The document discusses equations of planes in 3D space. It introduces the point-normal form of a plane equation: A(x-r) + B(y-s) + C(z-t) = 0, where <A,B,C> is a normal vector to the plane and <r,s,t> is a point on the plane. This equation is analogous to the point-slope form for a line, and can be derived by setting the dot product of the normal vector and a position vector from the point <r,s,t> to a generic point <x,y,z> on the plane equal to zero.

1. Equations of Planes

2. Equations of Planes
In general, graphs of equations of the form f(x, y, z) = 0
are surfaces.

3. Equations of Planes
In general, graphs of equations of the form f(x, y, z) = 0
are surfaces. Specifically, the graphs of the equations
Ax + By + Cz + D = 0, where A, B, C and D are
constants, are planes.

4. Equations of Planes
In general, graphs of equations of the form f(x, y, z) = 0
are surfaces. Specifically, the graphs of the equations
Ax + By + Cz + D = 0, where A, B, C and D are
constants, are equations
Graphs of the planes.
x = k, y = k, or z = k
are planes that are parallel to the
coordinate planes as shown here.

5. Equations of Planes
In general, graphs of equations of the form f(x, y, z) = 0
are surfaces. Specifically, the graphs of the equations
z
Ax + By + Cz + D = 0, where A, B, C and D are
constants, are equations
Graphs of the planes.
x = k, y = k, or z = k x=4 y
are planes that are parallel to the
coordinate planes as shown here. x

6. Equations of Planes
In general, graphs of equations of the form f(x, y, z) = 0
are surfaces. Specifically, the graphs of the equations
z
Ax + By + Cz + D = 0, where A, B, C and D are
constants, are equations
Graphs of the planes.
x = k, y = k, or z = k x=4
y=4 y
are planes that are parallel to the
coordinate planes as shown here. x

7. Equations of Planes
In general, graphs of equations of the form f(x, y, z) = 0
are surfaces. Specifically, the graphs of the equations
z
Ax + By + Cz + D = 0, where A, B, C and D are
z=4
constants, are equations
Graphs of the planes.
x = k, y = k, or z = k x=4
y=4 y
are planes that are parallel to the
coordinate planes as shown here. x

8. Equations of Planes
In general, graphs of equations of the form f(x, y, z) = 0
are surfaces. Specifically, the graphs of the equations
z
Ax + By + Cz + D = 0, where A, B, C and D are
z=4
constants, are equations
Graphs of the planes.
x = k, y = k, or z = k x=4
y=4 y
are planes that are parallel to the
coordinate planes as shown here. x
In R3, a line L (not in the plane P) is parallel to P
if L and P do not intersect, i.e. they do not have any
point in common.

9. Equations of Planes
In general, graphs of equations of the form f(x, y, z) = 0
are surfaces. Specifically, the graphs of the equations
z
Ax + By + Cz + D = 0, where A, B, C and D are
z=4
constants, are equations
Graphs of the planes.
x = k, y = k, or z = k x=4
y=4 y
are planes that are parallel to the
coordinate planes as shown here. x
In R3, a line L (not in the plane P) is parallel to P
if L and P do not intersect, i.e. they do not have any
point in common. In particular if the equation of a
plane has a missing variable, then the plane is parallel
to the corresponding axis of that variable.

10. Equations of Planes
In general, graphs of equations of the form f(x, y, z) = 0
are surfaces. Specifically, the graphs of the equations
z
Ax + By + Cz + D = 0, where A, B, C and D are
z=4
constants, are equations
Graphs of the planes.
x = k, y = k, or z = k x=4
y=4 y
are planes that are parallel to the
coordinate planes as shown here. x
In R3, a line L (not in the plane P) is parallel to P
if L and P do not intersect, i.e. they do not have any
point in common. In particular if the equation of a
plane has a missing variable, then the plane is parallel
to the corresponding axis of that variable. For
example, x = 4 is parallel to both the y and the z axes.

11. 3D Coordinate Systems
The Graphs of Linear Equations

12. 3D Coordinate Systems
The Graphs of Linear Equations
The graphs of linear equations ax + by + cz = d are
planes.

13. 3D Coordinate Systems
The Graphs of Linear Equations
The graphs of linear equations ax + by + cz = d are
planes. We may use the intercepts to graph the plane
a, b, c and d are all nonzero:

14. 3D Coordinate Systems
The Graphs of Linear Equations
The graphs of linear equations ax + by + cz = d are
planes. We may use the intercepts to graph the plane
a, b, c and d are all nonzero:
set x = y = 0 to obtain the z intercept,
set x = z = 0 to obtain the y intercept,
set y = z = 0 to obtain the x intercept.

15. 3D Coordinate Systems
The Graphs of Linear Equations
The graphs of linear equations ax + by + cz = d are
planes. We may use the intercepts to graph the plane
a, b, c and d are all nonzero:
set x = y = 0 to obtain the z intercept,
set x = z = 0 to obtain the y intercept,
set y = z = 0 to obtain the x intercept.
The three intercepts position the
plane in the coordinate system.

16. 3D Coordinate Systems
The Graphs of Linear Equations
The graphs of linear equations ax + by + cz = d are
planes. We may use the intercepts to graph the plane
a, b, c and d are all nonzero:
set x = y = 0 to obtain the z intercept,
set x = z = 0 to obtain the y intercept,
set y = z = 0 to obtain the x intercept.
The three intercepts position the
plane in the coordinate system.
Example A. Sketch 2x – y + 2z = 4.

17. 3D Coordinate Systems
The Graphs of Linear Equations
The graphs of linear equations ax + by + cz = d are
planes. We may use the intercepts to graph the plane
a, b, c and d are all nonzero:
set x = y = 0 to obtain the z intercept,
set x = z = 0 to obtain the y intercept,
set y = z = 0 to obtain the x intercept.
The three intercepts position the
plane in the coordinate system.
Example A. Sketch 2x – y + 2z = 4
Set x = y = 0  z = 2, we’ve (0, 0, 2),

18. 3D Coordinate Systems
The Graphs of Linear Equations
The graphs of linear equations ax + by + cz = d are
planes. We may use the intercepts to graph the plane
a, b, c and d are all nonzero:
set x = y = 0 to obtain the z intercept,
set x = z = 0 to obtain the y intercept,
set y = z = 0 to obtain the x intercept.
The three intercepts position the
plane in the coordinate system.
Example A. Sketch 2x – y + 2z = 4
Set x = y = 0  z = 2, we’ve (0, 0, 2),
set x = z = 0  y = –4, we’ve (0, –4, 0),

19. 3D Coordinate Systems
The Graphs of Linear Equations
The graphs of linear equations ax + by + cz = d are
planes. We may use the intercepts to graph the plane
a, b, c and d are all nonzero:
set x = y = 0 to obtain the z intercept,
set x = z = 0 to obtain the y intercept,
set y = z = 0 to obtain the x intercept.
The three intercepts position the
plane in the coordinate system.
Example A. Sketch 2x – y + 2z = 4
Set x = y = 0  z = 2, we’ve (0, 0, 2),
set x = z = 0  y = –4, we’ve (0, –4, 0),
set y = z = 0  x = 2, we’ve (2, 0, 0).

20. 3D Coordinate Systems
The Graphs of Linear Equations
The graphs of linear equations ax + by + cz = d are
planes. We may use the intercepts to graph the plane
a, b, c and d are all nonzero:
set x = y = 0 to obtain the z intercept,
set x = z = 0 to obtain the y intercept, +
z
set y = z = 0 to obtain the x intercept.
The three intercepts position the
plane in the coordinate system. y
Example A. Sketch 2x – y + 2z = 4
Set x = y = 0  z = 2, we’ve (0, 0, 2), x
x
set x = z = 0  y = –4, we’ve (0, –4, 0),
set y = z = 0  x = 2, we’ve (2, 0, 0). Plot these
intercepts and the plane containing them is the graph.

21. 3D Coordinate Systems
The Graphs of Linear Equations
The graphs of linear equations ax + by + cz = d are
planes. We may use the intercepts to graph the plane
a, b, c and d are all nonzero:
set x = y = 0 to obtain the z intercept,
set x = z = 0 to obtain the y intercept, z +
(0, 0, 2)
set y = z = 0 to obtain the x intercept.
The three intercepts position the (0,–4, 0)
plane in the coordinate system. y
Example A. Sketch 2x – y + 2z = 4 (2, 0, 0)
Set x = y = 0  z = 2, we’ve (0, 0, 2), x
x
set x = z = 0  y = –4, we’ve (0, –4, 0),
set y = z = 0  x = 2, we’ve (2, 0, 0). Plot these
intercepts and the plane containing them is the graph.

22. Equations of Planes
A vector N that is perpendicular to a plane is called
a normal vector of the plane.
N

23. Equations of Planes
A vector N that is perpendicular to a plane is called
a normal vector of the plane.
N

24. Equations of Planes
A vector N that is perpendicular to a plane is called
a normal vector of the plane. To specify a plane in
3D space, we specify the direction the plane is
facing and a point in the plane that gives the
location of the plane.
p
N

25. Equations of Planes
A vector N that is perpendicular to a plane is called
a normal vector of the plane. To specify a plane in
3D space, we specify the direction the plane is
facing and a point in the plane that gives the
location of the plane. A normal vector N defines the
direction the plane is facing.
p
N

26. Equations of Planes
A vector N that is perpendicular to a plane is called
a normal vector of the plane. To specify a plane in
3D space, we specify the direction the plane is
facing and a point in the plane that gives the
location of the plane. A normal vector N defines the
direction the plane is facing.
Similar to finding the equation of a line
p
in 2D where we find the slope of the
line and a point on the line then use
the point-slope formula,
N

27. Equations of Planes
A vector N that is perpendicular to a plane is called
a normal vector of the plane. To specify a plane in
3D space, we specify the direction the plane is
facing and a point in the plane that gives the
location of the plane. A normal vector N defines the
direction the plane is facing.
Similar to finding the equation of a line
p
in 2D where we find the slope of the
line and a point on the line then use
the point-slope formula,
to find the equation of a plane in 3D, N
we find a normal vector N and a point p in the
plane then use the point-normal formula
that’s based on the dot product.

28. Equations of Planes
Let N = <A,B,C> be a normal
vector to the plane P and p=(r,s,t)
p = <r, s, t> be a point in P.
N=<A,B,C>

29. Equations of Planes
Let N = <A,B,C> be a normal
vector to the plane P and p=(r,s,t)
p = <r, s, t> be a point in P.
Let q = (x, y, z) be a generic point q=(x,y,z)
in the plane, N=<A,B,C>

30. Equations of Planes
Let N = <A,B,C> be a normal
vector to the plane P and p=(r,s,t)
p = <r, s, t> be a point in P.
Let q = (x, y, z) be a generic point q=(x,y,z)
in the plane, then the vector N=<A,B,C>
pq = <x – r, y – s, z – t>

31. Equations of Planes
Let N = <A,B,C> be a normal
vector to the plane P and p=(r,s,t)
p = <r, s, t> be a point in P.
Let q = (x, y, z) be a generic point q=(x,y,z)
in the plane, then the vector N=<A,B,C>
pq = <x – r, y – s, z – t> is perpendicular to
N = <A,B,C>.

32. Equations of Planes
Let N = <A,B,C> be a normal
vector to the plane P and p=(r,s,t)
p = <r, s, t> be a point in P.
Let q = (x, y, z) be a generic point q=(x,y,z)
in the plane, then the vector N=<A,B,C>
pq = <x – r, y – s, z – t> is perpendicular to
N = <A,B,C>. Therefore the dot product N•pq = 0,

33. Equations of Planes
Let N = <A,B,C> be a normal
vector to the plane P and p=(r,s,t)
p = <r, s, t> be a point in P.
Let q = (x, y, z) be a generic point q=(x,y,z)
in the plane, then the vector N=<A,B,C>
pq = <x – r, y – s, z – t> is perpendicular to
N = <A,B,C>. Therefore the dot product N•pq = 0,
and we have the Point–Normal Equation of a plane:
A(x – r) + B(y – s) + C(z – t) = 0
where <A, B, C> is any normal vector and p = <r, s, t>
is any point in the plane.

34. Equations of Planes
Let N = <A,B,C> be a normal
vector to the plane P and p=(r,s,t)
p = <r, s, t> be a point in P.
Let q = (x, y, z) be a generic point q=(x,y,z)
in the plane, then the vector N=<A,B,C>
pq = <x – r, y – s, z – t> is perpendicular to
N = <A,B,C>. Therefore the dot product N•pq = 0,
and we have the Point–Normal Equation of a plane:
A(x – r) + B(y – s) + C(z – t) = 0
where <A, B, C> is any normal vector and p = <r, s, t>
is any point in the plane.
The problem of finding an equation of a plane P is
reduced to finding a normal vector N and a point p in P.

35. Equations of Planes
Example B. Find the equation of the plane that
contains the following given three points:
a(1, 0, 0), b(0, 2, 0), c(0, 0, 3).

36. Equations of Planes
Example B. Find the equation of the plane that
contains the following given three points: z
c(0, 0, 3)
a(1, 0, 0), b(0, 2, 0), c(0, 0, 3).
y
b(0,2, 0)
a(1, 0, 0)
x

37. Equations of Planes
Example B. Find the equation of the plane that
contains the following given three points: z
c(0, 0, 3)
a(1, 0, 0), b(0, 2, 0), c(0, 0, 3).
We need a normal vector to the plane.
y
b(0,2, 0)
a(1, 0, 0)
x

38. Equations of Planes
Example B. Find the equation of the plane that
contains the following given three points: z
c(0, 0, 3)
a(1, 0, 0), b(0, 2, 0), c(0, 0, 3).
We need a normal vector to the plane.
y
A normal vector is the cross product b(0,2, 0)
a(1, 0, 0)
ab x ac where
x

39. Equations of Planes
Example B. Find the equation of the plane that
contains the following given three points: z
c(0, 0, 3)
a(1, 0, 0), b(0, 2, 0), c(0, 0, 3).
We need a normal vector to the plane.
y
A normal vector is the cross product b(0,2, 0)
a(1, 0, 0)
ab x ac where
x
ab = <–1, 2, 0> and ac = <–1, 0, 3>

40. Equations of Planes
Example B. Find the equation of the plane that
contains the following given three points: z
c(0, 0, 3)
a(1, 0, 0), b(0, 2, 0), c(0, 0, 3).
We need a normal vector to the plane.
y
A normal vector is the cross product b(0,2, 0)
a(1, 0, 0)
ab x ac where
x
ab = <–1, 2, 0> and ac = <–1, 0, 3>
i j k
ab x ac = det –1 2 0
–1 0 3

41. Equations of Planes
Example B. Find the equation of the plane that
contains the following given three points: z
c(0, 0, 3)
a(1, 0, 0), b(0, 2, 0), c(0, 0, 3).
We need a normal vector to the plane.
y
A normal vector is the cross product b(0,2, 0)
a(1, 0, 0)
ab x ac where
x
ab = <–1, 2, 0> and ac = <–1, 0, 3>
i j k
ab x ac = det –1 2 0 = 6i + 3j + 2k = N.
–1 0 3

42. Equations of Planes
Example B. Find the equation of the plane that
contains the following given three points: z
c(0, 0, 3)
a(1, 0, 0), b(0, 2, 0), c(0, 0, 3). N=<6, 3, 2>
We need a normal vector to the plane.
y
A normal vector is the cross product b(0,2, 0)
a(1, 0, 0)
ab x ac where
x
ab = <–1, 2, 0> and ac = <–1, 0, 3>
i j k
ab x ac = det –1 2 0 = 6i + 3j + 2k = N.
–1 0 3
Using N = <6, 3, 2> and a(1, 0, 0) for the point–normal
formula,

43. Equations of Planes
Example B. Find the equation of the plane that
contains the following given three points: z
c(0, 0, 3)
a(1, 0, 0), b(0, 2, 0), c(0, 0, 3). N=<6, 3, 2>
We need a normal vector to the plane.
y
A normal vector is the cross product b(0,2, 0)
a(1, 0, 0)
ab x ac where
x
ab = <–1, 2, 0> and ac = <–1, 0, 3>
i j k
ab x ac = det –1 2 0 = 6i + 3j + 2k = N.
–1 0 3
Using N = <6, 3, 2> and a(1, 0, 0) for the point–normal
formula, we have an equation of the plane in question:
6(x – 1) + 3(y – 0) + 2(z – 0) = 0 or 6x + 3y + 2z = 6.

44. Equations of Planes
Conversely, the plane defined by Ax + By + Cz = D
has the vector N = <A, B, C > as a normal vector.

45. Equations of Planes
Conversely, the plane defined by Ax + By + Cz = D
has the vector N = <A, B, C > as a normal vector.
As the constant D varies, the equations
Ax + By + Cz = D define parallel planes that are
perpendicular to the normal vector N = <A, B, C >,
i.e. planes that face the same direction.

46. Equations of Planes
Conversely, the plane defined by Ax + By + Cz = D
has the vector N = <A, B, C > as a normal vector.
As the constant D varies, the equations
Ax + By + Cz = D define parallel planes that are
perpendicular to the normal vector N = <A, B, C >,
i.e. planes that face the same direction.
Example C. Using the intercept–
method, we draw
a. 6x + 3y + 2z = 6 = D
b. 6x + 3y + 2z = 12 = D

47. Equations of Planes
Conversely, the plane defined by Ax + By + Cz = D
has the vector N = <A, B, C > as a normal vector.
As the constant D varies, the equations
Ax + By + Cz = D define parallel planes that are
perpendicular to the normal vector N = <A, B, C >,
i.e. planes that face the same direction.
z
Example C. Using the intercept–
method, we draw
a. 6x + 3y + 2z = 6 = D (0, 0, 3)
b. 6x + 3y + 2z = 12 = D
a y
(0,2, 0)
(1, 0, 0)
x

48. Equations of Planes
Conversely, the plane defined by Ax + By + Cz = D
has the vector N = <A, B, C > as a normal vector.
As the constant D varies, the equations
Ax + By + Cz = D define parallel planes that are
perpendicular to the normal vector N = <A, B, C >,
i.e. planes that face the same direction.
z
Example C. Using the intercept– (0, 0, 6)
method, we draw
a. 6x + 3y + 2z = 6 = D (0, 0, 3)
b. 6x + 3y + 2z = 12 = D
a b y
(0,2, 0) (0,4, 0)
(1, 0, 0)
(2, 0, 0)
x

49. Equations of Planes
Conversely, the plane defined by Ax + By + Cz = D
has the vector N = <A, B, C > as a normal vector.
As the constant D varies, the equations
Ax + By + Cz = D define parallel planes that are
perpendicular to the normal vector N = <A, B, C >,
i.e. planes that face the same direction.
z
Example C. Using the intercept– (0, 0, 6)
method, we draw
a. 6x + 3y + 2z = 6 = D (0, 0, 3)
b. 6x + 3y + 2z = 12 = D
We see that as D varies, a b y
it layers out the corresponding (0,2, 0) (0,4, 0)
parallel planes along the (1, 0, 0)
(2, 0, 0)
normal vector N = <6, 3, 2>. x

50. Equations of Planes
Example D. Find an equation of the plane L that
contains the point p(2,1, –2) and is parallel to the
plane 2x – 3y + z = –3. p=(2, 1, -2)
2x – 3y + z = -3

51. Equations of Planes
Example D. Find an equation of the plane L that
contains the point p(2,1, –2) and is parallel to the
plane 2x – 3y + z = –3. p=(2, 1, -2)
We need a normal vector for L.
2x – 3y + z = -3

52. Equations of Planes
Example D. Find an equation of the plane L that
contains the point p(2,1, –2) and is parallel to the
plane 2x – 3y + z = –3. p=(2, 1, -2)
We need a normal vector for L.
2x – 3y + z = -3
Since L is parallel to 2x – 3y + z = –
3,
hence a normal vector is <2, –3, 1>. N = <2, -3, 1>

53. Equations of Planes
Example D. Find an equation of the plane L that
contains the point p(2,1, –2) and is parallel to the
plane 2x – 3y + z = –3. p=(2, 1, -2)
We need a normal vector for L.
2x – 3y + z = -3
Since L is parallel to 2x – 3y + z = –
3,
hence a normal vector is <2, –3, 1>. N = <2, -3, 1>
Use the point-normal formula,
an equation for L is 1(z + 2) = 0 or 2x – 3y + z = –1.
2(x – 2) – 3(y – 1) +

54. Equations of Planes
Example D. Find an equation of the plane L that
contains the point p(2,1, –2) and is parallel to the
plane 2x – 3y + z = –3. p=(2, 1, -2)
We need a normal vector for L.
2x – 3y + z = -3
Since L is parallel to 2x – 3y + z = –
3,
hence a normal vector is <2, –3, 1>. N = <2, -3, 1>
Use the point-normal formula,
an equation for L is 1(z + 2) = 0 or 2x – 3y + z = –1.
2(x – 2) – 3(y – 1) +
Skew lines
Two lines in 3D space are said to
be skew if they are not in the same
plane and they don't meet.

55. Equations of Planes
Example D. Find an equation of the plane L that
contains the point p(2,1, –2) and is parallel to the
plane 2x – 3y + z = –3. p=(2, 1, -2)
We need a normal vector for L.
2x – 3y + z = -3
Since L is parallel to 2x – 3y + z = –
3,
hence a normal vector is <2, –3, 1>. N = <2, -3, 1>
Use the point-normal formula,
an equation for L is 1(z + 2) = 0 or 2x – 3y + z = –1.
2(x – 2) – 3(y – 1) +
Skew lines
Two lines in 3D space are said to
z
be skew if they are not in the same
plane and they don't meet.
y
They are parallel if they don't meet
but they are in the same plane. Parallel lines x

56. Equations of Planes
Given two parallel lines,
there are infinite many
planes containing one but
not the other.

57. Equations of Planes
Given two parallel lines,
there are infinite many
planes containing one but
not the other.
Given two skew lines, there is P
a unique plane P that contains
one line but not the other.

58. Equations of Planes
Example: Show that
L: x(t) = 2t –1, y(t) = t + 3, z(t) = –t +1
M: X(t) = t + 2, Y(t) = 3t, Z(t) = 2t – 1 are skew lines.

59. Equations of Planes
Example: Show that
L: x(t) = 2t –1, y(t) = t + 3, z(t) = –t +1
M: X(t) = t + 2, Y(t) = 3t, Z(t) = 2t – 1 are skew lines.
The directional vectors for the lines are <2, 1, –1> and
<1, 3, 2>,hence the lines are not in the same direction.

60. Equations of Planes
Example: Show that
L: x(t) = 2t –1, y(t) = t + 3, z(t) = –t +1
M: X(t) = t + 2, Y(t) = 3t, Z(t) = 2t – 1 are skew lines.
The directional vectors for the lines are <2, 1, –1> and
<1, 3, 2>,hence the lines are not in the same direction.
Need to show they don't have intersection.

61. Equations of Planes
Example: Show that
L: x(t) = 2t –1, y(t) = t + 3, z(t) = –t +1
M: X(t) = t + 2, Y(t) = 3t, Z(t) = 2t – 1 are skew lines.
The directional vectors for the lines are <2, 1, –1> and
<1, 3, 2>,hence the lines are not in the same direction.
Need to show they don't have intersection. Assume
they meet at point p, i.e. for some value t = a such that
p = (x(a), y(a), z(a)), and some value t = b
such that p = (X(b), Y(b), Z(b)).
Therefore x(a) =X(b), y(a) =Y(b), z(a) = Z(b), or
2a – 1 = b + 2  E1
a + 3 = 3b  E2
–a + 1 = 2b – 1  E3

62. Equations of Planes
Example: Show that
L: x(t) = 2t –1, y(t) = t + 3, z(t) = –t +1
M: X(t) = t + 2, Y(t) = 3t, Z(t) = 2t – 1 are skew lines.
The directional vectors for the lines are <2, 1, –1> and
<1, 3, 2>,hence the lines are not in the same direction.
Need to show they don't have intersection. Assume
they meet at point p, i.e. for some value t = a such that
p = (x(a), y(a), z(a)), and some value t = b
such that p = (X(b), Y(b), Z(b)).
Therefore x(a) =X(b), y(a) =Y(b), z(a) = Z(b), or

63. Equations of Planes
Example: Show that
L: x(t) = 2t –1, y(t) = t + 3, z(t) = –t +1
M: X(t) = t + 2, Y(t) = 3t, Z(t) = 2t – 1 are skew lines.
The directional vectors for the lines are <2, 1, –1> and
<1, 3, 2>,hence the lines are not in the same direction.
Need to show they don't have intersection. Assume
they meet at point p, i.e. for some value t = a such that
p = (x(a), y(a), z(a)), and some value t = b
such that p = (X(b), Y(b), Z(b)).
Therefore x(a) =X(b), y(a) =Y(b), z(a) = Z(b), or
2a – 1 = b + 2  E1
a + 3 = 3b  E2
–a + 1 = 2b – 1  E3

64. Equations of Planes
2a – 1 = b + 2  E1
a + 3 = 3b  E2
–a + 1 = 2b – 1  E3
Add E2 and E3 we get b = 1.
Putting this in E2, we get a = 0 which is impossible.
Hence the lines do not intercept.

65. Equations of Planes
Example F.
Find the plane containing L and parallel to M where
L: x(t) = 2t –1, y(t) = t + 3, z(t) = –t +1
M: X(t) = t + 2, Y(t) = 3t, Z(t) = 2t – 1
M
L

66. Equations of Planes
Example F.
Find the plane containing L and parallel to M where
L: x(t) = 2t –1, y(t) = t + 3, z(t) = –t +1
M: X(t) = t + 2, Y(t) = 3t, Z(t) = 2t – 1
Need a normal vector and a point in the plane.
To get a point, set t = 0, we get (–1, 3, 1) a point on L.
M
L

67. Equations of Planes
Example F.
Find the plane containing L and parallel to M where
L: x(t) = 2t –1, y(t) = t + 3, z(t) = –t +1
M: X(t) = t + 2, Y(t) = 3t, Z(t) = 2t – 1
Need a normal vector and a point in the plane.
To get a point, set t = 0, we get (–1, 3, 1) a point on L.
A normal vector of the plane must be
perpendicular to both lines.
M
L

68. Equations of Planes
Example F.
Find the plane containing L and parallel to M where
L: x(t) = 2t –1, y(t) = t + 3, z(t) = –t +1
M: X(t) = t + 2, Y(t) = 3t, Z(t) = 2t – 1
Need a normal vector and a point in the plane.
To get a point, set t = 0, we get (–1, 3, 1) a point on L.
A normal vector of the plane must be
perpendicular to both lines. That is,
it must be perpendicular to the two
directional vectors <2, 1, –1>,<1, 3, 2>
of the lines. L
M

69. Equations of Planes
Example F.
Find the plane containing L and parallel to M where
L: x(t) = 2t –1, y(t) = t + 3, z(t) = –t +1
M: X(t) = t + 2, Y(t) = 3t, Z(t) = 2t – 1
Need a normal vector and a point in the plane.
To get a point, set t = 0, we get (–1, 3, 1) a point on L.
A normal vector of the plane must be
perpendicular to both lines. That is,
it must be perpendicular to the two
directional vectors <2, 1, –1>,<1, 3, 2>
of the lines. Hence their cross product L
M
<2, 1, –1> x <1, 3, 2> = <5, –5, 5>
is a normal vector for the plane.
Use N = <1, –1, 1> instead.

70. Equations of Planes
Use the point-normal form, we get:
1(x + 1) – 1(y – 3) + 1(z – 1) = 0
or x – y + z = 3 is an equation for the plane.

71. Equations of Planes
Use the point-normal form, we get:
1(x + 1) – 1(y – 3) + 1(z – 1) = 0
or x – y + z = 3 is an equation for the plane.
The (dihedral) angle AVB between two planes is the
measurement of an angle whose vertex V is on their
line of the intersection, with AV and BV
perpendicular to the line of intersection.
N
M V B
A
The dihedral angle

72. Equations of Planes
Use the point-normal form, we get:
1(x + 1) – 1(y – 3) + 1(z – 1) = 0
or x – y + z = 3 is an equation for the plane.
The (dihedral) angle AVB between two planes is the
measurement of an angle whose vertex V is on their
line of the intersection, with AV and BV
perpendicular to the line of intersection. There are
two such angles between two planes, one < 90 o and
one > 90o. N
M V B
A
The dihedral angle

73. Equations of Planes
Use the point-normal form, we get:
1(x + 1) – 1(y – 3) + 1(z – 1) = 0
or x – y + z = 3 is an equation for the plane.
The (dihedral) angle AVB between two planes is the
measurement of an angle whose vertex V is on their
line of the intersection, with AV and BV
perpendicular to the line of intersection. There are
two such angles between two planes, one < 90 o and
one > 90o. angle is also
The dihedral N
angle between the two
normal vectors of the two M V B
planes. A
The dihedral angle

74. Equations of Planes
Example G. Find the angles between
2x – y + 3z = 1 and –2x + 4y – 3z = 5.

75. Equations of Planes
Example G. Find the angles between
2x – y + 3z = 1 and –2x + 4y – 3z = 5.
The normal vectors are N = <2, –1, 3> and
M = <–2, 4, –3>.

76. Equations of Planes
Example G. Find the angles between
2x – y + 3z = 1 and –2x + 4y – 3z = 5.
The normal vectors are N = <2, –1, 3> and
M = <–2, 4, –3>. Let θ be the angle between them, then
M•N
cos(θ) = |M|*|N| = –17
√14*√29

77. Equations of Planes
Example G. Find the angles between
2x – y + 3z = 1 and –2x + 4y – 3z = 5.
The normal vectors are N = <2, –1, 3> and
M = <–2, 4, –3>. Let θ be the angle between them, then
M•N
cos(θ) = |M|*|N| = –17 , or θ ≈ 147.5o
√14*√29
The other angle is 180 – 147.5 ≈ 32.5o