Topic: Distance between parallel planes
Question: Find the distance between the parallel planes.
3x+ 2y−z= 3
9x+ 6y−3z= 2
Answer choices:
A 1
3 2
B −7
3 14
C 7
3 14
D −1
3 2
!
!
165
Solution: C
First we’ll confirm that the planes
3x+ 2y−z= 3
9x+ 6y−3z= 2
are parallel. To test whether the planes are parallel, we’ll take the ratio of
the components of the normal vectors to each plane.
a1
b1
=a2
b2
=a3
b3
where the planes are given in the form
a1x+a2y+a3z=c
b1x+b2y+b3z=d
If the ratios are the same, then the planes are parallel.
This means the two normal vectors are a⟨a1,a2,a3⟩ and b⟨b1,b2,b3⟩. First we
can determine our normal vectors. For the plane 3x+ 2y−z= 3, we’ll get
the normal vector a⟨3,2, −1⟩. For the plane 9x+ 6y−3z= 2, we’ll get the
normal vector b⟨9,6, −3⟩. Now we can set up the ratio
3
9=2
6=−1
−3
1
3=1
3=1
3
166
We can see that these ratios are all equal, which means that the planes are
parallel.
Next we can find a point on one of the planes. We can take the plane
3x+ 2y−z= 3 and set y= 0 and z= 0.
3x+ 2(0) −(0) = 3
3x= 3
x= 1
This means a point on the plane is (1,0,0).
Now we can find the distance from the point to a plane using the distance
formula
d=|ax1+by1+cz1+d|
a2+b2+c2
where the point is (x1,y1,z1) and the plane is ax +by +cz =−d.
The point (1,0,0) will give us x1= 1, y1= 0, and z1= 0. The plane
9x+ 6y−3z= 2 will give us a= 9, b= 6, c=−3, and d=−2.
d=|(9)(1) + (6)(0) + (−3)(0) + (−2) |
(9)2+ (6)2+ (−3)2
d=|9+0+0−2|
81 + 36 + 9
167
d=|7|
126
d=7
3 14
This is the distance between the planes.
168
Topic: Distance between parallel planes
Question: Find the distance between the parallel planes.
−2x+ 1y−2z= 6
−8x+ 4y−8z=−3
Answer choices:
A 9
4
B 7
4
C 3
2
D 7
2
!
!
169
Solution: A
First we’ll confirm that the planes
−2x+ 1y−2z= 6
−8x+ 4y−8z=−3
are parallel. To test whether the planes are parallel, we’ll take the ratio of
the components of the normal vectors to each plane.
a1
b1
=a2
b2
=a3
b3
where the planes are given in the form
a1x+a2y+a3z=c
b1x+b2y+b3z=d
If the ratios are the same, then the planes are parallel.
This means the two normal vectors are a⟨a1,a2,a3⟩ and b⟨b1,b2,b3⟩. First we
can determine our normal vectors. For the plane −2x+ 1y−2z= 6, we’ll get
the normal vector a⟨−2,1, −2⟩. For the plane −8x+ 4y−8z=−3, we’ll get
the normal vector b⟨−8,4, −8⟩. Now we can set up the ratio
−2
−8=1
4=−2
−8
1
4=1
4=1
4
170
We can see that these ratios are all equal, which means that the planes are
parallel.
Next we can find a point on one of the planes. We can take the plane
−2x+ 1y−2z= 6 and set y= 0 and z= 0.
−2x+ 1(0) −2(0) = 6
−2x= 6
x=−3
This means a point on the plane is (−3,0,0).
Now we can find the distance from the point to a plane using the distance
formula
d=|ax1+by1+cz1+d|
a2+b2+c2
where the point is (x1,y1,z1) and the plane is ax +by +cz =−d.
The point (−3,0,0) will give us x1=−3, y1= 0, and z1= 0. The plane
−8x+ 4y−8z=−3 will give us a=−8, b= 4, c=−8, and d= 3.
d=|(−8)(−3) + (4)(0) + (−8)(0) + (3) |
(−8)2+ (4)2+ (−8)2
d=|24 + 0 + 0 + 3 |
64 + 16 + 64
171
d=|27 |
144
d=27
12
d=9
4
This is the distance between the planes.
172
Topic: Distance between parallel planes
Question: Find the distance between the parallel planes.
−6x+ 2y+ 4z=−12
−9x+ 3y+ 6z= 2
Answer choices:
A −10
3 7
B −20
3 14
C 10
3 7
D 20
3 14
!
!
173
Solution: D
First we’ll confirm that the planes
−6x+ 2y+ 4z=−12
−9x+ 3y+ 6z= 2
are parallel. To test whether the planes are parallel, we’ll take the ratio of
the components of the normal vectors to each plane.
a1
b1
=a2
b2
=a3
b3
where the planes are given in the form
a1x+a2y+a3z=c
b1x+b2y+b3z=d
If the ratios are the same, then the planes are parallel.
This means the two normal vectors are a⟨a1,a2,a3⟩ and b⟨b1,b2,b3⟩. First we
can determine our normal vectors. For the plane −6x+ 2y+ 4z=−12, we’ll
get the normal vector a⟨−6,2,4⟩. For the plane −9x+ 3y+ 6z= 2, we’ll get the
normal vector b⟨−9,3,6⟩. Now we can set up the ratio
−6
−9=2
3=4
6
2
3=2
3=2
3
174
We can see that these ratios are all equal, which means that the planes are
parallel.
Next we can find a point on one of the planes. We can take the plane
−6x+ 2y+ 4z=−12 and set y= 0 and z= 0.
−6x+ 2(0) + 4(0) = −12
−6x=−12
x= 2
This means a point on the plane is (2,0,0).
Now we can find the distance from the point to a plane using the distance
formula
d=|ax1+by1+cz1+d|
a2+b2+c2
where the point is (x1,y1,z1) and the plane is ax +by +cz =−d.
The point (2,0,0) will give us x1= 2, y1= 0, and z1= 0. The plane
−9x+ 3y+ 6z= 2 will give us a=−9, b= 3, c= 6, and d=−2.
d=|(−9)(2) + (3)(0) + (6)(0) + (−2) |
(−9)2+ (3)2+ (6)2
d=|−18 + 0 + 0 −2|
81 + 9 + 36
175
d=|−20 |
126
d=20
3 14
This is the distance between the planes.
176
