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.

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:

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