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Chapter 3( 3 d space vectors)

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  • 1. BMM 104: ENGINEERING MATHEMATICS I Page 1 of 22 CHAPTER 3: 3-D SPACE VECTORSBasic Concept of VectorA vector is a quantity that having a magnitude/length (absolute) and a direction. → →Notation: i. AB or a is vector → → ii. AB or a is modulus/absolute value/length of the vector → → BA is opposite to AB → → BA = − AB → →Meanwhile AB = BA .Characteristics of vectors
  • 2. BMM 104: ENGINEERING MATHEMATICS I Page 2 of 22 → → → → → →1. If AB is parallel to c , then AB = k c OR c = t AB where k and t are the scalars or parameters. → →2. If k < 0 OR t < 0 , then AB and c are in the opposite directions.3. If k > 0 OR t > 0 , then they are in the same direction.Example: → → → 1 → i) AB = 2 c or c= AB 2 → ii) 1→ or → → AB = − d d = −2 AB 2Addition Law of Vectors → → → →Let c = AB , d = BC . Refer to below diagram. → → → → → AC = AB + BC = c + d → →If OA is the position vector for A and OB is the position vector for B then → → → OA+ AB = OB → → → → → → AB = OB − OA BUT AB ≠ OA −OB →Example: Find NS = ?Components of Vectors2-D Space
  • 3. BMM 104: ENGINEERING MATHEMATICS I Page 3 of 22 i j2 basic vectors : ~ and ~They are also called unit vectors as i =1 and j =1 ~ ~ →Position vector of point A( a ,b ) is given by OA = a ~ + b j . i ~ →Absolute/ modulus OA = a 2 + 2 b by Pythagoras TheoremExample: →i) OA = −3 i + 4 j ~ ~ → OA = 9 +16 =53-D Space
  • 4. BMM 104: ENGINEERING MATHEMATICS I Page 4 of 22 o All the x-axes, y-axes are perpendicular to each other. o There are 3 basic vectors: i , j , k . ~ ~ ~ o They are all unit vectors that parallel to the axes respectively and thus they also perpendicular to each other. a  →  Position vector of A( a ,b , c ) is OA = a ~ + b j + c k = ( a ,b , c ) =  b  and i ~ ~ c    →Length is OA = a 2 + 2 + 2 b c 3    → →Example: Given OA =  4  . Find OA =? 6   Addition of vectors for 3-D Space → →The sum of two vector OA = ( a1 , a 2 , a 3 ) and OB = ( b1 ,b2 ,b3 ) is the vector formed byadding the respective component;
  • 5. BMM 104: ENGINEERING MATHEMATICS I Page 5 of 22 → → OA+ OB = ( a1 + b1 , a 2 + b2 , a 3 + b3 )Subtraction of vectors for 3-D Space → →The subtraction of two vectors OA = ( a1 , a 2 , a 3 ) and OB = ( b1 ,b2 ,b3 ) is the vectorformed by adding the respective component; → → → AB = OB − OA → → OB − OA = ( b1 − a1 ,b2 − a 2 ,b3 − a 3 )Example: Given A( −1,1,4 ) , B ( 8 ,0 ,2 ) and C ( 5 ,− ,11) . Find 2 →(i) OA →(ii) OB →(iii) OC →(iv) AB →(v) AC →(vi) AB →(v) AC →Unit Vector v~ in the Direction of v → → → v v= ~ → → is a unit vector in the direction of v . v
  • 6. BMM 104: ENGINEERING MATHEMATICS I Page 6 of 22  − 1 →  →Example: Given v =  2  . Find unit vector in the direction of v ?  3   Dot/ Scalar Product → →Notation for Dot product: a • b ∈ ℜ → →The dot product of a = ( a1 , a 2 , a 3 ) and b = ( b1 ,b2 ,b3 ) is the real number a • b obtained → →by → → →→ a• = a b b cos θ where θ is the angle between a and b and 0 ≤ θ ≤ π . → →When we measuring angle between two vectors, the vectors must have the same initialpoint. → → → →Example: (i) Given a =6 and b =7. Find a • b . → → = and θ = π → → (ii) Given a =5 and b 4 . Find a • b . 4
  • 7. BMM 104: ENGINEERING MATHEMATICS I Page 7 of 22Example:i • i = i i cos 0~ ~ ~ ~i • i = 1 ×1 ×1 = 1~ ~Important:(a) j• j = k • k = i • i = 1 ~ ~ ~ ~ ~ ~(b) i• j = i ~ ~ ~ j cos 90  ~(c) i• j = 1×1×0 = 0 ~ ~(d) i • j = 0 = i • k = j• k ~ ~ ~ ~ ~ ~. → → → →If a ⊥ b then a • b = a b cos 90  = 0 . ~ ~Note:
  • 8. BMM 104: ENGINEERING MATHEMATICS I Page 8 of 22 ~ ~ → →(i) If a• b = 0 then a ⊥ b → → → →(ii) a× b ≠ b× aProperties of the dot product → → → →a) a• b = b• a : Cumulative → → → → → → b •a = b a cos θ=a •b → → → → → → →b) a •b + c  = a • b + a • c   : Distributive → →  → → →  →c) k a • b  = k a  • b = a •k b       Example:  5   8    →  →  → →Finding angle between the vectors a =  3  and b =  − 9  given that a • b = −9 . − 2  11     Formula to compute scalar product  a1   b1  → →     a • b =  a 2  •  b2  a  b   3  3 → → a • b = a1b1 + a 2 b2 + a 3 b3  3  − 2    →→   → → → →Example: Given a =  4  and b =  6  . Verify a • b = b • a .  − 1  3     Remark: By using addition law of vectors and the law of cosine, it can be showed that → → a b + b + b = a b cos θ. a 1 a 1 2 2 3 3 → → →Component of a in the direction of n : Denoted as comp n a →
  • 9. BMM 104: ENGINEERING MATHEMATICS I Page 9 of 22 OQcos θ = OPOQ = OP cos θ a cos θ → = → →OQ is called component of vector a in the direction of n →OQ = Comp→ a n = OP cos θ = a cos θ → ^ n=1 → ^ = a cos θ ⋅ n since ~ ~By the definition of product → → ^ comp → a = a • n n ~Example:Find 2  1  →   → →  (i) comp → a given a =  1  and n = 1  . n 1  1     
  • 10. BMM 104: ENGINEERING MATHEMATICS I Page 10 of 22  2  1  →  →  →  (ii) comp → a given a =  − 1  and n = 1  . n  −7  1     Application of Dot Product : WORK DONEWork done = Magnitude of force in the direction of motion times the distance it travels = F cos θPQ → → → → W =F •PQ  1  →  Example: A force F =  − 2  causes a body to move from P (1,− ,2 ) to Q (7 ,3 ,6 ) . 1  3   Find the work done by the force.Vector/ Cross Product → →Definition: a×b is defined as a vector that → → → →1. a×b is perpendicular to both a and b . → → → → → → ( a× b ) • a = 0 and ( a× b ) • b = 0
  • 11. BMM 104: ENGINEERING MATHEMATICS I Page 11 of 22 → → → →2. Direction of a×b follows right-handed screw turned from a to b→ → → →b× a = − a× b b sin θ →→ → →3. Modulus of a×b is aRemarks: a sin θ →→ → →Modulus of b×a is therefore b→ → → →a× b ≠ b× a→ → → → → → → →a× b = − b × a or b× a = − a× b
  • 12. BMM 104: ENGINEERING MATHEMATICS I Page 12 of 22 → → a×b ∧ =eUnit Vector : → → a×b ~ → → → → ∧ → → ∧ a×b = a×b e = a b sin θ e ~ ~4. i× i = 0 j× j = 0 k× k = 0 ~ ~ ~ ~ ~ ~i× j = i~ ~ ~ j sin 90  k =k ~ ~ ~ and ~ ~ ~ ~ ( ) j×i = j i sin 90  −k = −k ~ ~Similarly, j× k = i and k× j = −i ~ ~ ~ ~ ~ ~ k× i = j and i× k = − j ~ ~ ~ ~ ~ ~ → →Determinant Formula for a×b
  • 13. BMM 104: ENGINEERING MATHEMATICS I Page 13 of 22 i j k→ → ~ ~ ~a× b = a1 a2 a3 b1 b2 b3 a2 a3 a1 a3 a1 a2 = i− j+ k b2 a3 ~ b1 b3 ~ b1 b2 ~ = ( a 2 b3 − a 3 b2 ) i − ( a1b3 − a 3 b1 ) ~ + ( a1b2 − a 2 b1 ) k ~ j ~  3  → → → → → →  → → →Example: Find a×b , b×a and verify that a× b = −b× a given a =  − 4  and  2     9 →  b = −6  .  2   Applications of Cross Product1. The moment of a force → →A force F is applied at a point with position vector r to an object causing the object torotate around a fixed axis.
  • 14. BMM 104: ENGINEERING MATHEMATICS I Page 14 of 22As the magnitude of moment of the force at O isM 0 = F • d = ( Magnitude of force perpendicular to d ) × (Magnitude of displacement) → → → →Thus we have M 0 = F sin θ• r = r×FTherefore we define the moment of the force about O as the vector → → → M 0 = r×F → → → → → As r×F =r F sin θ= M 0 =M 0 →The magnitude, M0 , is a measure of the turning effect of the force in unit of Nm. 2   →Example: Calculate the moment about O of the force F =  3  that is applied at the point 1   with position vector 3j. Then calculate its magnitude.2. Calculate the area of a triangle
  • 15. BMM 104: ENGINEERING MATHEMATICS I Page 15 of 22By the sine rule: 1 1 → → Area of ∆ABC = 2 bc sin A = 2 AC× AB 1 1 → → = 2 ac sin B = 2 BC× BA 1 1 → → = 2 ab sin C = 2 CB×CAExample: Find area for a triangle with vertices A(0 ,7 ,1) , B (1,3 ,2 ) and C ( − 2 ,0 ,3 ) .Equations (Vector, parametric and Cartesian equations) of a line
  • 16. BMM 104: ENGINEERING MATHEMATICS I Page 16 of 22 x  v1  → →   →  Let r ( t ) = OP =  y  and v =  v 2  as a vector that parallel to the line L. z v     3 → → → →As P0 P // v thus P0 P = t v , t is a parameter (scalar). By the addition law of vectors, we obtain → → → OP = OP 0 + P0 Pi.e. the vector equation of a line passing through a fixed point P0 and parallel to a vector→ → → →v is r ( t ) = OP 0 + t v .  x   a   v1         y  =  b  + t  v2   z c v       3  x   a + tv1       y  =  b + tv 2   z   c + tv     3 ⇒ x = a + tv1 , y = b + tv 2 , z = c + tv 3 are the parametric equations of L. x −a y −b z −c⇒ = = is the Cartesian equation of L. v1 v2 v3
  • 17. BMM 104: ENGINEERING MATHEMATICS I Page 17 of 22Example: Find the vector equation of line passes through A( 3,2 ) and B(7 ,5 ) .Example: Find Cartesian equation of line passes through A( 5 ,− ,3 ) and B ( 2 ,1,− ) . 2 4Equation (Vector and Cartesian equations) of a plane  n1   x   n1   a          Vector equation:  n2  •  y  =  n2  •  b  n   z n   c  3    3   Cartesian equation: n1 x + n 2 y + n 3 z = n1 a + n 2 b + n 3 cRemark: In general, Ax + By + Cz = D OR Ax + By + Cz = 1 is the Cartesian equation of a A  plane with a normal vector  B . C   
  • 18. BMM 104: ENGINEERING MATHEMATICS I Page 18 of 22Example: A plane contains A(1,0 ,1) , B ( − 2 ,5 ,0 ) and C ( 3 ,1,1) . Find the vector andCartesian forms of the equation of the plane. → →AB = −3 i + 5 j − k and AC = 2 i + j ~ ~ ~ ~ ~Distance From A Point to A Line and to A PlaneDistance From A Point to A Line → → →Distance, d, from a fixed point P to a line: r ( t ) = OA+ t v where A is a point on the line →and v is a vector parallel to the line is given by → → ∧ → ∧d = AP sin θ = AP v sin θ = AP×v ~ ~  1−t  →  Example: Find the distance from point ( 4 ,3 ,2 ) to the line L : r ( t ) =  2 + 3t  .  3 +t   Distance From A Point to A Plane
  • 19. BMM 104: ENGINEERING MATHEMATICS I Page 19 of 22 → → → →Distance, D, from a fixed point P to a plane n • r = n • OA where A is a point on the plane →and n is a normal vector to the plane is given by → → → ∧ ∧ D = AP cos θ = AP n cos θ = AP •n . ~ ~Example: Find the distance from point ( 4 ,− ,3 ) to the plane x + 3 y − 6 z = 9 . 3 PROBLEM SET: 3-D SPACE VECTORS1. Points P , Q and R have coordinates ( 9 ,1,0 ) , ( 8 ,− ,5 ) and C ( 5 ,5 ,7 ) 3 respectively. Find (a) the position vectors of P, Q and R. → → (b) PQ and QR .
  • 20. BMM 104: ENGINEERING MATHEMATICS I Page 20 of 22 → → (c) PQ and QR .2. A triangle has vertices A(1,3 ,2 ) , B ( −1,5 ,9 ) and C ( 2 ,7 ,1) respectively. Calculate the vectors which represent the sides of the triangle. → → → → → →3. Find a • b and verify that a • b = b • a if  2  3  1  3  →   →   →   →   (a) a = − 5  , b = 2  (b) a = 8  , b = − 2   0  0  7   5         4. (a) Find the component of the vector 2 ~ + ~ + 7 k in the direction of the vector i j ~ i + j+ k . ~ ~ ~ (b) Find the component of the vector 7 ~ + 2~ j − k in the direction of the vector i ~ i − j+ 2 k . ~ ~ ~ →5. A force F = 3 i + 7 k causes a body to move from point A(1,1,2 ) to point ~ ~B (7 ,3 ,5 ) . Find the work done by the force.  3   5  →   →   → →6. (a) If a =  2  and b =  − 1 , find a×b .  − 1  − 1     → → → → (b) Verify that a× b = −b× a .7. (a) Find the area of the triangle with vertices P (1,2 ,3 ) , Q( 4 ,− ,2 ) and 3 R ( 8 ,1,1) . → (b) A force F of magnitude 2 units acts in the same direction of the vector 3 i − 2 j + 4 k . It causes a body to move from point S ( −2 ,− ,− ) to 3 4 ~ ~ ~ point T (7 ,6 ,5 ) . Find the work done by the force.8. Find the vector equation of the line passing through (a) A( 3 ,2 , ) and B ( −1,2 ,3 ) . 7 → → (b) the points with position vectors p = 3 i + 7 k − 2 k and q = −3 ~ + 2 j + 2 k . i ~ ~ ~ ~ ~ Find also the cartesian equation of this line.
  • 21. BMM 104: ENGINEERING MATHEMATICS I Page 21 of 22 1    (c) ( 9 ,1,2 ) and which is parallel to the vector 1  . 1   9. Given A( 9 ,1,1) , B ( 8 ,1,1) and C ( 9 ,0 ,2 ) . Find (a) the area of the triangle ABC. (b) the Cartesian equation of the plane containing A, B and C.  4 −t  →  10. (a) Find the distance from point (6 ,3 ,5 ) to the line L : r ( t ) =  − 3 + 2t  .  2 + 5t    (b) Find the distance from point ( 4 ,− ,6 ) to the plane 2 x − 4 y + 3 z = 8 . 3  −7    →11. (a) Find the distance from A( 2 ,1,− ) to the plane  5  • r ( t ) = 10. 3  6    y +4 z −3 (b) Find the distance from point B (6 ,3 ,− ) to the line L : 5 = ,x = 5 3 2. ANSWERS FOR PROBLEM SET: 3-D SPACE VECTORS 9   8  5  →   →   →  1. (a) OP =  1  , OQ =  − 3  , OR =  5  0   5  7         −1  − 3     (b) − 4 , 8  (c) 42 , 77  5  2     − 2   3   1  →   →   →  2. AB =  2  , BC =  2  , AC =  4   7  − 8   − 1      
  • 22. BMM 104: ENGINEERING MATHEMATICS I Page 22 of 223. (a) −4 (b) 22 10 34. (a) (b) 3 65. 39 Joule6. (a) − 3 i − 2 j − 13 k ~ ~ ~  3  1 → 2   907. (a) 1106 (b) F = − 2  , W = Joule 2 29   29  4  3  − 4  →    8. (a) r =  2  + t 0  7   − 4       3  − 6  →     x −3 y −3 z +2 (b) r =  7  + t − 5  , = = − 2  4  −6 −5 4      9  1 →     (c) r =  1  + t 1  2  1     19. (a) (b) y +z =2 2 845 3010. (a) ≈ 5.31 (b) ≈ 5.57 30 29 37 145711. (a) (b) ≈ 10.59 110 13