Question

Consider the plane${\mathit{x}}_{\mathbf{1}}\mathbf{+}\mathbf{2}{\mathit{x}}_{\mathbf{2}}\mathbf{+}{\mathit{x}}_{\mathbf{3}}\mathbf{=}\mathbf{0}$with basis B consisting of vectors $\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right]\mathbf{}\mathbf{and}\mathbf{}\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right]$. If${\left[\overrightarrow{x}\right]}_{\mathbf{B}}\mathbf{=}\left[\begin{array}{c}2\\ -3\end{array}\right]$

Short answer

If we have a plane$x_1+2x_2+x_3=0$ with basis B consisting of vectors$\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right]$and $\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right]$ and${\left[\overrightarrow{\mathrm{x}}\right]}_{\mathrm{B}}=\left[\begin{array}{c}2\\ -3\end{array}\right]$ then$\overrightarrow{x}=\left[\begin{array}{c}4\\ -3\\ 2\end{array}\right]$ .

Step-by-step solution

Mentioning the concept

Let $\mathrm{B}=\left\{\overrightarrow{\mathrm{v}},\overrightarrow{\mathrm{w}}\right\}$be the basis of the given plane $x_1+2x_2+x_3=0$
such that

$\overrightarrow{v}=\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right]$and $\overrightarrow{w}=\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right]$

Then any vector $\overrightarrow{\mathrm{x}}$in the plane can be uniquely written as:

$\overrightarrow{\mathrm{x}}={\mathrm{c}}_1\overrightarrow{\mathrm{v}}+{\mathrm{c}}_2\overrightarrow{\mathrm{w}}$ (1)

where, ${\mathrm{c}}_1\&{\mathrm{c}}_2$are the B -coordinates of $\overrightarrow{x}$.

i.e

${\left[\overrightarrow{\mathrm{x}}\right]}_{\mathrm{B}}=\left[\begin{array}{c}{\mathrm{c}}_1\\ {\mathrm{c}}_2\end{array}\right]$

$\Rightarrow \left[\left.\begin{array}{c}2\\ -3\end{array}\right]=\left[\begin{array}{c}{\mathrm{c}}_1\\ {\mathrm{c}}_2\end{array}\right.\right]$

$\Rightarrow {\mathrm{c}}_1=2\&\hspace{0.17em}{\mathrm{c}}_2=-3\left(2\right)$

Finding the vector x→

From equations (1) & (2), we have

_{$\begin{array}{l}\overrightarrow{x}=2\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right]-3\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right]\\ \Rightarrow \overrightarrow{x}=\left[\begin{array}{c}-2+6\\ 0-3\\ 2-0\end{array}\right]=\left[\begin{array}{c}4\\ -3\\ 2\end{array}\right]\\ \Rightarrow \overrightarrow{x}=\left[\begin{array}{c}4\\ -3\\ 2\end{array}\right]\end{array}$}

Final answer  

If we have a plane$x_1+2x_2+x_3=0$ with basis B consisting of vectors$\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right]$and $\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right]$and${\left[\overrightarrow{\mathrm{x}}\right]}_{\mathrm{B}}=\left[\begin{array}{c}2\\ -3\end{array}\right]$ then$\overrightarrow{\mathrm{x}}=\left[\begin{array}{c}4\\ -3\\ 2\end{array}\right]$ .