How do you find the parametric equations for the line through the point P = (3, 5, -2) that is perpendicular to the plane −5x + 1y + 3z = 1?

1 Answer
Aug 31, 2016

#vec l(t) = ((3),( 5), (-2)) + t ((-5),(1),(3))#

Explanation:

a plane in vector form can be written as

#(vec r - vec r_o)* vec n = 0#; or, as it is here, as #vec r * vec n = vec r_o* vec n = const#

where, in Cartesian, #vec r = ((x),(y),(x))# and normal vector #vec n = ((a),(b),(c))#

so, here, the plane's normal vector is #vec n = ((-5),(1),(3))#

that is to say, the plane is: # ((x),(y),(x)) * ((-5),(1),(3)) = 1#

and the line through P , in terms of parameter t, can be written as

#vec l(t) = ((3),( 5), (-2)) + t ((-5),(1),(3))#