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The BEng & Foundation Degrees and HNC/D in Engineering are designed to meet the needs of employers by placing the emphasis on the theoretical, practical and vocational aspects of engineering within the workplace and beyond. Engineering is becoming more high profile, and therefore more in demand as a skill set, in today’s high-tech world. This course has been designed to provide you with knowledge, skills and practical experience encountered in everyday engineering environments.

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- 1. Vectors Mathematics 1 Level 4 © University of Wales Newport 2009 This work is licensed under a Creative Commons Attribution 2.0 License .
- 2. <ul><li>The following presentation is on the basis of Vectors – one part for level 4 Mathematics. This resources is a part of the 2009/2010 Engineering (foundation degree, BEng and HN) courses from University of Wales Newport (course codes H101, H691, H620, HH37 and 001H). This resource is a part of the core modules for the full time 1 st year undergraduate programme. </li></ul><ul><li>The BEng & Foundation Degrees and HNC/D in Engineering are designed to meet the needs of employers by placing the emphasis on the theoretical, practical and vocational aspects of engineering within the workplace and beyond. Engineering is becoming more high profile, and therefore more in demand as a skill set, in today’s high-tech world. This course has been designed to provide you with knowledge, skills and practical experience encountered in everyday engineering environments. </li></ul><ul><li>Contents </li></ul><ul><li>Introduction </li></ul><ul><li>Vector Addition and Resolution </li></ul><ul><li>Vector Subtraction </li></ul><ul><li>The Unit Triad </li></ul><ul><li>The Scalar Product of Two Vectors </li></ul><ul><li>The Vector Product of Two Vectors </li></ul><ul><li>Credits </li></ul><ul><li>In addition to the resource below, there are supporting documents which should be used in combination with this resource. Please see: </li></ul><ul><li>KA Stroud & DJ Booth, Engineering Mathematics, 8 th Editon, Palgrave 2008. </li></ul><ul><li>http://www.mathcentre.ac.uk/ </li></ul><ul><li>Derive 6 </li></ul>Vectors
- 3. Introduction Definitions of Vectors and Scalars Physical quantities can be classified under two main headings -- Vectors and Scalars. A vector quantity is any quantity that has both magnitude (size) and direction. E.g., velocity, acceleration, force, momentum. A scalar quantity is any quantity that has magnitude only , while direction is not taken into account. E.g., speed, pressure, temperature, energy. Vectors
- 4. Vectors are represented by lines with arrow. The length of the line indicates the magnitude of the vector, and the direction of the line indicates the vector's direction. An arrow is used to denote the sense of the vector, i.e. for a horizontal vector, say, whether it acts from left to right or vice versa. The arrow is positioned at the end of the vector and its position is called the nose of the vector. Vector Addition and Resolution A vector of 20kN acting at an angle of 45 º to the horizontal may be depicted by: oa = 20kN at 45 º to the horizontal 45 º a o 20kN Vectors
- 5. To distinguish between vector and scalar quantities, different conventions are used. The one these notes will adopt is to denote vector quantities in bold print. (Note in these presentations they will also be in blue while the scalar will be in plain text and red – the colour will not help with the copied notes). Thus oa represents the vector quantity but oa is the magnitude of vector oa . Also the convention is that positive angles will be measured in an anticlockwise direction from the horizontal right facing line and negative angles clockwise from this line. Thus 90 º is a line vertically upwards and -90º is a line vertically downwards Vectors
- 6. Let us say we want to add two vectors together, say F 1 at angle θ 1 and F 2 at angle θ 2 as shown. θ 1 F 1 θ 2 F 2 The resultant can be obtained by drawing oa to represent F 1 and then drawing ar to represent F 2 . The resultant of F 1 and F 2 is given by or . This is called the nose-to-tail method of vector addition. F 1 F 2 o a r Vectors
- 7. Alternatively, by drawing lines parallel to F 1 and F 2 from the noses of F 2 and F 1 respectively, and letting the intersection of these lines be R, gives OR as the magnitude and direction of the resultant of adding F 1 and F 2 . This is called the parallelogram method of vector addition. F 1 F 2 O R These two methods are graphical and rely on the accuracy of the drawing of the lines. There is a purely mathematical method for adding these vectors and it is shown on the next slide. Vectors
- 8. A vector can be resolved into two component parts such that the two new vectors are equal to the original vector. The two components are normally a horizontal component and a vertical component. F θ o a F cos θ F sin θ Consider the vector F . If we need to sum a number of vectors then if each is resolved into two directions then the resultant vertical and horizontal components can then be conventionally summed as they are in the same direction. Vectors
- 9. Vectors F 1 and F 2 are to be summed. F 1 θ 1 F 1 cos θ 1 F 1 sin θ 1 The two horizontal vectors sum to give: H = F 1 cos θ 1 + F 2 cos θ 2 The two vertical vectors sum to give: V = F 1 sin θ 1 + F 2 sin θ 2 F 2 θ 2 F 2 sin θ 2 F 2 cos θ 2 Once we have the two resultant vectors V and H we can determine the single resultant by using Pythagoras and trig: Magnitude of the resultant Angle to the horizontal is given by
- 10. Note Resolving the vectors may result in vertical and horizontal components which are either up or down (for the vertical) or to the left or right (for the horizontal). To the right and up are taken as the positive direction and to the left and down as negative. If the angle is measured anticlockwise from the axis to the right then modern calculators will automatically generate the correct sign. 60 º 10N The angle that would be used is 120 º V = 10 sin 120º = 8.66 H = 10 cos 120º = -5 Vectors
- 11. It must be remembered that the subtraction F 1 – F 2 can be thought of as F 1 + (- F 2 ). So how do we find the negative of a vector? Vector Subtraction In the diagram F is represented by oa . The vector – oa can be obtained by drawing a vector from o in the opposite sense to have the same magnitude, shown as ob . ob = - oa F a b o -F Note -F is the same as F but with an angle increase of 180 º Vectors
- 12. For two vectors acting at a point the resulting vector addition is os = oa + ob o a b s If we now want ob + (- oa ) or ob – oa then we will have od = ob - oa o -a b d Comparing od with the first diagram, it is the same as the line ab . Therefore when we complete the parallelogram the two diagonals give us the sum and the difference vectors. Vectors
- 13. Example <ul><li>1. Vector F 1 has magnitude 8, with direction θ =30 °; </li></ul><ul><li>vector F 2 has magnitude 12, with direction θ =60 °. </li></ul><ul><li>Use (1) vector diagram, (2) vector resolution </li></ul><ul><li>to get the resultant force F=F 1 +F 2 </li></ul>2. Vector F 1 has magnitude 6, with direction θ =120 °; vector F 2 has magnitude 10, with direction θ =-30 °. Use (1) vector diagram, (2) vector resolution to get the resultant force F=F 1 +F 2 3. Vector F 1 has magnitude 9, with direction θ =240 °; vector F 2 has magnitude 18, with direction θ =-60 °. Use (1) vector diagram, (2) vector resolution to get the resultant force F=F 1 +F 2
- 14. When a vector x of magnitude x and direction θ º is divided by the magnitude of the vector the result is a vector of unit length at an angle θ º. The unit vector of a velocity 10 m/s at 50º is The Unit Triad. In general the unit vector for oa is: oa /| oa | oa being a vector with both magnitude and direction and | oa | being the magnitude of the vector only. Vectors
- 15. One method of completely specifying the direction of a vector in space relative to some reference point is to use three unit vectors mutually at right angles to each other. o i j k x y z This is called a unit triad. The next slide shows how this is used to specify a three dimensional vector. Vectors
- 16. In the diagram below one way of getting from o to r is to move x units in the i direction, to a point a, y units in the j direction to get to b and z units in the k direction to get to r. o i j k x y z a b r The vector or is specified as: x i + y j + z k α β Vectors
- 17. Example <ul><li>1. A spatial vector r has magnitude of 10 at direction α =60 ° and β =30 °. </li></ul><ul><li>Re-write the vector in the form of x i + y j + z k . </li></ul>2. A spatial vector r has magnitude of 8 at direction α =120 ° and β =-30 °. Re-write the vector in the form of x i + y j + z k . 3. A spatial vector r has magnitude of 12 at direction α =240 ° and β =-60 °. Re-write the vector in the form of x i + y j + z k . Vectors
- 18. When a vector oa is multiplied by a scalar quantity k , the magnitude of the resultant vector will be k times the magnitude of oa and its direction will remain the same. Thus 2 x 5N at 20 º results in a vector 10N at 20 º . One of the products of two vector quantities is called the scalar or dot product of the two vectors and is defined as the product of their magnitudes multiplied by the cosine of the angle between them The scalar product of oa and ob is shown as oa • ob . For vectors oa = oa at θ 1 º, and ob = ob at θ 2 º, where θ 2 > θ 1 , the scalar product is: oa • ob = oa ob cos( θ 2 - θ 1 ) The Scalar Product of Two Vectors Vectors
- 19. For the vectors v 1 and v 2 shown, the scalar product is v 1 • v 2 = v 1 v 2 cos θ The cumulative law of algebra, a x b = b x a, applies to the scalar product. θ v 1 v 2 If v 1 is oa and v 2 is ob then this is shown below: θ v 1 v 2 v 2 cos θ a o b By geometry it can be seen that the projection of ob on oa is v 2 cos θ . But we know that: v 1 • v 2 = v 1 v 2 cos θ = v 1 (v 2 cos θ ) v 1 • v 2 = v 1 times the projection of v 2 on v 1
- 20. Similarly: θ v 1 v 2 v 1 cos θ a o b By geometry it can be seen that the projection of oa on ob is v 1 cos θ . But we know that: v 1 • v 2 = v 1 v 2 cos θ = v 2 (v 1 cos θ ) v 1 • v 2 = v 2 times the projection of v 1 on v 2 This shows that the scalar product of two vectors is the product of the magnitude of one vector and the magnitude of the projection of the other vector on it. The angle between the two vectors can be expressed in terms of the vector constants as follows: Vectors
- 21. Three dimensional space: Let a = a 1 i + a 2 j + a 3 k and b = b 1 i + b 2 j + b 3 k a•b = (a 1 i + a 2 j + a 3 k ) • (b 1 i + b 2 j + b 3 k ) a•b = a 1 b 1 i•i + a 1 b 2 i•j + a 1 b 3 i•k + a 2 b 1 j•i + a 2 b 2 j•j + a 2 b 3 j•k + a 3 b 1 k•i + a 3 b 2 k•j + a 3 b 3 k•k The unit vectors i , j and k have length 1 and are at 90 º to each other and so any unit vector when scalar product combined with itself will give: i •i = 1 x 1 x cos 0 º = 1 Whilst any unit vector when scalar product combined with a different one will give: i •j = 1 x 1 x cos 90 º = 0 Therefore a•b = a 1 b 1 + a 2 b 2 + a 3 b 3 Vectors
- 22. Three dimensional space: O P A B a b c From the diagram the length of OP in terms of the side lengths can be determined as follows: OP 2 = OB 2 + BP 2 and OB 2 = OA 2 + AB 2 Thus OP 2 = OA 2 + AB 2 + BP 2 OP 2 = a 2 + b 2 + c 2 For our two vectors: Using, Vectors
- 23. Example <ul><li>1. For a = 2 i - 3 j + 4 k , b = 5 i + 2 j + 6 k , </li></ul><ul><li>find and . </li></ul>2. For a = -5 i + 3 j -6 k , b = 2 i - 2 j + 3 k , find and . 3. For a = 7 i - j + 3 k , b = i + 3 j - 4 k , find and . 4. For a = 2 i + 3 j + 5 k , b = 4 i + 2 j - 3 k , find and . Vectors
- 24. The second product of two vectors is called the vector product or cross product and is defined in terms of its modulus and the magnitudes of the two vectors and the sine of the angle between them. The vector product of vectors oa and ob is written as oa x ob and is defined by: | oa x ob | = oa ob sin θ , where θ is the angle between the two vectors. The direction of oa x ob is perpendicular to both oa and ob as shown: The Vector Product of Two Vectors θ a b θ a b oa x ob ob x oa Vectors
- 25. The direction is obtained by considering that a right handed screw is screwed along oa x ob with its head at the origin and if the direction of oa x ob is correct, the head should rotate from oa to ob (left hand diagram – previous slide). If the vector product is reversed then the direction of ob x oa is reversed (right hand diagram). This oa x ob ≠ ob x oa . The magnitudes are the same (oa ob sin θ ) but their directions are 180 º displaced i.e. oa x ob = - ob x oa Vectors
- 26. Three dimensional space: Once again let a = a 1 i + a 2 j + a 3 k and b = b 1 i + b 2 j + b 3 k axb = (a 1 i + a 2 j + a 3 k ) x (b 1 i + b 2 j + b 3 k ) axb = a 1 b 1 ixi + a 1 b 2 ixj + a 1 b 3 ixk + a 2 b 1 jxi + a 2 b 2 jxj + a 2 b 3 jxk + a 3 b 1 kxi + a 3 b 2 kxj + a 3 b 3 kxk The unit vectors i , j and k have length 1 and are at 90 º to each other and so any unit vector when vector product combined with itself will give: i xi = 1 x 1 x sin 0 º = 0 Whilst any unit vector when vector product combined with a different one will give: i xj = 1 x 1 x sin 90 º = 1 Vectors
- 27. Three dimensional space: The direction will be the same as the thirds unit vector: i x j = k j x i = -k j x k = i k x j = -i k x i = j i x k = -j Therefore axb = a 1 b 2 k - a 1 b 3 j - a 2 b 1 k + a 2 b 3 i + a 3 b 1 j - a 3 b 2 i axb = (a 2 b 3 - a 3 b 2 ) i + (a 3 b 1 - a 1 b 3 ) j + (a 1 b 2 - a 2 b 1 ) k The magnitude of the vector product of two vectors can be found by expressing it in scalar product form and then using the relationship a•b = a 1 b 1 + a 2 b 2 + a 3 b 3 Vectors i j k
- 28. Express the cross-product of two vectors in a determinant form Vectors i j k
- 29. Three dimensional space: Squaring both sides of the vector product equation gives: | a x b | = a b sin θ so (| a x b |) 2 = a 2 b 2 sin 2 θ sin 2 θ + cos 2 θ = 1 so sin 2 θ = 1 - cos 2 θ So (| a x b |) 2 = a 2 b 2 (1 - cos 2 θ ) (| a x b |) 2 = a 2 b 2 - a 2 b 2 cos 2 θ But we know that a•b = a b cos θ therefore a•a = a 2 cos0 = a 2 And square this then multiply by a 2 b 2 Vectors
- 30. Three dimensional space: Using (| a x b |) 2 = a 2 b 2 - a 2 b 2 cos 2 θ And substituting in gives us… (| a x b |) 2 = ( a • a )( b • b ) – ( a • b ) 2 Example for the vectors: a = i + 4j – 2k and b = 2i – j + 3k Determine a x b and |a x b| Vectors
- 31. Example <ul><li>1. For a = 2 i - 3 j + 4 k , b = 5 i + 2 j + 6 k , </li></ul><ul><li>find and . </li></ul>2. For a = -5 i + 3 j -6 k , b = 2 i - 2 j + 3 k , find and . 3. For a = 7 i - j + 3 k , b = i + 3 j - 4 k , find and . 4. For a = 2 i + 3 j + 5 k , b = 4 i + 2 j - 3 k , find and . Vectors
- 32. <ul><li>This resource was created by the University of Wales Newport and released as an open educational resource through the Open Engineering Resources project of the HE Academy Engineering Subject Centre. The Open Engineering Resources project was funded by HEFCE and part of the JISC/HE Academy UKOER programme. </li></ul><ul><li>© 2009 University of Wales Newport </li></ul><ul><li>This work is licensed under a Creative Commons Attribution 2.0 License . </li></ul><ul><li>The JISC logo is licensed under the terms of the Creative Commons Attribution-Non-Commercial-No Derivative Works 2.0 UK: England & Wales Licence. All reproductions must comply with the terms of that licence. </li></ul><ul><li>The HEA logo is owned by the Higher Education Academy Limited may be freely distributed and copied for educational purposes only, provided that appropriate acknowledgement is given to the Higher Education Academy as the copyright holder and original publisher. </li></ul><ul><li>The name and logo of University of Wales Newport is a trade mark and all rights in it are reserved. The name and logo should not be reproduced without the express authorisation of the University. </li></ul>Vectors

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