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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.

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Presentation on theme: "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."— Presentation transcript:

1 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:

2 Analytic Geometry of Space O P Q x y z

3 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

4 Analytic Geometry of Space Vector Algebra Equality Two vectors are equal if they have the same magnitude and direction Addition a b a+b

5 Analytic Geometry of Space Negation The vector –a to be the vector having the same magnitude,but the opposite direction a – a = 0 Subtraction a– b = a + (-b) Scalar multiplication The vector ka having the same direction as a, but magnitude k times that of a

6 Vector Algebra Consequence of these definitions: 1.a+b = b+a 2.a+(b+c)=a+(b+c) 3.k1(k2a)=k1k2a 4.(k1+k2)a=k1a+k2a 5.k(a+b)=ka +kb Where k1,k2,k are real numbers and a,b,c are vectors

7 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

8 Let,, In terms of these Cartesian components, we have 1.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

9 The vector equation of a straight line: –Passing through P0 –Direction V P = P0 + t V

10 Vector algebra: the scalar and vector products AB C a D c

11 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.

12 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:

13 1.It can be solved based on the theory of linear equation system: 2.The vector is denoted by

14 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


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