analytic geometry o f space 3d space (right-handed coordinate system) introduction to vectors –let...
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Analytic Geometry o f Space
• 3D Space (right-handed coordinate system)
• Introduction to Vectors– Let – We may to know the displacement from P to Q
• From P to Q denoted by PQ(or )
• From Q to P denoted by PQ(or )
– The relative coordinates of PQ is as follows:•
),,(),,,( 222111 zyxQzyxP
PQ
QP
),,( 121212 zzyyxxPQ
Analytic Geometry of Space
O
P
Q
x
yz
Analytic Geometry of Space
We call the directed displacement PQ the “relative position vector” of Q with respect to P
OP and OQ are the absolute vector of P and Q • OQ = OP +PQ• PO+OQ=PQ (PO= -OP)
O
P
Q
Analytic Geometry of Space
Vector Algebra
• EqualityTwo vectors are equal if they have the same
magnitude and direction
• Additioncba
a
b
a+b
Analytic Geometry of SpaceNegation• The vector –a to be the vector having the same
magnitude,but the opposite direction • a – a = 0Subtraction • a– b = a + (-b)Scalar multiplication• The vector ka having the same direction as a, but
magnitude k times that of a
Vector Algebra
• Consequence of these definitions:1. a+b = b+a 2. a+(b+c)=a+(b+c)3. k1(k2a)=k1k2a4. (k1+k2)a=k1a+k2a5. k(a+b)=ka +kb
Where k1,k2,k are real numbers and a,b,c are vectors
Magnitude of a vector. Unit vector 1. Let a=(x,y,z) be a vector, then the magnitude or
length of a is denoted by |a|
2. The unit vector u=a/|a|
Cartesian components of a vector
If the unit vectors of x,y and z axis are respectively be i,j,k , and vector a=(x,y,z) then
a=(x,y,z) = xi + yj + zk
222|| zyx a
Let , , In terms of these Cartesian components, we have1. a+b =
2. 3. 4. The components of a unit vector u give the
cosines of the angles between the vector directions of the x,y and z axes
kjia 321 aaa
),,( 321 aaaa ),,( 321 bbbb
kji )()()(
),,(
332211
332211
bababa
bababa
23
22
21|| aaa a
)(|| kji uu
• The vector equation of a straight line:– Passing through P0– Direction V
P = P0 + t V
Vector algebra: the scalar and vector products
332211. bababa ba
A B
C
a
bD
c
cba
cos||||. baba
The properties of scalar product
1. a.b = b.a
2. a.(b+c)=a.b+a.c
3. (ka).b=a.(kb)=k(a.b)
4. 2|| aa.a
Vector product
• Find a vector v, which perpendicular to both of two given vector a and b:
v.a=v.b=0
We thus have:
0
0
332211
332211
vbvbvb
vavava
1.It can be solved based on the theory of linear equation system:
2.The vector is denoted by
1221
3
3113
2
2332
1
baba
v
baba
v
baba
v
ba
kjiba )()()( 122131132332 babababababa
3. For convenient, it can be written as follows: (determinant of order 3)
4. We have the following result for the modulus of cross product
321
321
bbb
aaa
kji
ba
222
22
sin||||
)cos1(
ba
|b||a||ba| 22