cartesian equation

Apr 2012
6
0
New Zealand
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
Jan 2006
5,854
2,553
Germany
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:

\(\displaystyle P: (5,6,8)((x,y,z)-(3,1,5))=0~\implies~\boxed{5x+6y+8z-61=0}\)
 
Apr 2012
15
6
Ottawa
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