CMIIh

2021-09-01

Find a basis for the set of vectors in ${\mathbb{R}}^{3}$ in the plane $x-6y+9z=0$.[Hint: Think of the equation as a​ "system" of homogeneous​ equations.]
A basis for the set of vectors in ${\mathbb{R}}^{3}$ in the plane $x-6y+9z=0$

brawnyN

Consider the plane $x-6y+9z=0$ as system of homogeneous equation $Ax=0$.

Thus, the null space of the above matrix  is the basis for the set of vectors ${\mathbb{R}}^{3}$ in the given plane $x-6y+9z=0$. Solve the equation $x-6y+9z=0$ for x as follows.

$x-6y+9z=0$

Therefore, the vector x in terms of free variables can be written as,

$x=\left[\begin{array}{c}6y-9z\\ y\\ z\end{array}\right]$

$=y\left[\begin{array}{c}6\\ 1\\ 0\end{array}\right]+z\left[\begin{array}{c}-9\\ 0\\ 1\end{array}\right]$

That is, the set of vectors $\left[\begin{array}{c}6\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-9\\ 0\\ 1\end{array}\right]$a basis for null space of A.

Thus, the vasis for the set of vectors ${\mathbb{R}}^{3}$ in the given plane

$x-6y+9z=0$ is $\left[\begin{array}{c}6\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-9\\ 0\\ 1\end{array}\right]$.

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