How to find a line that intersects two planes?

Start with the vector form for the equation of a line:

#(x,y,z)= (x_1,y_1,z_1)+tvecv" [1]"#

The orthogonal vector for the plane #x = 0# is:

#1hati+0hatj+0hatk#

The orthogonal vector for the plane #z = -5# is:

#0hati+0hatj+1hatk#

The vector along the intersection of the two planes must be orthogonal to both planes, therefore, we may compute the vector using the cross product:

#vec v=(1hati+0hatj+0hatk) xx (0hati+0hatj+1hatk)#

#vecv= 0hati-1hatj+0hatk#

Written in your preferred form:

#vecv = (0,-1,0)#

Substitute the vector into equation [1]:

#(x,y,z)= (x_1,y_1,z_1)+t(0,-1,0)" [1.1]"#

For the line to be in the plane #x=0#, #x# must always equal 0:

#(0,y,z)= (x_1,y_1,z_1)+t(0,-1,0)" [1.2]"#

This forces #x_1=0#:

#(0,y,z)= (0,y_1,z_1)+t(0,-1,0)" [1.3]"#

For the line to be in the plane #z = -5#, #z# must always equal #-5:

#(0,y,-5)= (0,y_1,z_1)+t(0,-1,0)" [1.5]"#

This forces #z_1=-5#

#(0,y,-5)= (0,y_1,-5)+t(0,-1,0)" [1.5]"#

Please observe that the value of y is the only coordinate that changes on this line and because the domain of #t# is #{tinRR}# the value of #y_1# does not matter, therefore, we shall choose #y_1=0#

#(0,y,-5)= (0,0,-5)+t(0,-1,0)" [1.6]"#

For the same reason, we can, also, change the -1 to 1:

#(0,y,-5)= (0,0,-5)+t(0,1,0)" [1.7]"#

Equation [1.7] is the vector equation of the line.

The parametric equations are:

#x = 0, y = t, z = -5#

The symmetric form is:

#y#

with restrictions #x = 0, z = -5#