# cartesian equation

#### rcarron

Find a cartesian equation for the plane containing the point (3, 1, 5) and perpendicular to the line with
parametric form x = 5t + 1, y = 6t − 2, z = 8t, t 2 R.

#### earboth

MHF Hall of Honor
Find a cartesian equation for the plane containing the point (3, 1, 5) and perpendicular to the line with
parametric form x = 5t + 1, y = 6t − 2, z = 8t, t 2 R.
The direction vector of the line is the normal vector of the plane: $$\displaystyle \overrightarrow{n_P}= (5,6,8)$$

Let A(3,1,5) denote a point located in the plane with it's stationary vector $$\displaystyle \vec a$$ and $$\displaystyle \vec p = (x,y,z)$$ the stationary vector of any point in the plane then the equation of the plane P is:

$$\displaystyle P: \overrightarrow{n_P} \cdot ( \vec p - \vec a) = 0$$

Using the given values you have:

$$\displaystyle P: (5,6,8)((x,y,z)-(3,1,5))=0~\implies~\boxed{5x+6y+8z-61=0}$$

#### TheIntegrator

Again, find the direction vector first => <5,6,8>
C = <5,6,8> dot product <x,y,z>
C = <5,6,8> dot product <3,1,5> = 61

Therefore our plane is 5x + 6y + 8z = 61

thanks guys