Question

How do you find parametric equation for the line through the point P(-7,-3,-6) and perpendicular to the plane -2x + 2y + 6z =3?

Answer 1

We can write the parametric equations as;

`{: (x=-7-lamda), (y=-3+lamda), (z=-6+3 lamda) :}`

Explanation:

If we examine the plane equation `-2x+2y+6z=3` then the normal vector is given by the coefficients of `x`, `y`, and `z` respectively. So the normal vector is:

`vec (n) = ( (-2), (2), (6) ) = 2( (-1), (1), (3) )`

So then the vector equation of the line that has this direction and passes through `P(-7,-3,-6)` is given by:

`vec(r)=( (-7), (-3), (-6) ) + lamda ( (-1), (1), (3) )`

So we can write the generic coordinates for any point on the lines as:

`( (x), (y), (z) ) = ( (-7), (-3), (-6) ) + lamda ( (-1), (1), (3) ) = ( (-7-lamda), (-3+lamda), (-6+3lamda) )`

So we can write the parametric equations as;

`{: (x=-7-lamda), (y=-3+lamda), (z=-6+3 lamda) :}`

We can verify this with a 3D graph: