Question

Consider the line $\mathit{L}$spanned by$\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$in${\mathbf{ℝ}}^{\mathbf{3}}$. Find a basis of ${\mathit{L}}^{\mathbf{\perp }}$. See Exercise 53.

Short answer

The basis for $L^{\perp }$is, $\left\{\left[\begin{array}{c}3\\ 0\\ -1\end{array}\right],\left[\begin{array}{c}2\\ -1\\ 0\end{array}\right]\right\}$.

Step-by-step solution

Consider the set

The vectors in ${\mathrm{ℝ}}^n$are linearly dependent if and only if there are nontrivial relation among them.

If linearly dependent, then, the vectors are said to be redundant.

Consider the vector$\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$

Consider the condition

Let $\overrightarrow{w}·\overrightarrow{v}=0,\overrightarrow{w}\in L^{\perp },\overrightarrow{v}\in V$

Then, $L^{\perp }$ must be orthogonal to $L$.

If, localid="1664203964228" $L=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$then, localid="1664203969891" $L^{\perp }=\left[1 2 3\right]$

localid="1664203957436" $$\begin{gathered} \left[1 2 3\right]·\left[x_1 \\ {x}_2 \\ {x}_3\right]=0 \\ {x}_1+2x_2+3x_3=0 \end{gathered}$$

Compute the values

If, $x_3=0$

Then, $x_1=-2x_2$

Thus, $x_1=2,x_2=-1,x_3=0$

$$\begin{gathered} x=\left[2 \\ -1 \\ 0\right] \end{gathered}$$

If, $x_2=0$

Then, $x_1=-3x_3$

Thus, $x_1=3,x_2=0,x_3=-1$.

$$\begin{gathered} x=\left[3 \\ 0 \\ -1\right] \end{gathered}$$

Final answer

The basis for $L^{\perp }$is$\left\{\left[\begin{array}{c}3\\ 0\\ -1\end{array}\right],\left[\begin{array}{c}2\\ -1\\ 0\end{array}\right]\right\}$