Download 12. Vectors and the geometry of space 12.1. Three dimensional

12. Vectors and the geometry of space
12.1. Three dimensional coordinate systems. To locate a point in space, one needs to
choose a point first as the reference, which is usually called the origin. Then, select three
directed lines and labelled the x-axis, y-axis and z-axis. These three lines are named as the
coordinate axes. In general, the z-axis is determined by the right-hand-rule.
The three axes determine three coordinate planes including the xy-plane (the plane that
contains the x-axis and y-axis), the yz-plane and the xz-plane. These three coordinate planes
separate the space into eight regions, which are called octant. In general, the first octant
denotes the octant determined by the positive axes.
According to the origin and these three axes (with specified unit lengths), we can identify
each point in space with a unique triple (a, b, c), where a denotes the directed distance of the
point to the yz-plane and so on. Here, a, b, c are called respectively the x-, y- and z-coordinates
of the point. Then, the projection of the point P = (a, b, c) to the xy-plane, yz-plane and
xz-plane are respectively the points (a, b, 0), (0, b, c) and (a, 0, c).
One can see that there is a one-to-one correspondence between the space and the Cartesian
product
R × R × R = R3 = {(x, y, z)|x, y, z ∈ R}.
This system is then called a three dimensional rectangular coordinate system.
Example 12.1. Find the set determined by the following equations.
(i)z = 5,
(ii)x = 3,
(iii)y = 2z.
Definition 12.1 (Distance formula in three dimensions). The distance between P1 (x1 , y1 , z1 )
and P2 (x2 , y2 , z2 ) is denoted by |P1 p2 | and defined by
√
|P1 P2 | = (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2 .
Example 12.2. The set
{(x, y, z)|(x − 1)2 + (y + 2)2 + z 2 = 9}
is sphere of radius 3 with center (1, −2, 0). The set
{(x, y, z)|1 ≤ x2 + y 2 + z 2 ≤ 4, x ≥ 0}
is the difference of the balls centered at the origin with radii 1 and 2 that lies on the right side
(the region with positive x-coordinate) of the yz-plane.
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Remark 12.1. It is known that a ball of radius r has volume 4π
3 r . Also, the way to determine
whether three points are in a line is to see whether the segments connecting any two of these
three points form a triangle.
−−→
12.2. Vectors. A vector with initial point A and end point B denotes the replacement AB. It
is clear that for any three points A, B, C in the space, the displacement from A to C is a sum
−→ −−→ −−→
of the displacement from A to B and the displacement from B to C. That is, AC = AB + BC.
Definition 12.2 (Vector addition and scalar multiplication). If u and v are vectors where the
initial point of u is the end point of v, then u + v denotes the vector from the initial point of v
to the end point of u. If c is a scalar and v is a vector, then cv denotes a vector whose length
is |c| times the length of v and the direction is the same same as v if c > 0 and opposite if
c < 0. If c = 0 or v = 0, then cv = 0.
Remark 12.2. For any two vectors u, v, u + v = v + u. Parallelogram law. Any two vectors are
parallel if one of them is a scalar multiple of the other. The difference u − v means u + (−v).
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For some purposes, it is better to introduce a coordinate system and treat vectors algebraically. By removing the initial point of a vector v to the origin, the terminal point
has coordinate (v1 , v2 ) or (v1 , v2 , v3 ) depending on whether the space is 2-dimensional or 3dimensional. To the convention, write
v = ⟨v1 , v2 ⟩,
v = ⟨v1 , v2 , v3 ⟩.
Here, v1 , v2 , v3 are components of v. Consider the example, let O be the origin in R3 and
−−→
P (1, 3, 2). Then, OP = ⟨1, 3, 2⟩. Also, one can shift the vector to the other with a different
initial point and either of them is thought of a representation of ⟨1, 3, 2⟩. Thus, a vector
can have infinitely many different representations depending on whether the initial point is
the same. Among all representations of a vector, the one having the origin as the initial point
is called a position vector. Obviously, if A(x1 , y1 , z1 ) and B(x2 , y2 , z2 ) are points such that
−−→
AB is a representation of a vector v = ⟨a, b, c⟩, then x2 = x1 + a, y2 = y1 + b and z2 = z1 + c.
Definition 12.3. For any two points A(x1 , y1 , z1 ) and B(x2 , y2 , z2 ), define
−−→
AB = ⟨x2 − x1 , y2 − y1 , z2 − z1 ⟩.
For any vector v = ⟨a, b, c⟩, the length or magnitude is defined to be
√
|v| = ∥v∥ = a2 + b2 + c2 .
The definition for vectors in R2 is similar.
−−→
Remark 12.3. For any two point A, B, |AB| = |AB| where the latter term denotes the length
of the segment connecting A and B.
Definition 12.4. For any two vectors v1 = ⟨a1 , b1 , c1 ⟩ and v2 = ⟨a2 , b2 , c2 ⟩,
v1 + v2 = ⟨a1 + a2 , b1 + b2 , c1 + c2 ⟩.
If α is a scalar, then
αv1 = ⟨αa1 , αb1 , αc1 ⟩.
Remark 12.4. Note that all discussions in the above can be generalized to the n-dimensional
case, whereas a vector v in an n-dimensional vector space is an ordered n-tuple
v = ⟨a1 , a2 , ..., an ⟩.
Properties of vectors If u, v, w are n-dimensional vectors and c, d are scalars, then
(1) u + v = v + u,
(3) u + 0 = 0 + u,
(5) c(u + v) = cu + cv,
(7) (cd)u = c(du),
(2) u + (v + w) = (u + v) + w,
(4) v + (−v) = 0,
(6)(c + d)u = cu + du,
(8) 1u = u.
Example 12.3. Let u = ⟨3, 5⟩, v = ⟨4, −1⟩. Then,
2u − 3v = 2u + (−3v) = ⟨6, 10⟩ + ⟨−12, 3⟩ = ⟨−6, 13⟩
and
|u + v| = |⟨7, 4⟩| =
√
72 + 42 =
√
65.
Remark 12.5. Let
i = ⟨1, 0, 0⟩, j = ⟨0, 1, 0⟩, k⟨0, 0, 1⟩.
Then, for any vector v = ⟨a, b, c⟩,
v = ai + bj + ck.
Remark 12.6. A unit vector is a vector of length 1. For any vector v,
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v
|v|
is a unit vector.
Example 12.4. A woman walks due west on the deck of a ship at 5 km/h. The ship is moving
north at a speed of 35 km/h. Find the speed and
√ the direction√of the woman relative to the
surface of the water. (Answer: The speed is 52 + 352 = 25 2 km/h and the direction is
⟨−1, 7⟩.)
Example 12.5. A load with mass 100 kg hangs from two wires where the angles between the
ceiling and these wires are 60◦ and 37◦ and the tensions are T1 and T2 respectively. Find the
magnitudes of T1 and T2 .
Note that |T1 | and |T2 | satisfy the following two equations.
|T1 | cos 60◦ = |T2 | cos 37◦ ,
|T1 | sin 60◦ + |T2 | sin 37◦ = 100 · 9.8
In detail, since F = T1 + T2 and
T1 = −|T1 | cos 60◦ i + |T1 | sin 60◦ j,
Thus, |T1 | =
√
7840(4 3−3)
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and |T2 | =
T2 = |T2 | cos 37◦ i + |T2 | sin 37◦ j,
√
4900(4 3−3)
.
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F = −100 · 9.8j.