Question

How do you find parametric equations for the line through (2, 4, 6) that is perpendicular to the plane x − y + 3z = 7?

Answer 1

The parametric equation of our line is
`x=2+t`
`y=4-t`
`z=6+3t`

Explanation:

A vector perpendicular to the plane `ax+by+cz+d=0`
is given by `〈a,b,c〉`
So a vector perpendiculat to the plane `x-y+3z-7=0`
is `〈1,-1,3〉`
The parametric equation of a line through `(x_0,y_0,z_0)`
and parallel to the vector `〈a,b,c〉` is
`x=x_0+ta`
`y=y_0+tb`
`z=z_0+tb`

So the parametric equation of our line is
`x=2+t`
`y=4-t`
`z=6+3t`

The vector form of the line is `vecr=〈2,4,6〉+t〈1,-1,3〉`