# cartesian equation

• Apr 27th 2012, 09:01 PM
rcarron
cartesian equation
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.
• Apr 28th 2012, 12:34 AM
earboth
Re: cartesian equation
Quote:

Originally Posted by 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.

The direction vector of the line is the normal vector of the plane: $\overrightarrow{n_P}= (5,6,8)$

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

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

Using the given values you have:

$P: (5,6,8)((x,y,z)-(3,1,5))=0~\implies~\boxed{5x+6y+8z-61=0}$
• Apr 28th 2012, 10:56 AM
TheIntegrator
Re: cartesian equation
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
• Apr 28th 2012, 08:16 PM
rcarron
Re: cartesian equation
thanks guys