The Cartesian equation is $x-3y-4z=1$. Here is what I have tried:
Finding three points on the plane by setting two variables equal to 0:
$x=0$, $y=0$; $z=\frac{-1}{4}$
$y=0$, $z=0$; $x=1$
$x=0$, $z=0$; $y=\frac{-1}{3}$
Vector equation: $(x,y,z)=(1,0,0)+\lambda[(0,0,\frac{-1}{4})-(0,\frac{-1}{3},0)] + t((0,0,\frac{-1}{4})-(1,0,0)]$
$(x,y,z)=(1,0,0)+\lambda(0,\frac{1}{3},\frac{-1}{4})+t(-1,0,\frac{-1}{4})$
However, the solution gives the vector equation as: $(x,y,z)=(1,0,0)+\lambda(3,1,0)+t(4,0,1)$. I understand that there are multiple ways to find the vector equation of a plane. But since I am doing this for transformation purposes, the vector equation I found is a little more complicated than the solution's equation. How can I derive the solution's vector equation?
I would appreciate if someone can assist me with this. Thanks
Best Answer
I'll show you how to do a similar one (yours goes exactly the same way), starting with the plane $$3x+4y+5z=6.$$ First, one of the variables has a non-zero coefficient. Pick one, I'll choose $z$. Now I'll solve for $z$. $$z = \frac{1}{5}(6-3x-4y)$$ So a given point on our plane looks like $$\begin{align*} (x,y,z) &= \left(x,y,\frac{1}{5}(6-3x-4y)\right) \\ &= \frac{1}{5}\left(5x,5y,6-3x-4y\right) \\ &= \frac{1}{5}\left[(5x,0,-3x)+(0,5y,-4y)+(0,0,6)\right]\\ &=\frac{x}{5}(5,0,-3)+\frac{y}{5}(0,5,-4)+\left(0,0,\frac{6}{5}\right) \end{align*}$$
Now, to clean things up, I'll change variables to $s=x/5$ and $t=y/5$ to get $$s(5,0,-3)+t(0,5,-4)+\left(0,0,\frac{6}{5}\right)$$ as a parametrization of the plane.