Linear algebra
2.2 Three-dimensional space
Every point in three-dimensional space has a unique set of x-y-z coordinates defi ned in a fi xed Cartesian reference frame, as shown in Figure 2.1. The location of a particle at any instant of time is defi ned by the Cartesian coor-dinates of the point it occupies in space. A position vector, r, defi ning the location of the particle is a line segment drawn from the origin (0,0,0) of the Cartesian system to the particle. The position vector has a unique direction with respect to the Cartesian frame and a calculable length denoted by |r|.
Let r1 and r2 represent two vectors defi ned in a Cartesian system. Vector addition in three-dimensional space is defi ned geometrically, as illustrated in Figure 2.2a. A vector parallel to v is drawn such that its tail coincides with the head of the second vector u. If w = r1 + r2, then w is the vector drawn from the origin to the head of the vector parallel to r2. Figure 2.2b shows the process repeated, but with a vector parallel to r1 drawn with its tail coincid-ing with the head of v. The resultcoincid-ing sum is the same as that obtained in Figure 2.2a. Thus vector addition is commutative,
r1+ = + (2.1)r2 r2 r1
Figure 2.2 illustrates the triangle rule for vector addition. The vectors being added and the resultant (sum) are depicted as sides of a triangle. Since the
x
y z
r
(x,y,z)
Figure 2.1 A point in a Cartesian coordinate system is referenced by three coordi-nates (x,y,z). A position vector is drawn from the origin to the point.
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length of any side of a triangle must be less than the sum of the lengths of the other two sides,
r1+ ≤ +r2 r1 r2 (2.2) Equation 2.2 is called the triangle inequality.
The concept of multiplication of a vector by a scalar is illustrated in Figure 2.3. Let α be any real value. The vector αr is the vector parallel to r whose length is α times the length of r. If α is positive, then the vector α r lies in the same direction as r. If α is negative, the vector α r lies in the direction opposite that of r. If α equals zero, then αr is a vector whose length is zero and is called the zero vector, 0.
The following properties follow from the defi nitions of vector addition and multiplication of a vector by a scalar:
i. Associative law of addition: (r1 + r2) + r3 = r1 + (r2 + r3) ii. Commutative law of addition: r1 + r2 = r2 + r1
iii. Addition of zero vector: 0 + r1 = r1
iv. Multiplication by one: (1)r1 = r1
v. Negative vector: −r1 = (−1)r1 r1 + (−r1) = 0
vi. Distributive law of scalar multiplication: (α β+ )r1=αr1+βr1
vii. Associative law of scalar multiplication: (αβ)r1=α β( r1) viii. Distributive law of addition: α(r1+r2)=αr1+αr2
A unit vector is a vector whose length is one. Let i , j, and k be a set of unit vectors parallel to the x, y, and z coordinate axes respectively. Using the defi nitions of vector addition and multiplication of a vector by a sca-lar, the position vector for the particle can be written in terms of the unit vectors as:
r= + +xi yj z (2.3)k
r1
(a) (b)
r1+r2
r1+r2 r2
r2
r1
Figure 2.2 (a) Vector addition is illustrated using the triangle rule. The resultant is drawn from the tip of r1to the head of r2. (b) Vector addition is commutative.
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Any vector in three-dimensional space may be written as a linear combina-tion of the trio of unit vectors. The vectors i, j, and k form a basis for three-dimensional space.
The Pythagorean theorem is used to determine the length of the position vector as
r= x2+ +y2 z2 (2.4)
A scalar function of two vectors, called the dot product, is defi ned by r r1⋅ =2 r r1 2cosθ (2.5) where θ is the angle made between r1 and r2. The dot product has the geomet-ric interpretation that it is equal to the length of the projection of the vector r1 onto r2. This leads to the calculation of the dot product as:
r r1⋅ =2 x x1 2+y y1 2+z z (2.6)1 2
It should be noted that when computed using Equation 2.6, the dot product has the following properties:
Commutative property: r1 . r2 = r2 . r1
Multiplication by scalar: (αr1)⋅ =r2 α(r r1⋅ 2) Distributive property: (r1 + r2) . r3 = r1 . r3 + r2 . r3
Non-negative property: r1 . r1≥ 0 and r1 . r1 = 0 if and only if r1 = 0
r
αr,α>1 (a)
r
αr,α<0 (b)
Figure 2.3 (a) Multiplication of a vector by a scalar leads to a vector lying in the same direction as the vector, but whose length is multiplied by the scalar. (b) If the scalar is negative, the product vector opposes the original vector.
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It should also be noted that from Equation 2.5 and Equation 2.6,
r r⋅ =r2 (2.7)
Three-dimensional space, the defi nition of a vector, and the operations of vector addition and multiplication by a scalar are generalized in section 2.3 into the concept of vector spaces. The properties of vectors in a general vector space are defi ned in the same way as the properties satisfi ed by vectors in three-dimensional space. The scalar function of the dot product is generalized in section 2.5 into the concept of inner products.
Example 2.1 A force, F, acting on a particle is represented as a vector in three-dimensional space. The force has a magnitude |F| and lies in the direc-tion of a unit vector e. The resultant force acting on a particle is the vector sum of all forces acting on the particle. A free-body diagram of a particle at a given instant is shown in Figure 2.4. Determine the resultant force acting on the particle at this instant. Specify its magnitude and a unit vector in the direction of the resultant.
Solution The forces are represented as:
F1= −300j N (a)
F2=600e2 N (b)
F3=500e3 N (c)
z
y
x 300N
(0, 1, 0)
600N
600N
(4, 2, 2) (–2, 4, 0)
500N
Figure 2.4 Forces on a free-body diagram are drawn in the direction of their line of action with magnitude shown. The forces are thus modeled as vectors.
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where the unit vectors are
The resultant force acting on the particle is F= + +F
The magnitude of the resultant force is
F= ( ) + +
The resultant force is in the direction of the unit vector,
e i j k
Example 2.2 The motion of a particle is tracked by its position vector r, a vector drawn from the origin of a Cartesian coordinate system to the position of the particle. Since the particle moves in space, the particle’s position vector is a function of time as well as of the initial position of the particle. The veloc-ity is defi ned as the time rate of change of the position vector,
v r
=d
dt (a)
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By defi nition, the velocity vector is always tangent to the spatial path tra-versed by the particle. The acceleration is defi ned as the time rate of change of the velocity vector,
a v r
=d = dt
d dt
2
2 (b)
Equation a and Equation b represent the velocity and acceleration of the par-ticle as seen by an observer attached to the parpar-ticle. Modeling of a system using the views of observers attached to particles is called a Lagrangian formulation.
The position vector of a particle moving in space, referenced to a Carte-sian frame, is determined to be
r=[cos( )2t i+sin( )2tj+0 5. tk]m (c) where t is in seconds.
Determine a unit vector tangent to the path of motion at any instant of time; (b) Determine the component of the acceleration normal to the path of motion.
Solution (a) The velocity vector is determined by applying Equation a to the position vector of Equation c, leading to
v= −[ 2sin( )2ti+2cos( )2t j+0 5. k] m
s (d)
Since the velocity vector is always tangent to the path of motion, it can be expressed as
v=v et (e)
where et is a unit vector tangent to the path. To this end,
v = −[ ] +[ ] +
=
2sin( )2t 2 2cos( )2t 2 ( . )0 52 m s 2.87 m
s
(f)
Using Equation d and Equation f, Equation e can be rearranged to give
e v
v
i j k
t=
= −0 693. sin( )2t +0 693. cos( )2t +0 174.
(g)
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The acceleration vector can be calculated from Equation b and Equation d as a= −[ 4cos( )2ti−4sin( )2t j] m
s2 (h)
The tangential component of the acceleration is
at= ⋅a e (i)t
which using Equation g and Equation h, becomes
at= −[ 4cos( )2ti−4sin( )2tj] −i 0 693. sin( )2ti+0.6693 2 0 174 0
cos( )t j+ . k
[ ]
= (j)
Thus, since the tangential component of the acceleration is zero, the accelera-tion is in a direcaccelera-tion normal to the surface.
Example 2.3 A Eulerian formulation is usually used in modeling the fl ow of a fl uid. In a Eulerian formulation, the properties used are those observed at a fi xed location in the fl ow fi eld as different fl uid particles pass through the point. In a Eulerian formulation, the properties are functions of spatial coordinates and time. If v(x,y,z,t) is the Eulerian velocity vector at a point (x,y,z) in the fl ow fi eld, then the acceleration is comprised of two terms. The local acceleration is the rate of change of the velocity and is simply calculated al= ∂ ∂t. The convective acceleration is the rate of change of the velocity v/ at the point (x,y,z) due to the rate at which fl uid particles pass through the point. The convective acceleration is calculated as ac = v . ∇v, where
v⋅∇ =v ∂ i The velocity vector for a fl ow fi eld is
v=⎡⎣xye−2ti+y e2 −2tj⎤⎦ m
s (a)
(a) Determine the acceleration vector for the fl ow
(b) The fl ow rate Q through a surface area is calculated as
Q dA
A
=
∫
v n⋅ (b)where n is a unit vector normal to the surface. Determine the fl ow rate through a square whose vertices are (0,2,0),(0,2,− 2), (2,2,− 2), and (2,2,0).
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Solution (a) Noting that the total acceleration is
(b) The unit normal to the square is n = j, and everywhere on the square, y = 2. Thus