How do you give parametric equations for the line through (1, 3, -5) that is perpendicular to the plane #2x -3y +4z = 11#?

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Answer 1 · 2016-09-20T11:06:40

#{ (x = 1+2lambda), (y =3-3lambda ), (z=-5+4lambda):}#

The plane

#Pi->2x -3y +4z = 11# can be represented as

#<< vec v, p-p_0 >> = 0#

where

#vec v = (2,-3,4)#
#p = (x,y,z)#
#p_0=(0,0,11/4)#

Here, #vec v# is the normal to all the points #p in Pi#

Given now #p_1 = (1,3,-5)# the line passing by #p_1# and orthogonal to #Pi# is by construction

#L->p = p_1+lambda vec v# where #lambda in RR#.

So #L# has as parametric equations:

#{ (x = 1+2lambda), (y =3-3lambda ), (z=-5+4lambda):}#