How to find a line that intersects two planes?

We have two planes
x=0x=0 and z=-5z=−5
I know the line should be something like (0,0) +t(1,0,0). i just dont know how to prove that

1 Answer
Douglas K.
May 15, 2018

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

(x,y,z)= (x_1,y_1,z_1)+tvecv" [1]"(x,y,z)=(x1,y1,z1)+t→v [1]

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

1hati+0hatj+0hatk1ˆi+0ˆj+0ˆk

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

0hati+0hatj+1hatk0ˆi+0ˆj+1ˆk

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)→v=(1ˆi+0ˆj+0ˆk)×(0ˆi+0ˆj+1ˆk)

vecv= 0hati-1hatj+0hatk→v=0ˆi−1ˆj+0ˆk

Written in your preferred form:

vecv = (0,-1,0)→v=(0,−1,0)

Substitute the vector into equation [1]:

(x,y,z)= (x_1,y_1,z_1)+t(0,-1,0)" [1.1]"(x,y,z)=(x1,y1,z1)+t(0,−1,0) [1.1]

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

(0,y,z)= (x_1,y_1,z_1)+t(0,-1,0)" [1.2]"(0,y,z)=(x1,y1,z1)+t(0,−1,0) [1.2]

This forces x_1=0x1=0:

(0,y,z)= (0,y_1,z_1)+t(0,-1,0)" [1.3]"(0,y,z)=(0,y1,z1)+t(0,−1,0) [1.3]

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

(0,y,-5)= (0,y_1,z_1)+t(0,-1,0)" [1.5]"(0,y,−5)=(0,y1,z1)+t(0,−1,0) [1.5]

This forces z_1=-5z1=−5

(0,y,-5)= (0,y_1,-5)+t(0,-1,0)" [1.5]"(0,y,−5)=(0,y1,−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 tt 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

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