Question

Question: Consider the vector

$\overrightarrow{\mathbf{V}}\mathbf{=}\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$

In R⁴consisting of all vectors perpendicular to $\overrightarrow{\mathbf{V}}$.

Short answer

Any vector in the form$\overrightarrow{x}=\left[\begin{array}{c}-\left\{2{\overrightarrow{x}}_2+3{\overrightarrow{x}}_3+4{\overrightarrow{x}}_4\right\}\\ {\overrightarrow{x}}_2\\ {\overrightarrow{x}}_3\\ {\overrightarrow{x}}_4\end{array}\right]$ is perpendicular to$\overrightarrow{V}=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$ which is spanned by spanned by $\left\{\left(\begin{array}{cccc}-2& 1& 0& 0\end{array}\right),\left(\begin{array}{cccc}-3& 0& 1& 0\end{array}\right),\left(\begin{array}{cccc}-4& 0& 0& 1\end{array}\right)\right\}$.

Step-by-step solution

Determine orthogonal vector.

Consider a vector $\overrightarrow{V}=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$and $\overrightarrow{x}=\left[\begin{array}{c}{\overrightarrow{x}}_1\\ {\overrightarrow{x}}_2\\ {\overrightarrow{x}}_3\\ {\overrightarrow{x}}_4\end{array}\right]$.

If the vector $\overrightarrow{\mathbf{x}}$ is perpendicular $\overrightarrow{\mathbf{y}}$then $\overrightarrow{\mathbf{x}}\mathbf{\cdot }\overrightarrow{\mathbf{y}}\mathbf{=}\mathbf{0}$.

As$\overrightarrow{v}$ is perpendicular to $\overrightarrow{x}$, by the definition of orthogonal $\overrightarrow{x}\cdot \overrightarrow{v}=0$.

Simplify the equation$\overrightarrow{x}\cdot \overrightarrow{v}=0$ as follows.

$\begin{array}{c}\overrightarrow{x}\cdot \overrightarrow{v}=0\\ \left[\begin{array}{cccc}{\overrightarrow{x}}_1& {\overrightarrow{x}}_2& {\overrightarrow{x}}_3& {\overrightarrow{x}}_4\end{array}\right]\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]=0\\ {\overrightarrow{x}}_1+2{\overrightarrow{x}}_2+3{\overrightarrow{x}}_3+4{\overrightarrow{x}}_4=0\\ {\overrightarrow{x}}_1=-\left\{2{\overrightarrow{x}}_2+3{\overrightarrow{x}}_3+4{\overrightarrow{x}}_4\right\}\end{array}$

Substitute the value $-\left\{2{\overrightarrow{x}}_2+3{\overrightarrow{x}}_3+4{\overrightarrow{x}}_4\right\}$for $\overrightarrow{x_1}$in the equation $\overrightarrow{x}=\left[\begin{array}{c}{\overrightarrow{x}}_1\\ {\overrightarrow{x}}_2\\ {\overrightarrow{x}}_3\\ {\overrightarrow{x}}_4\end{array}\right]$as follows.

$\begin{array}{l}\overrightarrow{x}=\left[\begin{array}{c}{\overrightarrow{x}}_1\\ {\overrightarrow{x}}_2\\ {\overrightarrow{x}}_3\\ {\overrightarrow{x}}_4\end{array}\right]\\ \overrightarrow{x}=\left[\begin{array}{c}-\left\{2{\overrightarrow{x}}_2+3{\overrightarrow{x}}_3+4{\overrightarrow{x}}_4\right\}\\ {\overrightarrow{x}}_2\\ {\overrightarrow{x}}_3\\ {\overrightarrow{x}}_4\end{array}\right]\end{array}$

Therefore, any vector in the form$\begin{array}{l}\\ \overrightarrow{x}=\left[\begin{array}{c}-\left\{2{\overrightarrow{x}}_2+3{\overrightarrow{x}}_3+4{\overrightarrow{x}}_4\right\}\\ {\overrightarrow{x}}_2\\ {\overrightarrow{x}}_3\\ {\overrightarrow{x}}_4\end{array}\right]\end{array}$ is perpendicular to $\overrightarrow{V}=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$.

Determine basis.

Simplify the equation$\overrightarrow{x}=\left[\begin{array}{c}-\left\{2{\overrightarrow{x}}_2+3{\overrightarrow{x}}_3+4{\overrightarrow{x}}_4\right\}\\ {\overrightarrow{x}}_2\\ {\overrightarrow{x}}_3\\ {\overrightarrow{x}}_4\end{array}\right]$ as follows.

$\begin{array}{c}\overrightarrow{x}=\left[\begin{array}{c}-\left\{2{\overrightarrow{x}}_2+3{\overrightarrow{x}}_3+4{\overrightarrow{x}}_4\right\}\\ {\overrightarrow{x}}_2\\ {\overrightarrow{x}}_3\\ {\overrightarrow{x}}_4\end{array}\right]\\ =\left[\begin{array}{c}-2{\overrightarrow{x}}_2\\ {\overrightarrow{x}}_2\\ 0\\ 0\end{array}\right]+\left[\begin{array}{c}-3{\overrightarrow{x}}_3\\ 0\\ {\overrightarrow{x}}_3\\ 0\end{array}\right]+\left[\begin{array}{c}-4{\overrightarrow{x}}_4\\ 0\\ 0\\ {\overrightarrow{x}}_4\end{array}\right]\\ \overrightarrow{x}={\overrightarrow{x}}_2\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\end{array}\right]+{\overrightarrow{x}}_3\left[\begin{array}{c}-3\\ 0\\ 1\\ 0\end{array}\right]+{\overrightarrow{x}}_4\left[\begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\right]\end{array}$

Therefore, the vector $\overrightarrow{x}\in span\left\{\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-3\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\right]\right\}$.

Hence, the vector$\overrightarrow{x}$ is spanned by $\left\{\left(\begin{array}{cccc}-2& 1& 0& 0\end{array}\right),\left(\begin{array}{cccc}-3& 0& 1& 0\end{array}\right),\left(\begin{array}{cccc}-4& 0& 0& 1\end{array}\right)\right\}$which is perpendicular to$\overrightarrow{V}=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$ .