We'll determine first the intercepting line of the planes
x-3y+7=0 and x-2z-4=0.
For this reason, we'll find out x
from both equations of the planes:
x-3y+7=0 => x =
3y - 7 => x = (y - 7/3)/(1/3) (1)
x-2z-4=0 =>
x = 2z + 4 => x = (z + 2)/(1/2) (2)
We'll equate (1)
and (2):
x = (y - 7/3)/(1/3) = (z +
2)/(1/2)
The equation of the intercepting line is: x = (y -
7/3)/(1/3) = (z + 2)/(1/2)
Now, we'll write the parametric
form of this equation:
x = t
y
= 7/3 + t/3
z = -2 + t/2
We'll
verify if the intercepting line is perpendicular to the plane
3x-6y-2z-12=0.
For this reason, we'll have to prove that
the director vector of the intercepting line is collinear to the normal vector n, to the
plane 3x-6y-2z-12=0.
The director vector of the line is v(
1 , 1/3 , 1/2).
The normal vector to the plane
3x-6y-2z-12=0 is n(3 , -6 , -2).
The vectors v and n are
said to be collinear, if v = a*n
( 1 , 1/3 , 1/2) = (3a ,
-6a , -2a)
3a = 1 => a =
1/3
1/3 = -6a => a =
-1/18
It is obvious that the values of a are different,
therefore the vectors v and n are not collinear, so the intercepting line is not
perpendicular to the plane 3x-6y-2z-12=0.
If the
intercepting line is parallel to the plane 3x-6y-2z-12=0, then the dot product of the
vectors n and v is cancelling.
v*n = 1*3 + (1/3)*(-6) +
(1/2)*(-2)
v*n = 3 - 2 - 1
v*n
= 0
We notice that the dot product of the
vectors n and v is cancelling, therefore, the intercepting line of the planes x-3y+7=0
and x-2z-4=0 is parallel to the plane
3x-6y-2z-12=0.