How you determine the parametric equation of a line with the cartesian equation #ax+by+cz + d = 0 ; a'x+b'y+c'z + e = 0# and vice versa ?

1 Answer
Feb 25, 2017

See below.

Explanation:

Given a plane

#Pi->a x + b y + c z + d=0#

this plane can be also be represented as

#Pi-> << p-p_1, n >> = 0# with

#p = (x,y,z)#
#n = (a,b,c)#
#p_1 | << p_1, n >> = -d# so giving

#Pi_1 -> << p-p_1, n_1 >> = 0#

#Pi_2 -> << p-p_2, n_2 >> = 0#

The line #L = Pi_1 nn Pi_2# can be established as

#L -> p = p_0 + lambda (n_1 xx n_2)#

This line must be contained in both planes so

#<< p_0-p_1+lambda (n_1 xx n_2),n_1 >> = 0#
#<< p_0-p_2+lambda (n_1 xx n_2),n_2 >> = 0#

The condition is then

#<< p_0-p_1, n_1 >> = 0 -> << p_0, n_1 >> = -d#
#<< p_0-p_2, n_2 >> = 0-> << p_0,n_2 >> = -e#

Here #<< cdot, cdot >># represent the scalar product of two vectors and
#cdot xx cdot# represents the vector product.

Concluding, giving the two planes

#Pi_1->a_1 x + b_1 y + c_1 z + d_1=0#

#Pi_2->a_2 x + b_2 y + c_2 z + d_2=0#

the parametric equation of #L# is

#L -> p = p_0 + lambda (n_1 xx n_2)#. (Supposing that #n_1 xx n_2 ne \vec 0#)

if #n_1=(a_1,b_1,c_1)# and #n_2=(a_2,b_2,c_2)# then

#n_1 xx n_2 = (b_1 c_2-b_2 c_1, a_2 c_1 - a_1 c_2, a_1 b_2-a_2 b_1 )#

and #p_0 = (p_(0x),p_(0y),p_(0z))# is any of the infinite solutions for

#{( << p_0, n_1 >> = -d_1),( << p_0,n_2 >> = -d_2):}#

and finally

#{(x = p_(0x)+lambda(b_1 c_2-b_2 c_1) ),(y=p_(0y)+lambda(a_2 c_1 - a_1 c_2)),(z=p_(0z)+lambda(a_1 b_2-a_2 b_1 )):}#