LECTURE NOTES OF CALCULUS III
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1.3. Planes in Three Dimensional Space
Although a line in space is determined by a point and direction, a plane
in space is more difficult to describe. A single vector parallel to a plane is not
enough to convey the ‘’ direction’’ of the plane, but a vector perpendicular to
the plane does completely specify its direction. Thus, a plane in space is
determined by a point 𝑃𝑜 (𝑥0 , 𝑦0 , 𝑧0 ) in the plane and a vector n that is
orthogonal to the plane. This orthogonal vector n is called a normal vector.
Let 𝑃(𝑥, 𝑦, 𝑧) be an arbitrary point in the plane, and let 𝒓𝟎 and 𝒓 be the
position vector of 𝑃𝑜 and P. Then the vector 𝒓 − 𝒓𝟎 is represented by ̅̅̅̅̅
𝑃𝑃𝑜 . The
normal vector n is orthogonal to every vector in the given plane. In
particular, n is orthogonal to 𝒓 − 𝒓𝟎 and so we have
𝒏. (𝒓 − 𝒓𝟎 ) = 0
Then, we called the above equation as a vector equation of the plane
To obtain a scalar equation for the plane, we write
𝒏 = ⟨𝑎, 𝑏. 𝑐⟩, 𝒓 = ⟨𝑥, 𝑦, 𝑧⟩, and 𝒓𝟎 = ⟨𝑥0 , 𝑦0 , 𝑧0 ⟩.
Then, we get the following equation :
𝑎(𝑥 − 𝑥0 ) + 𝑏(𝑦 − 𝑦0 ) + 𝑐(𝑧 − 𝑧0 ) = 0
The above equation is the scalar equation of the plane through
𝑃0 (𝑥0 , 𝑦𝑜 , 𝑧0 ) with normal vector 𝒏 = ⟨𝑎, 𝑏, 𝑐⟩.
In general, a linear equation in x, y, z, i.e.ax+by+cz+d=0 is an equation of a
plane in ℝ3 .
Example 1.3.1. Find an equation of the plane passing through the points
P(1,3,2), Q(3,-1,6) and R(5,2,0).
Lecture Notes of Calculus III by Rubono Setiawan, S.Si.,M.Sc.
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̅̅̅̅ and 𝑃𝑅
̅̅̅̅ are :
Solution. The vector a and b corresponding to 𝑃𝑄
𝒂 = ⟨2, −4,4⟩
𝒃 = ⟨4, −1, −2⟩
Since both a and b lie in the plane, their cross product 𝒂 × 𝒃 is
orthogonal ( i.e. the direction of 𝒂 × 𝒃 is perpendicular to the respective
plane ) to the plane and can be taken as the normal vector. Thus
𝒊
𝒋
𝒌
𝒏 = 𝒂 × 𝒃 = |2 −4 4 | = 12𝒊 + 20𝒋 + 14𝒌 = ⟨12,20,14⟩.
4 −1 −2
Therefore, with the point P(1,3,2) and the normal vector n, an equation of the
plane is given by :
⟨12, 20,14⟩. ⟨𝑥 − 1, 𝑦 − 3, 𝑧 − 2⟩ = 0
12(𝑥 − 1) + 20(𝑦 − 3) + 14(𝑧 − 2) = 0
or
6𝑥 + 10𝑦 + 7𝑧 = 50.
Example 1.3.2. Find the point at which the line with parametric equations
𝑥 = 3 − 𝑡, 𝑦 = 2 + 𝑡, 𝑧 = 5𝑡, where t is the parametric value, intersect the plane
𝑥 − 𝑦 + 2𝑧 = 9.
Solution. We subtitute the expressions for x, y and z from parametric
equations into the equations of the plane :
(3 − 𝑡) − (2 + 𝑡) + 2(5𝑡) = 9.
So, we get 𝑡 = 1. Therefore, the point of intersection occurs when parameter
value is 𝑡 = 1. Then 𝑥 = 2, 𝑦 = 3 and 𝑧 = 5, hence the point of intersection is
(2,3,5).
Definition 1.3.3. Two planes are parallel if their normal vector are parallel.
For instance, let’s see the following example :
Example 1.3.4. The planes 𝑥 + 2𝑦 − 3𝑧 = 4 and 2𝑥 + 4𝑦 − 6𝑧 = 3 are parallel
because their normal vectors are 𝒏𝟏 = ⟨1,2, −3⟩ and 𝒏𝟐 = ⟨2,4, −6⟩ and 𝒏𝟐 =
2𝒏𝟏 .
If two planes are not parallel, then they intersect in a straight line and the
angle between the two planes is defined as the acute angle between their
normal vectors. See the following figure :
Lecture Notes of Calculus III by Rubono Setiawan, S.Si.,M.Sc.
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Figure 1.3.1
Example 1.3.5.
a. Find the angle between the planes 𝑥 + 𝑦 + 𝑧 = 1 and 𝑥 − 2𝑦 + 3𝑧 = 1.
b. Find the symmetric equations for the line of intersection L of these two
planes.
Solution.
a. The normal vectors of these planes are
𝒏𝟏 = ⟨1,1,1⟩
𝒏𝟐 ⟨1, −2,3⟩
and so, if 𝜃 is the angle between the planes, then
cos 𝜃 =
𝒏𝟏 . 𝒏𝟐
1(1) + 1(−2) + 1(3)
2
=
=
|𝒏𝟏 ||𝒏𝟐 | √1 + 1 + 1√1 + 4 + 9 √42
2
𝜃 = 𝑐𝑜𝑠 −1 (
) = 720
√42
b. We first need to find a point on L. For instance, we can find the point
where the line intersect the xy-plane by setting 𝑧 = 0 in the equations of
both planes. This gives the equations 𝑥 + 𝑦 = 1 and 𝑥 − 2𝑦 = 1, whose
solution is 𝑥 = 1, 𝑦 = 0. So the point (1, 0, 0 ) lies on L. Now, we observe
that, since L lies in both planes, it is perpendicular to both of the
normal vectors. Thus, a vector 𝒗 parallel to L is given by the cross
product
𝒊
𝒋 𝒌
𝒗 = 𝒏𝟏 × 𝒏𝟐 = |1 1 1| = 5𝒊 − 2𝒋 − 3𝒌
1 −2 3
So, the symmetric equation of line L can be written as :
𝑥−1
𝑦
𝑧
=
=
5
−2 −3
The following figure shows the plane in Example 1.3.5 and their line of
intersection L
Lecture Notes of Calculus III by Rubono Setiawan, S.Si.,M.Sc.
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Figure 1.3.2.
In general, when we write the equations of a line in the symmetric form
𝑥 − 𝑥0 𝑦 − 𝑦0 𝑧 − 𝑧0
=
=
𝑎
𝑏
𝑐
We can regard the line as the line of intersection of two planes
𝑥 − 𝑥0 𝑦 − 𝑦0
=
𝑎
𝑏
and
𝑦 − 𝑦0 𝑧 − 𝑧0
=
𝑏
𝑐
Proposition 1.3.6. The distance from 𝑃1 (𝑥1 , 𝑦1 , 𝑧1 ) to the plane
𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 + 𝑑 = 0 is
|𝑎𝑥1 + 𝑏𝑦1 + 𝑐𝑧1 + 𝑑|
√𝑎2 + 𝑏 2 + 𝑐 2
Example 1.3.7. Find the distance between the parallel planes 10𝑥 +
2𝑦 − 2𝑧 = 5 and 5𝑥 + 𝑦 − 𝑧 = 1.
Solution.
Then planes are parallel because the vector ⟨10,2, −2⟩ and ⟨5,1, −1⟩ are
parallel. Pick any point on the plane 10𝑥 + 2𝑦 − 2𝑧 = 5. For example
(1/2,0,0). Then the distance between the two planes is :
|5(1⁄2) + 0(1) + 0(−1) − 1|
√52
+
12
+
(−1)2
=
√3
6
Example 1.3.8. Find the distance between the skew lines
𝐿1 : 𝑥 = 1 + 𝑡, 𝑦 = −2 + 3𝑡, 𝑧 = 4 − 𝑡
𝐿2 : 𝑥 = 2𝑠, 𝑦 = 3 + 𝑠, 𝑧 = −3 + 4𝑠
Solution.
As 𝑳𝟏 and 𝑳𝟐 are skew, they are contained in two parallel planes
𝑷𝟏 and 𝑷𝟐 respectively. The common normal vector to both planes
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must be orthogonal to both 𝒗𝟏 = ⟨1,3, −1⟩ ( the direction of 𝐿1 ) and 𝒗𝟐 =
⟨2,1,4⟩ ( The direction of 𝐿2 ). A normal to these two parallel planes is
given by
𝒊 𝒋 𝒌
𝒏 = 𝒗𝟏 × 𝒗𝟐 = |1 3 −1| = ⟨13, −6, −5⟩
2 1 4
Let, 𝑠 = 0 in 𝐿2 . We get the point (0,3,-3) on 𝐿2 . Therefore, an equation
of the plane containing 𝐿2 is ⟨𝑥 − 0, 𝑦 − 3, 𝑧 + 3⟩. ⟨13, −6, −5⟩ = 0. That is
13𝑥 − 6𝑦 − 5𝑧 + 3 = 0. Let 𝑡 = 0 in 𝐿2 , then we get the point (1,-2, 4) in
𝐿1 . Hence, the distance between 𝐿1 and 𝐿2 is given by
|13(1) − 6(−2) − 5(4) + 3|
√132 + (−6)2 + (−5)2
=
8
√230
.
1.4. Cylinders and Quadric Surfaces
In this section, we have already look at two special types of surfaces –
planes and spheres. Here we investigate two other types of surfaces –
cylinders and quadric surfaces. In order to sketch the graph of a surfaces,
it is useful to determine the curves of intersection of the surface with planes
parallel to the coordinate planes. These curves are called traces ( or cross –
section ) of the surface.
Definition 1.4.1. A Cylinder is a surface that consist all line ( called rulling
that are parallel to a given line and pass through a plane curve.
Example 1.4.1. Parabolic Cylinder 𝑧 = 𝑥 2
Figure 1.4.1. A parabolic cylinder
Example 1.4.2. Circular Cylinders
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Figure 1.4.2. Circular Cylinder 𝑥 2 + 𝑦 2 = 1 and 𝑦 2 + 𝑧 2 = 1
When you dealing with surfaces, it is important to recognize that an equation
like 𝑥 2 + 𝑦 2 = 1 represent a cylinder not a circle. The trace of the cylinder
𝑥 2 + 𝑦 2 = 1 in the xy-plane is the circle with equations 𝑥 2 + 𝑦 2 = 1, 𝑧 = 0.
Definition 1.4.2. Quadric surface is the graph of the second degree equation
in 𝑥, 𝑦, 𝑧 :
𝐴𝑥 2 + 𝐵𝑦 2 + 𝐶𝑧 2 + 𝐷𝑥𝑦 + 𝐸𝑦𝑧 + 𝐹𝑥𝑧 + +𝐺𝑥 + 𝐻𝑦 + 𝐼𝑧 + 𝐽 = 0.
where A,B,C,D,E,F,G,H,I and J are real constant
Using translation and rotation, the equation can be expressed is one of the
following two standard form :
𝐴𝑥 2 + 𝐵𝑦 2 + 𝐶𝑧 2 + 𝐽 = 0 and 𝐴𝑥 2 + 𝐵𝑦 2 + 𝐼𝑧 = 0
Example 1.4.3. Use the trace to sketch the quadric surface with equation
𝑦2 𝑧2
𝑥 +
+ =1
9
4
2
Solution
By subtituting 𝑧 = 0, we find that the trace in xy-plane is 𝑥 2 +
𝑦2
9
= 1. which
we recognize as an equation of an ellipse. In general, the horizontal trace in
the plane 𝑧 = 𝑘 is
𝑦2
𝑘2
𝑥2 +
=1−
𝑧=𝑘
9
4
which is an ellipse , provided that 𝑘 2 < 4, −2 < 𝑘 < 2.
Similiarly the vertical traces are also ellipses :
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LECTURE NOTES OF CALCULUS III
𝑦2
9
+
𝑥2 +
𝑧2
4
𝑧2
4
= 1 − 𝑘2, 𝑥 = 𝑘
=1−
𝑘2
9
,𝑦 = 𝑘
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( if −1 < 𝑘 < 1)
( if −3 < 𝑘 < 3)
Figure 1.4.3. The ellipsoid 𝑥 2 +
𝑦2
9
+
𝑧2
4
=1
Figure 1.4.3. is called an ellipsoid because all of its traces are ellipse. Notice
that it is symmetric with respet to each coordinate plane; this is a reflection
of the fact that its equation involves only even powers of x,y, and z.
Example 1.4.4. The graph of the equation 𝑧 = 4𝑥 2 + 𝑦 2 is an elliptical
paraboloid
Figure 1.4.4. Elliptical paraboloid
The horizontal traces in 𝑧 = 𝑘 are ellipses : 4𝑥 2 + 𝑦 2 = 𝑘, where 𝑘 > 0
The vertical traces in 𝑥 = 𝑘 are parabolas 𝑧 = 𝑦 2 + 4𝑘 2 .
Similarly, the vertical traces in 𝑦 = 𝑘 are parabolas 𝑧 = 4𝑥 2 + 𝑘 2
Example 1.4.5. Skecth the surface
Solution
The trace in 𝑧 = 𝑘 are ellipses :
𝑥2
4
𝑥2
4
+ 𝑦2 −
+ 𝑦2 = 1 +
The trace in 𝑥 = 𝑘 are hiperbolas : 𝑦 2 −
The trace in 𝑦 = 𝑘 are hiperbolas :
𝑥2
4
−
𝑧2
4
=1
𝑘2
4
𝑧2
𝑘2
=
1
−
4
4
𝑧2
= 1 − 𝑘2
4
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Figure 1.4.5.a. The surface
𝑥2
4
+ 𝑦2 −
Figure 1.4.5.b. Vertical traces in 𝑦 = 𝑘 of
𝑥2
4
𝑧2
4
=1
+ 𝑦2 −
𝑧2
4
=1
Example 1.4.6. Identity and skecth the surface 4𝑥 2 − 𝑦 2 + 2𝑧 2 + 4 = 0
Solution
The equation can be rewritten in the standard form −𝑥 2 +
𝑦2
4
−
𝑧2
2
= 1. It is
therefore a hyperboloid of 2 sheets along of the y – axis. The traces in the xyand yz-planes are the hyperbolas.
−𝑥 2 +
𝑦2
4
= 1 𝑧 = 0 and
𝑦2
4
−
𝑧2
2
=1 𝑥=0
The surface has no trace in the xz-plane, but traces in the vertical planes 𝑦 =
𝑘 for |𝑘| > 2 are the ellipses.
𝑥2 +
𝑧2 𝑘2
=
− 1, 𝑦 = 𝑘
2
4
which can be written as :
Lecture Notes of Calculus III by Rubono Setiawan, S.Si.,M.Sc.
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𝑥2
𝑧2
+
= 1,
𝑘2
𝑘2
4 − 1 2 ( 4 − 1)
𝑦=𝑘
Figure 1.4.6. The planes 4𝑥 2 − 𝑦 2 + 2𝑧 2 + 4 = 0
Example 1.4.7 Classify the quadric surfaces 𝑥 2 + 2𝑧 2 − 6𝑥 − 𝑦 + 10 = 0.
Solution
By the method of completing squares, the equation can be written as
𝑦 − 1 = (𝑥 − 3)2 + 2𝑧 2
If we make change coordinates : 𝑥 ′ = 𝑥 − 3, 𝑦 ′ = 𝑦 − 1, 𝑧 ′ = 𝑧 so that the new
2
2
origin is at ( 3, 1, 0 ), then the equations becomes 𝑦 ′ = 𝑥 ′ + 2𝑧 ′ . Therefore it
is an elliptic paraboid with vertex at (3,1,0).
Figure 1.4.7. quadric surfaces 𝑥 2 + 2𝑧 2 − 6𝑥 − 𝑦 + 10 = 0.
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The graph of the quadric surfaces are summarized in the following table
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