 Linda Seales

2022-01-05

Show that the xy plane $W=\left(x,y,0\right)$ in ${\mathbb{R}}^{3}$ is generated by
(i) $u=\left[\begin{array}{c}1\\ 2\\ 0\end{array}\right]$ and $v=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$ (ii) $u=\left[\begin{array}{c}2\\ -1\\ 0\end{array}\right]$ and $v=\left[\begin{array}{c}1\\ 3\\ 0\end{array}\right]$ Louis Page

Expert

Let $\left(x,y,0\right)$ be any element from W. For a and b in $\mathbb{R}$, we have
$\left(x,y,0\right)=a\left(1,2,0\right)+b\left(0,1,0\right)$
$=\left(a,2a+b,0\right)$
Comparing both side we get
$x=a$ and $y=2a+b$
Now from $y=2a+b$
$b=y-2a$
$=-y-2x$
Therfore, for any $\left(x,y,0\right)$ in W, we can write
$\left(x,y,0\right)=x\left(1,2,0\right)+\left(y-2x\right)\left(0,1,0\right)$
$⇒\left(x,y,0\right)\left\{\left(1,2,0\right),\left(0,1,0\right)\right\}$
Therefore, W is generated by u and v.
Let $\left(x,y,0\right)$ be any element from W. For a and b in $\mathbb{R}$, we have
$\left(x,y,0\right)=a\left(2,-1,0\right)+b\left(1,3,0\right)$
$=\left(2a+b,-a+3b,0\right)$
Comparing both side we get
$x=2a+b$ ...(1)
and
$y=-a+3b$ ...(2)
Now we solve equationn (1) and (2) to find the value of a and b. From (1) and (2), we get
$x+2y=2a+b-2a+6b=7b$
$⇒b=\frac{x+2y}{7}$
Now from (2)
$a=y+3b$
$=y+3\left(\frac{x+2y}{7}\right)$
$=y+\frac{3x+6y}{7}$
$=\frac{7y+3x+6y}{7}$
$=\frac{3x+13y}{7}$
$⇒a=\frac{3x+13y}{7}$
Therfore, for any $\left(x,y,0\right)$ in W, we can write
$\left(x,y,0\right)=a\left(2,-1,0\right)+b\left(1,3,0\right)$
$=\frac{3x+13y}{7}\left(2,-1,0\right)+\frac{x+2y}{7}\left(1,3,0\right)$
$⇒\left(x,y,0\right)\left\{\left(2,-1,0\right),\left(1,3,0\right)\right\}$
Therefore, W is generated by u and v.

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