Find the equation of the plane through the line of intersection of the planes #ax+by+cz+d=0, a_1x+b_1y+c_1z+d_1=0# and perpendicular to the #XY# plane?

1 Answer
Cesareo R.
Jul 9, 2017

See below.

Explanation:

NOTE:

We will represent the scalar product of two vectors as #<< cdot, cdot >># and the cross product as #cdot xx cdot#

With #p = (x,y,z) # and calling

#Pi_1-> << vec n_1, p-p_1 >> = 0#
#Pi_2-> << vec n_2, p-p_2 >> = 0#
#L->p = p_0 + lambda vec v#

if #L sub Pi_1 nn Pi_2# then

#{(<< n_1, p_0+lambda vec v -p_1 >> =0),(<< n_2, p_0+lambda vec v -p_2 >> =0):}#

or

#<< vec n_1, vec v >> = << vec n_2, vec v >> = 0 rArr vec v = vec n_1 xx vec n_2#

and

#{(<< vec n_1, p_0 >> = << vec n_1, p_1 >>),(<< vec n_2, p_0 >> = << vec n_2, p_2 >>):}# ---- (*)

thus determining completely #L#

Given now the sought plane

#Pi-> << vec n, p-p_3 >> = 0# we have that #L sub Pi# and also

#<< vec n, hat e_3 >> = 0# where #hat e_3 = (0,0,1)#

then

#<< vec n, p_0+lambda vec v-p_3 >> = 0# or

#lambda << vec n, vec v >>+ << vec n, p_0-p_3 >> = 0#

but this implies on

#<< vec n, vec v >>=0->{(vec n xx (vec n_1 xx vec n_2)=vec 0),(vec n xx hat e_3 = vec 0):}#

and also we can choose #p_3 = p_0# so the final conditions are:

#{(vec n xx (vec n_1 xx vec n_2)=vec 0),(vec n xx hat e_3 = vec 0),(<< vec n_1, p_0 >> = << vec n_1, p_1 >>),(<< vec n_2, p_0 >> = << vec n_2, p_2 >>):}#

defining the plane

#Pi-> << vec n, p-p_0 >> = 0#

EXAMPLE:

Here supposing #c# and #c_1# non-null

#vec n_1 = (a,b,c)#
#p_1 = (0,0,-d/c)#

#vec n_2 = (a_1,b_1,c_1)#
#p_1 = (0,0,-d_1/c_1)#

and

#vec v = (b c_1-b_1 c, a_1 c - a c_1,a b_1 -a_1 b )#

The determination of #vec n# follows:

Giving #vec n = (u,v,w)# we have

#vec n xx (vec n_1 xx vec n_2) = (a b_1 v - a_1 c w + a c_1 w-a_1 b v, a_1 b u - a b_1 u - b_1 c w + b c_1 w, a_1 c u - a c_1 u + b_1 c v - b c_1 v)#

and from

#vec n xx hat e_3 = vec 0# we obtain

#{((a_1 b - a b_1) u + (b c_1 - b_1 c) w=0),((a_1 b - a b_1) v + (a_1 c - a c_1) w=0):}#

or for any #w#

#{(u = -( (b c_1 - b_1 c) /(a_1 b - a b_1))w),(v = -( (a_1 c - a c_1)/(a_1 b - a b_1))w):}#

then

#vec n = -( ( (b c_1 - b_1 c) /(a_1 b - a b_1)),( (a_1 c - a c_1)/(a_1 b - a b_1)),-1)w#

now #p_0# is determined by solving (*)

#a x_0+by_0+cz_0 = -c d/c= -d#
#a_1 x_0+b_1y_0+c_1z_0 = -c_1 d_1/c_1=-d_1#

or choosing #z_0 = 0# the solution of

#{(a x_0+by_0 = -d),(a_1 x_0+b_1y_0 = -d_1):}#

or

#p_0 = ((b_1 d - b d_1)/(a_1 b - a b_1),(a d_1-a_1 d)/(a_1 b - a b_1),0)#

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