Let l be the line that passes through the point P=(−5,−7,−1) and is perpendicular to the plane 8x−6y−1z=15, how do you find the parametric equations?

Question
Answer 1

# { (x=-5+8lamda), (y=-7-6lamda), (z=-1-lamda) :} #

Consider the plane equation:

# 8x-6y-1z=15 #

The Normal vector to this plane is given by the coefficient of #x#, #y# and #z#, and so we have;

# vec(n) = ( (8), (-6), (-1) ) #

And so the equation of the line #L# that passes through #P(-5,-7,-1) and has the same directional as the above normal is give by;

# L: \ \ \ vec(r) = ( (-5), (-7), (-1) ) + lamda ( (8), (-6), (-1) ) #

If we assign #vec(r)# a generic coordinate #(x,y,z)# the we have;

# ( (x), (y), (z) ) = ( (-5), (-7), (-1) ) + lamda ( (8), (-6), (-1) ) #
# \ \ \ \ \ \ \ \ \ \ = ( (-5+8lamda), (-7-6lamda), (-1-lamda) ) #

Leading to the parametric equations:

# { (x=-5+8lamda), (y=-7-6lamda), (z=-1-lamda) :} #

Which we can see graphically;

Here the plane is drawn in red and the line #L# in green

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