Vectors Part 2: Three Dimensions

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1 Vectors Part 2: Three Dimensions Last modified: 23/02/2018

2 Links Vectors Recap Three Dimensions: Cartesian Form Three Dimensions: Standard Unit Vectors Three Dimensions: Polar Form Basic Vector Maths Three Dimensions: Adding Vectors Scalar Product Recap Mathematical Properties 3D Cartesian Vectors Finding the Magnitude of a 3D Vector Finding the Angle Between 2 3D Vectors 3D Vector Example Vector Product Definition Right Hand Rule Example Mathematical Properties Example Revisited 3D Cartesian Vectors Determinant Method Determinant Example Another Example

3 Vectors: Recap Many quantities of interest in Physics are VECTORS: they have not only a magnitude, but also a DIRECTION. A vector is represented in bold type: v (in print) or with either a twiddle underneath: ṽ or an arrow above: v (in handwriting). Using the regular font, without a twiddle or arrow refers to the magnitude (or length, size etc.) of the vector: v = v. In two dimensions, a vector can be described in polar form: giving magnitude r and direction, usually as the angle θ anti-clockwise from the x-axis. y r θ x

4 Alternatively, using the definition of vector addition, this vector can be expressed in terms of its components in the x and y directions: y r θ ai bj x as the Cartesian form ai + bj where the unit vectors i and j indicate the directions of the x and y axes. Note: the co-ordinates (a, b) give the position of the tip of the vector.

5 Using basic trigonometry, we can convert between the two forms: y r = a 2 + b 2 bj = r sin θj θ = tan 1 ( b a ) ai = r cos θi x Calculating the sum of vectors is much easier when the vectors are expressed in Cartesian form.

6 Three Dimensions: Cartesian Form Of course, the world around us is three-dimensional, and generally vectors are also three-dimensional. For example, a vector can be drawn from the origin to a point with co-ordinates (a, b, c): y (a, b, c) z x

7 Extending the idea of components we used in two dimensions, gives us: y ck bj z ai x The Cartesian form of the vector is: ai + bj + ck where the new unit vector k is in the direction of the z-axis.

8 3 Dimensions: Standard Unit Vectors Three standard unit vectors are defined in the direction of the axes: i x-direction j y-direction k z-direction y z k j i x

9 Three Dimensions: Polar Form Several different systems are used to extend the 2D polar form into three dimensions, each with their own advantages and disadvantages. One of the most widely used is the spherical polar system, consisting of the length r of the vector and two angles θ and φ, shown below: z φ r x θ y Calculations using these systems can get complicated! In this course, we will only use the CARTESIAN form of 3D vectors.

10 Three Dimensions: Adding Vectors As we saw in 2D, adding 3D vectors is easy if they are given in Cartesian form. For example: u = 4i + 3j + ( 2)k v = ( 2)i + 2j + 3k u + v = (4 + 2)i + (3 + 2)j + ( 2 + 3)k = 2i + 5j + k

11 Scalar Product: Recap We have previously seen the definition of the scalar (or dot) product: The scalar product of two vectors a and b is: a b = a b cos θ = ab cos θ where θ is the angle between the two vectors. b θ a Though we have previously only applied this definition to 2D vectors, nothing in the definition requires the vectors to be 2D. It will also apply to vectors in three dimensions.

12 Scalar Product: Mathematical Properties Last lecture, we discussed some useful properties of the scalar product. a b = ab cos θ The scalar product is commutative: a b = b a The scalar product is distributive: a (b + c) = a b + a c The product of a vector with itself: a a = a 2 cos 0 = a 2 The product of two perpendicular vectors: a b = ab cos 90 = 0

13 3D Cartesian Vectors Calculate the scalar product of the two vectors u = 4i + 3j 2k and v = 2i + 2j + 3k The method is the same as previously seen with 2D vectors, but with a few extra numbers... u v = (4i + 3j 2k) ( 2i + 2j + 3k) = (4)( 2)i i + (4)(2)i j + (4)(3)i k + (3)( 2)j i + (3)(2)j j + (3)(3)j k + ( 2)( 2)k i + ( 2)(2)k j + ( 2)(3)k k = (4)( 2) + (3)(2) + ( 2)(3) = = 8 where we have used i i = j j = k k = 1 and i j = i k = j k = 0

14 For two general vectors, a = a x i + a y j + a z k and b = b x i + b y j + b z k then a b = a x b x + a y b y + a z b z From the definition, a a = a 2, so we have: a = a a = ax 2 + ay 2 + az 2 which is the 3D version of Pythagoras theorem. This formula allows us to calculate the magnitude of any 3D Cartesian vector.

15 Finding the Magnitude of a 3D Vector For example: u = 4i + 3j 2k u 2 = u u = ( 2) 2 = 29 u = 29 And for v = 2i + 2j + 3k v 2 = v v = ( 2) = 17 v = 17

16 Finding the Angle Between 2 3D Vectors Calculating the angle between two 3D Cartesian vectors is done using the scalar product (the same method we used for 2D vectors). For the example vectors u = 4i + 3j 2k and v = 2i + 2j + 3k, use the scalar product formula, with the results of our previous calculations: cos θ = u v uv = = θ = 111 For 3D vectors this is really the only practical way to determine this angle, as in the polar form there are two angles to consider.

17 Example Two students arrive by train at a railway station. They have to wait two hours until their next train departs. Student A is feeling adventurous, and decides to spend this time exploring the city. Student B meanwhile is feeling lazy and decides to wait at the station and play games on his ipad. Student A walks 500 m east along East St, then 400 m south along South St, then 300 m southwest along South-West St before finally taking the lift to an observation deck at the top of a 200 m tall building. Calculate the position vector of Student A relative to Student B. (Give the answer in Cartesian form using i = east, j = north, k = up.) Student A has a laser pointer which she uses to surprise Student B, by shining a red dot on his ipad screen. What is the distance travelled by this beam of light? What is the angle between the laser beam and the ground?

18 To find the position vector, we need to add all of the individual motions of Student A. The first three segments of the journey are along the ground, so only involve i and j. The position vector of the base of the building from the railway station is: 500i+( 400j)+( 212i 212j) = 288i 612j m W N S 300 sin 45 = 212 E cos 45 = 212 The final motion is up (i.e. the z-direction) so the complete position vector is: r AB = 288i 612j + 200k m We are asked to calculate the distance travelled by the light, which is the magnitude of the above vector: r AB = ( 612) = 705 m

19 To find the angle of the laser to the ground, we need to find the angle between the position vector r AB and the position vector of the base of the building to the railway station. i.e. the vectors: r AB = 288i 612j + 200k m and 288i 612j m This can be done using the dot product, as previously: cos θ = = θ = 16 Or a quicker way, using the previously calculated distance: A 705 m 200 m θ = sin 1 ( B θ ) = 16

20 Vector Product Another method of vector multiplication can be defined and is useful. This is the vector or cross product. The vector product of two vectors a and b is a third vector c = a b. The magnitude of c is given by: c = c = a b sin θ = ab sin θ where θ is the angle between the two vectors. b θ a The direction of c is given by the Right Hand Rule.

21 Vector Product: Right Hand Rule To find direction of the vector c = a b: Use your RIGHT HAND! Place 1st finger in the direction of the 1st vector (a)... a c then 2nd finger in the direction of the 2nd vector (b) b The vector c is in the direction of the thumb. c is perpendicular to BOTH of the vectors a and b.

22 Example Calculate the vector product of the two vectors a = 4i + 3j and b = 4j y 4 Clearly a = 5 and from the diagram we can work out that sin θ = 4 5. So using the definition: z b θ a 3 x a b = ab sin θ = = 16 Using the Right Hand Rule gives the direction to be along the z- axis, thus: a b = 16k

23 Vector Product Mathematical Properties Properties that follow from the definition: The vector a b is perpendicular to both a and b. a (a b) = b (a b) = 0 The vector product is NOT commutative: b a = (a b) The magnitude is unchanged: order doesn t matter in c = ab sin θ but in the Right Hand Rule, fingers are swapped and so the direction is reversed. The vector product is distributive: a (b + c) = a b + a c The vector product of a vector with itself: a a = 0

24 Example Revisited Calculate the vector product of the two vectors a = 4i + 3j and b = 4j Using the distributive property of the vector product: a b = (4i + 3j) (4j) = (4)(4)(i j) + (3)(4)(j j) The vector product of any vector with itself is zero, so j j = 0, and the Right Hand rule gives i j = k: a b = 16k This is the same result found previously. Note that when we calculate using Cartesian vectors like this, the answer is also in Cartesian form, so the magnitude AND direction are included.

25 Vector Product: 3D Cartesian Vectors How can we calculate the vector product of two 3D vectors given in Cartesian form? For the example vectors u = 4i + 3j 2k and v = 2i + 2j + 3k what is u v? Using the distributive property of the cross product: u v = (4i + 3j 2k) ( 2i + 2j + 3k) = (4)( 2)(i i) + (4)(2)(i j) + (4)(3)(i k) + (3)( 2)(j i) +(3)(2)(j j) + (3)(3)(j k) + ( 2)( 2)(k i) + ( 2)(2)(k j) +( 2)(3)(k k) = (8 + 6)(i j) + (12 4)(i k) + (9 + 4)(j k) = 14(i j) + 8(i k) + 13(j k) where i i = j j = k k = 0 and i j = j i etc.

26 Each of the remaining vectors is a unit vector (because sin 90 = 1), and using the Right Hand Rule gives: Thus: i j = k i k = j j k = i u v = 14(i j) + 8(i k) + 13(j k) = 13i 8j + 14k This was a long calculation, so it is a good idea to check our answer: u (u v) = (4i + 3j 2k) (13i 8j + 14k) = (4)(13) + (3)( 8) + ( 2)(14) = 0 Which is expected, since u u v (also v u v) by definition.

27 Taking two general vectors: a = a x i + a y j + a z k and b = b x i + b y j + b z k and following the same procedure as above, we get the general form for the vector product: a b = (a y b z a z b y )i + (a z b x a x b z )j + (a x b y a y b x )k The full process of calculating the cross product can be lengthy, and this expression isn t easy to remember - which terms have the the minus signs? - so a systematic recipe for performing the calculation would be useful. There is such a recipe, using a mathematical concept called the determinant.

28 Vector Product: Determinant Method The determinant is a widely used mathematical operation on matrices. We won t go into the details of its definition etc, but will just borrow the method of calculation. A determinant is represented by a square array of numbers between absolute value bars :... In terms of the determinant, the vector product of our two general vectors is (remember the order of a and b is important) : i j k a b = a x a y a z b x b y b z = i a y a z j a x a z + k a x a y b y b z b x b z = (a y b z a z b y )i + (a z b x a x b z )j + (a x b y a y b x )k Which is the same result found previously. b x b y

29 How do we apply these steps? For each element in the top row, find the remainder when the row and column containing that element are ignored, and write down the element times the remainder, being careful to keep the same order: and i j k a x a y a z b x b y b z i j k a x a y a z b x b y b z i j k a x a y a z b x b y b z i a y b y j a x b x k a x b x a z b z a z b z a y b y

30 Next is the tricky bit - these 3 are added together, but with a MINUS sign on the j. (Don t worry about why, it s just a recipe) i j k a x a y a z b x b y b z = + i a y a z b y b z j a x a z b x b z + k a x a y b x b y Now each of the 2 2 determinants can be evaluated using the rule: product of diagonals minus product of off-diagonals a b c d = ad bc This process may seem very complicated at first, but with a little practice you should be able to apply it quite quickly.

31 Vector Product: Determinant Example Let s apply the determinant method to re-calculate u v, for the same two vectors seen earlier. The previous result was: u v = 13i 8j + 14k. i j k u v = = i j k = [(3)(3) ( 2)(2)]i [(4)(3) ( 2)( 2)]j + [(4)(2) (3)( 2)]k = 13i 8j + 14k Which is the same answer as before, but obtained much quicker.

32 Another Example Two forces act at the same point O as shown below. F 1 is 10 N and acts in the direction OA, and F 2 is 14.1 N in the direction OB. y 3 m B F 1 3 m F 2 O 4 m A 3 m x Calculate: (a) The sum of the two forces F 1 + F 2 in Cartesian form. (b) The angle θ between the two forces z (c) The cross product F 1 F 2

33 (a) Before adding the two forces, we first need, as always, to express them in Cartesian form. F1 has magnitude 10 and is in the direction of OA. From the diagram, we find OA = 4i + 3j, so ÔA = OA OA (4i + 3j) = = (4i + 3j)/ and thus: F 1 = 10 ÔA = 8i + 6j N Similarly, F2 has magnitude 14.1 in the direction of OB, so OB F 2 = 14.1 ÔB = 14.1 OB (3j + 3k) = 14.1 = 10j + 10k N and the sum F1 + F 2 is: F 1 = 8i + 6j + 0k + F 2 = 0i + 10j + 10k F 1 + F 2 = 8i + 16j + 10k N

34 (b) To determine the angle between the vectors we must use the scalar product: thus F 1 F 2 = (8i + 6j + 0k) (0i + 10j + 10k) = = 60 cos θ = F 1 F 2 60 = = 0.43 θ = 65 F 1 F (c) We now know the magnitudes of the two vectors and the angle between them, so can calculate the magnitude of the vector product: F 1 F 2 = F 1 F 2 sin θ = sin 65 = 128 BUT trying to use the Right Hand Rule to get the direction is not really practical!

35 We need to use the Cartesian forms of the vectors: F 1 F 2 = (8i + 6j) (10j + 10k) = 80(i j) + 80(i k) + 60(j j) + 60(j k) = 60i 80j + 80k N 2 or, using the determinant: i j k F 1 F 2 = = i j = 60i 80j + 80k N 2 + k Except for very simple vectors, the Right Hand Rule is not very helpful in calculating cross products.