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?

Question

Christopher Allers · Accepted Answer

We can write the parametric equations as;

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

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:

Isabella Ackerman · Answer

undefined To find the parametric equation for the line through the point P(-7,-3,-6) and perpendicular to the plane -2x + 2y + 6z = 3, follow these steps:

1. Determine the direction vector of the line by finding the normal vector to the plane. The coefficients of x, y, and z in the equation of the plane represent the components of the normal vector. So, the normal vector to the plane is n = (-2, 2, 6).

2. Use the direction vector of the line to find the parametric equation. Since the line is perpendicular to the plane, its direction vector is parallel to the normal vector of the plane. Hence, the direction vector of the line is also (-2, 2, 6).

3. Use the point-slope form of the equation of a line to write the parametric equations. The point-slope form is given by r(t) = r0 + tv, where r0 is a point on the line, v is the direction vector of the line, and t is a parameter.

4. Substitute the given point P(-7,-3,-6) as r0 and the direction vector (-2, 2, 6) as v into the point-slope form.

5. The parametric equations for the line are:
   x(t) = -7 - 2t
   y(t) = -3 + 2t
   z(t) = -6 + 6t