Browse Questions

# Find the cartesian form of following planes: $\overrightarrow{r}=(1+s+t) \overrightarrow{i}+(2-s+t)\overrightarrow{j}+(3-2s+2t)\overrightarrow{k}$

Toolbox:
Step 1
Let $\overrightarrow r=x\overrightarrow i+y\overrightarrow j+z\overrightarrow k$
We have $x\overrightarrow i+y\overrightarrow j+z\overrightarrow k=(1+s+t)\overrightarrow i+(2-s+t)\overrightarrow j+(3-2s+2t)\overrightarrow k$
Equating the components in the $x-, y-, z-$ directions.
$x = 1+s+t\: \: \Rightarrow x-1=s+t$
$y = 2-s+t\: \: \Rightarrow y-2=-s+t$
$z = 3-2s+2t\: \: \Rightarrow z-3=-2s+2t$
Step 2
Eliminating $t, s$ from the above, we have
$\begin{vmatrix} x-1 & 1 & 1 \\ y-2 & -1 & 1 \\ z-3 & -2 & 2 \end{vmatrix}=0 \Rightarrow (x-1)(-2+2)-(y-2)(2+2)+(z-3)(1+1)=0$
$\therefore -4y+4+2z-6=0\: or \: 4y-2z=-2$
$\Rightarrow 2y-z=-1$ is the equation of the plane.