# Thread: Another Vector Parametric Problem

1. ## Another Vector Parametric Problem

How do I find equations of the planes for two similar yet different problems such as:

Find an equation of the plane through the point and parallel to the given plane:

(-5, 9, -4)
x + y + z + 4 = 0

and

Find an equation of the plane that contains the line and parallel to the given plane

x = 2 + 2t, y = t, z = 6 - t
2x + 4y + 8z = 17

2. For the first one, you must find the normal vector of the plane.

Observe that $\displaystyle x+y+z+4=0\Rightarrow (x,y,z+4)\cdot (1,1,1)=0\Rightarrow \left((x,y,z)+(0,0,4)\right)\cdot(1,1,1)=0$. So the plane looked for is $\displaystyle ((x,y,z)+(-5,9,-4))\cdot(1,1,1)=0\Rightarrow x+y+z=0$.

For the second one, observe that the line can be written as $\displaystyle \displaystyle\frac{x-2}{2}=\frac{y}{1}=\frac{z+6}{-1}$ so the directional vector is $\displaystyle (2,1,-1)$. From Linear Algebra, you know that the plane given belongs to $\displaystyle (17/2,0,0)+<(-2,1,0),(-4,0,1)>$. Also, observe that $\displaystyle (2,1,-1)=(-2,1,0)-(-4,0,1)$ what means the directional vector can be contained in a parallel plane to the given one. Finally, if you choose a positional vector (so the line can really be contained and not just be parallel to it) you get $\displaystyle \Pi: \ (-2,0,6)+<(-2,1,0),(-4,0,1)>$.