2
$\begingroup$

I have question abous parametrization and plane equation.How can I write the equation of a plane in the form $Ax+By+Cz=D$ if it is given as

$$x=2s+3t$$ $$y=s-2t$$ $$z=3s+t$$.

3 Answers 3

2

hint: $x - 2y = 7t, 3y - z = -7t \implies (x-2y) + (3y-z) = 0$

1

You have to explicit $s$ and $t$ from two of the three equations, then substitute in the remaining one. Pratically, explicit $s$ from the second equation $$s=y+2t$$ so, we get $$\begin{cases} x=2(y+2t)+3t \\ s=y+2t \\ z=3(y+2t)+t\end{cases}\quad \Longrightarrow \quad \begin{cases} x=2y+7t \\ s=y+2t \\ z=3y+7t \end{cases}.$$ Now, you have to explicit $t$ from the first or the third equation. For example, take $t$ from the first: $$t=\frac{x-2y}{7}$$ and substitute it in the remaining equation. So you get $$z=3y+7\frac{x-2y}{7} \quad \Longrightarrow \quad z=3y+x-2y \quad \Longrightarrow \quad x+y-z=0.$$ Finally you are in the form $Ax+By+Cz=D$ where $A=1$, $B=1$, $C=-1$ and $D=0$.

  • 0
    Thanks so much, I've two more question about this :(, so how can I determine if the point P=(2,3,7) is on the plane defined as x+y-z=0 ? – 2017-01-09
  • 0
    You have to substitute the coordinate of $P$ in the equation. If it exit $0=0$, then $P$ is on the plane. In this case we get $$2+3-7=-2\ne 0$$ so $P$ is not on the plane. – 2017-01-09
0

$$x=2s+3t$$ $$y=s-2t$$ $$z=3s+t$$

$x+y=3s+t$ Then $x+y=z$

Thus $$x+y-z=0$$