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?

1 Answer
Feb 14, 2017

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:

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