How do you find the parametric and symmetric equation of the line passing through the point (2,3,4) and perpendicular to the plane 5x+6y-7z=20?

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Answer 1 · 2017-11-22T18:20:05

Use the fact that a vector normal to a plane #Ax+By+Cz = D# is:

#Ahati+Bhatj+Chatk#

The vector perpendicular to the plane, #5x+6y-7z=20#, is:

#5hati+6hatj-7hatk#

This allows us to write the point-vector form of the line passing through the point #(2,3,4)#:

#(x,y,z) = (2,3,4) + t(5hati+6hatj-7hatk)#

From the point-vector form we can extract the 3 parametric equations by observation:

#x = 5t+2#
#y = 6t+3#
#z = -7t+4#

To find the symmetric form we solve each of the parametric equations for t and then set them equal:

#t = (x-2)/3#
#t = (y-3)/6#
#t = (z-4)/-7#

Setting them equal gives us the symmetric form:

#(x-2)/3 = (y-3)/6= (z-4)/-7#