\mathrm{x},\mathrm{y},\mathrm{z}|\mathrm{x}+3\mathrm{y}-\mathrm{z}=0\right\}$.