BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 1 3.1 Vector Concepts Magnitude and Direction of a Vector Measurements in sciences are of two types, i.e. scalar and vector quantities. A scalar quantity has magnitude, but not direction. For example, a pen may have length "10 cm". The length 10 cm is a scalar quantity - it has magnitude, but no direction is involved. Other examples of scalar quantities are volume, density, temperature, mass, speed, time, length, distance, work and energy. Each of these quantities has magnitude only, and do not involve direction. A vector is a quantity that has both magnitude and direction. (Magnitude just means 'size'.). Examples of Vector Quantities: - I travel 30 km in a Northerly direction (magnitude is 30 km, direction is North - this is a displacement vector) - The train is going 80 km/h towards Sydney (magnitude is 80 km/h, direction is 'towards Sydney' - it is a velocity vector) - The force on the bridge is 50 N acting downwards (the magnitude is 50 Newton and the direction is down - it is a force vector) Other examples of vectors include acceleration, momentum, angular momentum, magnetic and electric fields. Each of these examples involves magnitude and direction. 3.1.1 The Fundamentals of Vectors a) Vector Notation To distinguish between vector and scalar quantities, various ways are used. These include: - bold capital letter. For example, a force vector could be written as F , AB. - An arrow above the vector name, F, AB. - A line over the top or under the vector name ; F, F, F, F, AB, AB, AB - Cartesian component = 2 i +3 j . - You will also see vectors written using matrix-like notation. For example, the vector acting from (0, 0) in the direction of the point (2, 3): Column vector = 23| | |\ . or 23 ( ( ; Row vector = (2,3) or 2,3 ; b) Vector Representation Graphically, a vector is represented by an arrow, defining the direction, and the length of the arrow defines the vector's magnitude. This is shown in Figure 1. If we denote one end of the arrow by the origin O and the tip of the arrow by Q. Then the vector may be represented algebraically by q or OQ. BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 2 c) Equality of Vector Two vectors are equal if they have the same magnitude and direction. In Figure 2 vectors AB and CD are equal even though their positions differ. We can write: AB = CD. d) Negative Vectors A vector having the same magnitude as AB but in the opposite direction to AB is denoted by - AB . We can write: AB = - CD as shown in Figure 3. e) Unit Vectors A unit vector has length 1 unit and can take any direction. A one-dimensional unit vector is usually written i. If u is a vector with magnitude u , then uuis a unit vector having the same direction as u . If u has length 5, for example, then a unit vector in the direction of u is clearly 1u5 . Figure 1 OQ= q CD Figure 2 A B AB C D AB Figure 3 A B AB A B BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 3 f) Magnitude of a Vector We indicate the magnitude of a vector using vertical lines on either side of the vector name. The magnitude of RSis written RS . So for example, vector a above has magnitude 5 units. We would write the magnitude of vector a as: | a | = 5. If vector xAB xi yjy| |= + = |\ .then the magnitude of AB is 2 2AB x y = + g) Position Vector If O is the origin, OA is known as the position vector of A (relative to O). 3.2 The operation of Vectors 3.2.1 Addition and Subtraction of Vectors a) Adding 1-dimensional Vectors Subtraction of Vectors The 2 vectors are now acting in opposite directions. v u u + v B O A From the diagram; OB - OA = AB OA + AB = OB v u v u u - v BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 4 b) Adding Vectors Using a Triangle Law The resultant vector of the vectors a and b can be obtained by connecting the starting point of b with the endpoint of a as shown in Figure 4. c) Adding Vectors Using a Parallelogram Law The resultant vector of the vectors a and b can be obtained by connecting both vectors of a and b from the common starting point. Complete the parallelogram OABC as shown in Figure 5. d) Adding Vectors Using a Polygon method The resultant vector of three or more vectors can be using the polygon law. The resultant vector of a, b,c and d can be determined by constructing a polygon as shown in Figure 6: a A O a b + b b C O a A A Figure 5 C B C A O A A O C a b + a b b a Figure 4 BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 5 Resultant vector = a b c d + + + = ( )AB BC CD DE + + + = (AC CD) DE + + = AD DE AE + = d) Adding Vectors Using Method of Components We can write any 2-dimensional vector in terms of the unit vectors i and j. Example In an earlier example, we had the following vector: 3.2.2 Scalar Multiplication We can increase or decrease the magnitude of a vector by multiplying the vector by a scalar. When a vector a is multiplied by a scalar k, the product is ka. Scalar Multiplication - Example 1 C B A D E a b c Figure 6 d We can write these vector components using subscripts as follows: Vx = 6 units; Vy = 3 units We have: Magnitude of V = | V | = (62 + 32)= 45= 6.71 units We could write the components of the vector V as follows. Vx = 6 i ; Vy = 3 j ,So we can write the vector V using unit vectors as follows: V = 6 i + 3 j BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 6 In the examples we saw earlier, vector B (2 units) is half the size of vector A (which is 4 units) . We can write: B = 0.5 A This is an example of a scalar multiple. We have multiplied the vector A by the scalar 0.5. Properties of addition and scalar multiples of vector For any vectors p, q ,r and scalar and . 1. p + q = q + p Commutative Law for Addition 2. p + ( q + r) = ( q + p ) + r Associative Law for Addition 3. ( p) = ( ) p Associative Law for muliplication 4. (p + q) = p + q Distributive Law for muliplication 5. ( + ) p = p+ p Distributive Law for muliplication 3.3 Dot Product (Scalar Product) of Two Vectors 3.3.1 Definition of scalar product If we have any 2 vectors P and Q, the dot product of P and Q is given by: Where | P | and | Q| are the magnitudes of P and Q respectively, and is the angle between the two vectors. The dot product of the vectors P and Q is also known as the scalar product since it always returns a scalar value. 3.3.2 The properties for scalar products are given in the following list. If P and Q are parallel vectors, P Q P Qcos0 P Q - = = The scalar product is commutative The distributive rule Where m is a scalar u P Q P Q P Qcos - = u BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 7 If P and Q are perpendicular (orthogonal) vectors, 0 P Q P Qcos90 - = = 3.3.3 Calculation Of The Scalar Product Dot Products of Unit Vectors For the unit vectors i (acting in the x-direction) and j (acting in the y-direction), we have the following dot (i.e. scalar) products (since they are perpendicular to each other): i j i j 0 - = - = and i i j j 1 - = - = Recall that vectors can be written using scalar products of unit vectors. If we have 2 vectors P and Q defined as: P = a i + b j ; Q = c i + d j , Where a, b, c, d are constants; i is the unit vector in the x-direction; and j is the unit vector in the y-direction, then it can be shown that the dot product (scalar product) of P and Q is given by: P Q ac bd - = + P Q - = (a i + b j ) (c i + d j )= ( )ac i i - + ( ) ( ) ( )ad i j bc j i bd j j - + - + - = ac + bd This result can be generalized for P and Q in any orientation. Example Find P Q - if P = 6 i + 5 j and Q = 2 i 8 j Answer P Q - = (6 i + 5 j ) (2 i 8 j ) = (6 2) + (5 -8) = 12 40 = 28 Now we see another use for the dot product finding the angle between vectors. Angle Between Two Vectors P Q P Qcos - = u P QcosP Q-u = P Qar cosP Q-u = Example Find the angle between the vectors P = 3 i - 5 j and Q = 4 i +6 j BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 8 We use the formula we just derived: 2 2 2 2(3x4) ( 5x6) 18115.334 523 ( 5) 4 6P Qarccos arccos arccosP Q + = = = | | | |+ + | |\ . \ .-u = Therefore the angle between the vectors P and Q is, = 115.3 The 3-dimensional Co-ordinate System We can expand our 2-dimensional (x-y) coordinate system into a 3-dimensional coordinate system, using x-, y-, and z-axes. We normally use the 'right-hand orientation' for the 3 axes as shown below. 2 2 2distance OP OP a b c = = + + Vectors in 3-D Space The vector OP has initial point at the origin O (0, 0, 0) and terminal point at P (2, 3, 5). We can draw the vector OP as follows: 2OP 2i 3j 4k 34| | |= + + = | |\ . BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 9 Magnitude of a 3-Dimensional Vector For the vector OP above, the magnitude of the vector is given by: 2 2 2OP 2 3 5 38 6.164 units = + + = = Direction Cosines Direction of OPdetermined by cos , cos and cos are known as direction cosines and , and are direction angles. Adding 3-dimensional Vectors Example Two anchors are holding a ship in place and their forces acting on the ship are represented by vectors A and B as follows: A = 2i + 5j 4k and B = 2i 3j 5k Answer The problem just requires us to add the vectors to get the single resultant vector. A +B = (2 + 2) i + (5 3)j + (4 5)k = 0 i + 2 j 9 k = 2 j 9 k Dot Product of 3-dimensional Vectors Example - Dot Product Using Magnitude and Angle Find the dot product of the vectors P and Q given that the angle between the two vectors is 35 and | P | = 25 units and | Q | = 4 units Answer P Q = |P| |Q| cos ; we have: is the angle between OP and the x-axis, is the angle between OP and the y-axis is the angle between OP and the z-axis. OP = cos i + cos j + cos k y x zcos , cos , cosOP OP OPo = | = = P BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 10 P Q = |P| |Q| cos = 25 4 cos 35 = 81.92 Example - Dot Product if Vectors are Multiples of Unit Vectors Find the dot product of the vectors A and B. A = 2i + 5j 4k and B = 2i 3j 5k Answer A B = (2i + 5j 4k) (2i 3j 5k) = (2 -2) + (5 -3) + (-4 -5)= -4 + -15 + 20 = 1 Example a) Find the angle between the vectors P= 4i + 7k and Q = -2i + j + 3k. Answer Using the formula ; P QarccosP Q-u = 2 2 2 2 2(4x( 2)) (0x1) (7x3) 1364.4765 144 7 ( 2) 1 3P Qar cos ar cos ar cosP Q + += = = | | | |+ + + | |\ . \ .-u = Therefore the angle between the vectors P and Q is = 64.47 b) Find the angle between the vectors P = 3i + 4j 7k and Q = -2i + j + 3k. Answer 3.4 Cross Product (Vector Product) of 2 Vectors 3.4.1 Definition of Vector Product Suppose we have 2 vectors A and B. These 2 vectors lie on a plane and the unit vector n is normal (at right angles) to that plane. The cross product (also known as the vector product) of A and B is given by: A B A B sin = u n The right hand side represents a vector at right angles to the plane containing vectors A and B. Note: Some textbooks use the following notation for the cross product: AB. BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 11 3.4.2 Properties Of Vector Product - The length of the cross product of two vectors : - Anticommutativity: - Multiplication by scalars: - Distributivity: 3.4.4 The Area of a parallelogram 3.5 Triple Product In mathematics, the triple product is a product of three vectors. The name "triple product" is used for two different products, the scalar-valued scalar triple product and, less often, the vector-valued vector triple product. 3.5.1 Scalar Triple Product Area of parallelogram = OC x OA OC x OA sin = u Area of triangle OCA = 1OC x OA2 Unit vector that is perpendicular to both vector OC x OAOC and OAOC x OA= = H=OC sinu u O C B A Three vectors defining a parallelepiped The scalar triple product (also called the mixed or box product) is defined as the dot product of one of the vectors with the cross product of the other two. Properties Switching the two vectors in the cross product negates the triple product, i.e.: . BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 12 b~ h | a~ Geometric interpretation Geometrically, the scalar triple product a~- ( b~ x c~) = ( a~ x b~)- c~=b~- ( c~x a~) is the (signed) volume of the parallelepiped defined by the three vectors given. volume of the parallelepiped The area of base of =l b~ x c~l, and the height , h = ac x bc x b~~~~~- Volume of the parallelepiped = |b~ x c~ |x ac x bc x b~~~~~- = |( b~ x c~) - a~| unit3 Example Given a~ = 5k j, b~ = 3i + j and c~= i + j + k. Find the volume of the parallelepiped Solution a~x b~ = 0 1 35 1 0 k j i = (0 5)i (0 15)j + (0 + 3)k = -5i + 15j + 3k |a~x b~|- c~ = 3131113155unit =|||.|

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\| z y x c~ u A n c a b BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 13 Example Find the volume of the parallelepiped formed by the points A(5,4,5), B(4, 10, 6), C(1,8,7), and D(2,6,9) Solution The position vectorin unit vector; OA = 5 i + 4 j + 5k , OB = 4 i + 10 j +6k , OC = i + 8 j + 7k , OD = 2 i + 6 j + 9k AB = AO + OB = - OA + OB = - i + 6 j + k AC = AO + OC = - OA + OC = -4 i + 4 j + 2k AD = AO + OD = - OA + OD = -3 i + 2 j + 4k Volume = The area of base x height = |( a~ x b~) - c~ | = |( AB x AC) - AD| ABx AC = i j k1 6 14 4 2 = 8 i 2 j + 20k ( AB x AC) - AD = 524232028=((((

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unit3 EXERCISE 1. If A~ = i~+ 2 j~+ 3k~ , B~ = 2 i~- 3 j~+k~, C~ = 3 i~+ j~- 2Find A~ - ( B~ x C~) , B~ - ( A~x C~) and A~ - ( B~ x A~) Answer A~ - ( B~ x C~) = 2 1 31 3 23 2 1 = 1(6 - 1) 2(- 4 - 3) + 3(2 + 9) = 52 B~ - ( A~x C~) = 2 1 33 2 11 3 2 =2(- 4 - 3) + 3(-2 - 9) + 1(1- 6) = -52 A~ - ( B~ x A~) = 3 2 11 3 23 2 1 =1(-9 - 2) - 2(6 - 1) + 3(4 + 3) = 0 BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 14 2. If A~ = 2 i~-3 j~+ k~ , B~ = i~+ 2 j~- k~, C~ = 3 i~+ j~+ 3k~ . Find A~x (B~ x C~) . Answer B~ x C~ = 3 1 31 2 1k~j~i~ = 1 32 1k~3 31 1j~3 11 2i~ + = i~k~(6 + 1) - j~(3+3) + k~(1 6) = 7 i~- 6 j~ - 5 k~ Then A~x (B~ x C~) = = 5 6 71 3 2k~j~i~ 6 73 2k~5 71 2j~5 61 3i~+ = i~k~(15 + 6) - j~(-10 - 7) + k~(-12 + 21) = 21 i~+17 j~ + 9k~ 3. If a 2,1, 1 = and b 3,4,1 = compute each of the following. (a) a x b (b) b x a Answer (a) Here is the computation for this one. (b) And here is the computation for this one. 4. Given that P (-3, 6), Q (2, 8) and R (2, 7), find 3PQ QR in terms of unit vectors iand j . 5. Givena 2i j 3k and b=3i-2j+k = + + , find the angle between a and b . 6. The points A, B and C have position vectors 2i j 3k , 3i-j+k and i-2j+k + + respectively, Find: a. The length of AB and AC b. The angle BAC, correct to the nearest degree. BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 15 c. The area of triangle ABC. 7. If a~= k~j~5 i~ , b~ = k~j~5 i~2 + + and k~2 j~i~3 c~ + = . Find : i. a~ + c~ + b~ ii. |a~ + c~ - b~| 8. Given k j i OA + = 3 2 and k j i OB 3 2 + = ; find the direction angle of AB. 9. For what values or value of h will the vectors u 2i 4 j 5k = + and v 4 i 8 j hk = + be parallel? 10. The triangle ABC has vertices at (2,3,4) , (-2,1,0) and ( 4,0,2). Find; i. BAC Z ii. The area of the triangle 11. If a 2 = , b 3 = , and the angle between a and b is 3t, find a 2b . 12. Simplify; ii. a~ - ( a~ Xb~) ii. ( a~ - b~) X a~ 13. The angle between P and Q is 3t. Find x if P= k~j~i~x + + and Q = k~j~x i~ + + . b) Given k~2 j~i~3 A + = , k~2 j~3 i~B + = , k~2 j~2 i~9 C + + = , Find; i. Magnitude and direction cosine of B A ii. The angle between B A and C . 14. If OA = k~3 j~4 i~2 + and OB = k~2 j~3 i~ + + . Find :- i. OA - OB ii. OA x OB iii. The angle between OA and OB 15. If a~ = i~ - j~ + 2 k~, b~ = 3 i~ + 4 j~ - 2k~ and c~ = -5 i~ + j~ - 4 k~, find:- i. Direction cosine of c~ iv. A unit vector parallel to a~ - b~ + 2 c~ ii. The angle between ( a~ + b~) and ( a~ - b~) v. ( a~ x b~) - ( b~ x c~) iii. a - ( b x c~) 16. If a~= k j i~ ~5~ , b~ = k j i~ ~5~2 + + and k j i c~2~ ~3~ + = . Find b~ ( a~ - c~). 17. Find the value(values) of o, i. a~ = k~j~i~2 + , b~ = k~j~i~2 + + o and b~a~ = ii. a~ = j~i~2 , b~ = j~3 i~ + o and 5 b~a~ = + iii. a~ = k~j~i~ + + o , b~ = k~j~i~ + o + and the angle between a~ and b~is 3t 18. Given A~ = 2 i~- j~+ 3k~ , B~ = 3 i~+ 2 j~+ k~, C~ = i~+ p j~+ 4k~, if A~ - ( B~ x C~) = 0, find the value of p. 19. Given ( ) w~v~u~ - = 3. Find; i. ( ) v~w~u~ - iii. ( ) w~u~v~ - v. ( ) u~w~v~ - BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 16 ii. ( ) v~w~u~ - iv. ( ) v~u~w~ - vi. ( ) w~w~v~ - 20. Given a~= kji + 2 , b~ = ki4 3 , c~ = ji+ and kjid~3 2 + = . Find; i. ( ) c~b~a~ ii. ( ) c~b~a~ - iii. ( ) ( ) d~b~c~a~ - ANSWER a. (i) 75.96 ii) 12 unit2 b. |||.|

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