2022-07-06

How do you find the parametric equation through the point (6, 0, 10) and parallel to the plane $-3x+2y-4z=0$ ?

Alexis Fields

Expert

Step 1
Any plane parallel to the plane $-3x+2y-4z=0$ will have an equation of the same form but equal to a different constant:
$-3x+2y-4z=C\phantom{\rule{1ex}{0ex}}\text{[1]}$
To force the point $\left(6,0,10\right)$ to be in the plane, substitute the point into equation [1] and solve for C:
$-3\left(6\right)+2\left(0\right)-4\left(10\right)=C$
$C=-58$
The scalar equation of the plane parallel to the given plane is:
$-3x+2y-4z=-58\phantom{\rule{1ex}{0ex}}\text{[2]}$
The vector equation of the plane is
$\left(x,y,z\right)=\left(6,0,10\right)+s\left(\overline{u}\right)+t\left(\overline{v}\right)$
Where $\overline{u}\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}\overline{v}$ are any two vectors in the plane.
Let's use equation [2] to find two other points in the plane:
Let $x=6\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}y=2$
$-3\left(6\right)+2\left(2\right)-4z=-58$
$-4z=-44$
$z=11$
The vector from the point $\left(6,0,10\right)$ to the point $\left(6,2,11\right)$ is:
$\overline{u}=2\stackrel{^}{j}+\stackrel{^}{k}$
Let $x=10\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}y=0$
$-3\left(10\right)+2\left(0\right)-4z=-58$
$-4z=-28$
$z=7$
The vector from point $\left(6,0,10\right)$ to the point $\left(6,0,7\right)$ is:
$\overline{v}=4\stackrel{^}{i}-3\stackrel{^}{k}$
A vector equation of the plane is:
$\left(x,y,z\right)=\left(6,0,10\right)+s\left(2\stackrel{^}{j}+\stackrel{^}{k}\right)+t\left(4\stackrel{^}{i}-3\stackrel{^}{k}\right)$
From the above equation, we can write a set of parametric equations:
$x=4t+6$
$y=2s$
$z=s-3t+10$
NOTE: These are not unique equations, because we could have chosen other points and obtained different vectors.

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