Thread: 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.

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:

Let A(3,1,5) denote a point located in the plane with it's stationary vector and the stationary vector of any point in the plane then the equation of the plane P is:



Using the given values you have:

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