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.

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- Apr 27th 2012, 09:01 PMrcarroncartesian 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 AMearbothRe: cartesian equation
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:

$\displaystyle P: (5,6,8)((x,y,z)-(3,1,5))=0~\implies~\boxed{5x+6y+8z-61=0}$ - Apr 28th 2012, 10:56 AMTheIntegratorRe: 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 PMrcarronRe: cartesian equation
thanks guys