SCHOOL OF ENGINEERING & BUILT ENVIRONMENT

Mathematics

Scalars and Vectors

1. What are Scalars and Vectors? 2. Representing a Vector Mathematically – Polar Form 3. Vector Addition and Subtraction 4. Multiplying a Vector by a Scalar 5. The Cartesian or Rectangular Component Form of a Vector 6. The Relationship Between Polar and Cartesian (Rectangular)

Forms 7. Relative Position Vectors in Rectangular Form 8. The Scalar Product of Two Vectors (Dot Product) 9. Extension to Three Dimensions 10. The Vector Product of Two Vectors (Cross Product) Summary Examples in 3D

Tutorial Exercises

Dr Derek Hodson

Contents Page 1. What are Scalars and Vectors? ........................................................................... 1 2. Representing a Vector Mathematically – Polar Form ........................................ 2 3. Vector Addition and Subtraction ........................................................................ 3 4. Multiplying a Vector by a Scalar ........................................................................ 5 5. The Cartesian or Rectangular Component Form of a Vector ............................. 5 6. The Relationship Between Polar and Cartesian (Rectangular) Forms ................ 8 7. Relative Position Vectors in Rectangular Form .................................................. 10 8. The Scalar Product of Two Vectors (Dot Product) ............................................. 11 9. Extension to Three Dimensions ........................................................................... 14 10. The Vector Product of Two Vectors (Cross Product) .......................................... 15 Summary Examples in 3D ................................................................................... 16 Tutorial Exercises ................................................................................................ 19 Answers to Tutorial Exercises ............................................................................. 23

1 Vectors1 / D Hodson Scalars and Vectors 1) What are Scalars and Vectors? As you are all aware, we can use numbers to measure or quantify physical quantities. For example, a thermometer may tell us that the temperature at a particular point in a room at a particular time is 24°C. A voltmeter may indicate a voltage of 9V across the terminals of a battery. Many quantities, such as temperature and voltage, require just a single number to specify or measure them; others may require more than one number. Relative position is a physical quantity that we can measure. We might say that Jack is 6.5m from Jill. If we do, then we haven’t told the whole story. Is Jack 6.5m to Jill’s right, to her left, behind her, above her, or what? To specify relative position completely we must give not only a distance, but a direction as well. This would require at least two numbers. Basically, physical quantities that are specified by single numbers are termed scalars, while those that require more than one number are called vectors. Examples of Scalars mass energy temperature length area (sometimes) volume density voltage current entropy ............ Examples of Vectors relative position area (sometimes) displacement velocity acceleration force moment of force (torque) linear momentum angular momentum ............

2 2) Representing a Vector Mathematically – Polar Form In most practical cases a vector will be a quantity specified by a magnitude and a direction. The simplest way to represent such a composite quantity is by an arrow whose length is scaled to correspond to the magnitude and whose orientation represents the vector’s direction of action. In general the arrow will lie in 3-dimensional space, but to keep things simple (just for the moment) we shall work in 2 dimensions, like a map. There are various ways to specify the direction of action, for example compass points or map bearings if we are dealing with relative position or displacement. We shall use the mathematical convention of drawing an arrow on an Oxy axes system and recording the angle it makes with the positive direction of the x-axis (with positive angles measured anticlockwise and negative angles, clockwise). Notationally, we usually denote vectors by an underlined letter to distinguish them from scalars, e.g. etc.,,, 1 Fvv Algebraically, we group the magnitude and direction as a bracketed ordered pair of the form ),( °θr , the so-called polar form of a two-dimensional vector. Note: We sometimes use v to denote the magnitude of a vector v . Examples

O

y

x

4

45°

)45,4( °=v

O

y

x

5

120°

)120,5( °=v

O

y

x 3 −45°

)45,3( °−=v

The arrows don’t necessarily have to emanate from the origin:

O

y

x

4

45°

)45,4( °=v

O

y

x 4

45°

)45,4( °=v

3 3) Vector Addition and Subtraction Consider two displacement vectors )30,4(1 °=v and )60,2(2 °=v . We define the sum of these two vectors as the net resultant displacement of one following on from the other.

y

x O

1v

2vv

21 vvv +=

It is easily demonstrated that the order of addition does not matter. That is, 1221 vvvvv +=+= . This definition, sometimes called the triangle law for the addition of two vectors, is fairly natural when thinking in terms of displacement, but the same definition proves to be applicable for other types of vector. Consider to forces, 1F and 2F acting at the origin. The net effect of the two forces is given by the vector sum, as defined above.

y

x O

1F

2F

F

21 FFF +=

2F

1F

This is the parallelogram rule for adding two vectors, which is clearly equivalent to the triangle law.

4 Let’s look again at the first diagram illustrating vector addition:

y

x O

1v

2vv

21 vvv +=

Algebraically, we would like the vector notation to behave like scalar notation. That is, 21 vvv += should lead to 21 vvv −= . Writing this as )( 21 vvv −+= and noting from the diagram that 1v can be interpreted as the net result of “moving” along v , then backwards along 2v , it should be obvious that )( 2v− is simply 2v with its direction reversed. This means that vector subtraction is just a special case of vector addition. All this defines vector addition and subtraction geometrically, but does not give us an easy way of evaluating vector sums and differences. We can do such calculations using trigonometry and Pythagoras, but they aren’t straightforward calculations. Later we shall look at an alternative representation of two-dimensional vectors, one that makes the addition and subtraction of vectors dead easy. Before that, let us look at one other piece of vector arithmetic.

5 4) Multiplying a Vector by a Scalar If we have a vector v and a positive scalar k, then we can define the product vk as a vector oriented in the same direction as v with a magnitude k times that of v :

y

x O

v

v3

v5.0

Note: Multiplying by a negative scalar would reverse the direction. 5) The Cartesian or Rectangular Component Form of a Vector Using the definitions of vector addition and multiplication by a scalar, we can now develop an alternative algebraic representation of a two-dimensional vector that is easier to work with than the polar form. A unit vector is a vector with magnitude equal to one unit. We introduce two unit vectors, one parallel to the x-axis and one parallel to the y-axis; these we denote by i and j respectively:

O

y

x

i

)0,1( °=i

O

y

x

j

)90,1( °=j

6 This means that any vector parallel to one of the axes can be expressed as a scalar multiple of either i or j :

O

y

x

i2

O

y

x

j3

O

y

x

i5.2−

O

y

x

j5.1−

Bearing this and the definition of vector addition in mind, we can take any two-dimensional vector v and decompose it into the sum of two component vectors, one parallel to the x-axis and one parallel to the y-axis:

y

x O

v

ix

jy

Algebraically we have jyixv += . The vector is completely specified by the two values x and y . These are the Cartesian or rectangular components of the vector and we can use these as an alternative to the polar components r and θ .

7 The advantage of the rectangular form over the polar is that vector arithmetic is easier. If we have two vectors, jyixv 111 += and jyixv 222 += , then it is quite easy to see that jyyixxvvv )()( 212121 +++=+=

y

x O

1v

ix1

jy1

2v

ix2

jy2v

ixx )( 21 +

jyy )( 21 +

Multiplying a vector by a scalar is also easy in rectangular form: jyixv += jykixkvk )()( += . Convince yourself of this by sketching it out on paper; think of similar triangles! Examples (1) Given jiv 471 += and jiv 622 += determine (i) 21 vv + [Ans: ji 109 + ] (ii) 21 vv − [Ans: ji 25 − ] (iii) 13v [Ans: ji 1221 + ] (iv) 21 23 vv + [Ans: ji 2425 + ]

8 Instead of always carrying the i and j , we can use an abbreviated notation for the Cartesian form: ),( yxv = . We have to be a little careful when using this notation since is also used for the coordinates of a point on the Oxy axes system. In most cases, the correct interpretation should be clear from the context. In this abbreviated form we have

),( yx

),( 212121 yyxxvv ++=+ ),( 212121 yyxxvv −−=− ),( ykxkvk = . Using this notation, Example (1) would be set out as: (1) Given )4,7(1 =v and )6,2(2 =v determine (i) 21 vv + [Ans: ] )10,9( (ii) 21 vv − [Ans: )2,5( − ] (iii) 13v [Ans: )12,21( ] (iv) 21 23 vv + [Ans: )24,25( ] 6) The Relationship Between Polar and Cartesian (Rectangular) Forms Polar Form: ),( °= θrv Rectangular Form: ),( yxv = A combination of basic trigonometry and Pythagoras’ Theorem gives the following conversion formulae: Polar → Rectangular: °= θcosrx °= θsinry

Rectangular → Polar: 22 yxvr +== ⎟⎠⎞

⎜⎝⎛= −

xy1tanθ .

The conversion “Polar → Rectangular” is quite straightforward, but care must be taken when applying “Rectangular → Polar”, since the quadrant in which θ lies must be determined before evaluating the inverse tangent.

9 Examples (2) (a) A vector has magnitude 2 and direction 210° . Express the vector in its

Cartesian component form. )210,2(),( °=°= θrv

732.1

)210(cos2cos

−=°=

°= θrx

1)210(sin2

sin

−=°=

°= θry

)1,732.1(1732.1 −−=−−= jiv (2) (b) Express the vector )2,4(−=v in polar form: )2,4(),( −== yxv

472.420

2)4( 22

22

=

=

+−=

+= yxr

Determine the quadrant for the angle ( 2,4 =−= yx ) :

°=

⎟⎠⎞

⎜⎝⎛−

=

⎟⎠⎞

⎜⎝⎛=

−

−

43.153

angle]quadrant 2nd [4

2tan

tan

1

1

xyθ

)43.153,472.4( °=v Most calculators have these conversion formulae pre-programmed. Please refer to your own calculator’s instruction booklet for information on how to implement these conversion processes or, if that fails, ask in the tutorial classes.

10 7) Relative Position Vectors in Rectangular Form Suppose we have two points on an Oxy-axes system, and . ),( 11 yxP ),( 22 yxQ

y

x O

1x

1y

2x

2y

P

Q

The position of Q relative to P is given by a two-dimensional vector QP (note notation) whose Cartesian components are easily determined from the coordinates of the points: ),( 1212 yyxxQP −−= . Example (3) Determine the position of the point )2,1(−Q relative to )5.2,3( −P .

y

x O

P

Q

)5.4,4(−

))5.2(2,31(

=

−−−−=QP

11 8) The Scalar Product of Two Vectors (Dot Product) We now extend the arithmetic of vectors and introduce a way of multiplying two vectors together. There are, in fact, two ways of multiplying vectors together. In this section we shall look at the way that results in a scalar product, i.e. a number rather than another vector. Later we shall look at the other way which does result in another vector. Consider two vectors 1v and 2v whose orientations differ by an angle θ :

y

x O

1v

2v

θ

We define their scalar product as θcos2121 . vvvv = . Because we usually make a clear dot between the two vectors to distinguish this from the other form of multiplication that we shall see later, this is sometimes referred to as the dot product. For two (two-dimensional) vectors expressed in polar form, this scalar product is given by )(cos 212121 . θθ −= rrvv . Working in rectangular form it can be shown that 212121 . yyxxvv += . Examples (4) (a) Determine the scalar (or dot) product of )4,2(1 =v and )5,3(2 =v 26543221 . =×+×=vv (b) Determine the scalar (or dot) product of )1,3(1 −=v and )2,6(2 −−=v 16)2()1()6(321 . −=−×−+−×=vv

12 (5) (a) Determine the angle of separation of the two vectors )4,2(1 =v and

)5,3(2 =v . 26543221 . =×+×=vv 2042 22

1 =+=v 3453 22

2 =+=v . From the definition of the dot product

99705.03420

26

cos21

21 .

=

=

=vv

vvθ

and so the angle of separation is °= 40.4θ . Note: Usually we have to take care with quadrants when we invert a trig function.

However, the angle of separation of two vectors will always be between 0° and 180° (inclusive) and this is what a calculator cosine inversion will always give us.

(b) Determine the angle of separation of the two vectors )1,3(1 −=v and

)2,6(2 −−=v . 16)2()1()6(321 . −=−×−+−×=vv 10)1(3 22

1 =−+=v 40)2()6( 22

2 =−+−=v . From the definition of the dot product

8.04010

16

cos21

21 .

−=

−=

=vv

vvθ

and so the angle of separation is °= 13.143θ .

13 The scalar product has a number of applications. Here are just a few: a) Test for orthogonality: Two vectors are said to be orthogonal if they act at right-angles to each other. For 1v and 2v at right-angles we have 090cos2121 . =°= vvvv . This gives us a test for orthogonality; we look for a scalar product equal to zero. b) Component of a vector in a given direction: We have already spoken about vectors having components parallel to the x and y axes. Using the dot product, we can determine the component of a vector in any given direction. Consider a vector v and a direction specified by a unit vector u ( i.e. 1=u ).

y

x O

u

v

θ

The component of v in the direction of u is given by the length of the double-headed arrow in the above diagram. This is exactly the dot product θθ coscos. vuvuv == since the magnitude of u equals 1 . c) Work done by a force: Suppose an object is acted upon by several forces, nFFF ,...,, 21 , and undergoes a displacement d . The work done by each individual force is given by the dot product nkdF k ,...,2,1,. = .

14 9) Extension to Three Dimensions So far we have only considered two-dimensional vectors. The Cartesian form of a vector provides an easy and natural way of extending vectors into three dimensions. We take the Oxy axes system and include a third axis, a z-axis, perpendicular to both the x and y axes:

O x

y

z

i j

k

v

Denoting the unit vector parallel to the new z-axis by k , any three-dimensional vector v can be decomposed into the sum of three component vectors, one in each of the axes directions: ),,( zyxkzjyixv =++= . Vector magnitudes, addition, subtraction, multiplication by scalars, and dot products follow naturally. For ),,( zyxv = , ),,( 1111 zyxv = and ),,( 2222 zyxv = we have:

Magnitude: 222 zyxv ++=

Addition: ),,( 21212121 zzyyxxvv +++=+ Subtraction: ),,( 21212121 zzyyxxvv −−−=− Mult. by scalar: ),,( zkykxkvk = Scalar (dot) product: 21212121 . zzyyxxvv ++= . Note: The polar form of a vector does not extend as naturally into three dimensions as

the rectangular form. There do exist alternative three-dimensional forms of vectors related to the polar form, however we do not consider them here.

15 10) The Vector Product of Two Vectors (Cross Product) Having now extended into 3D, we can now consider the other form of vector multiplication, the so-called vector product of two vectors. This may seem a bit strange, but it does have practical applications. For two vectors ),,( 1111 zyxv = and ),,( 2222 zyxv = we define the vector product as follows:

222

11121

zyxzyxkji

vv =× ,

where the right-hand-side is the determinant of a 3 × 3 matrix. Recall that this is determined using cofactors:

22

11 ofcofactor zyzy

i +=

22

11 ofcofactor zxzx

j −=

22

11 ofcofactor yxyx

k += .

That means the vector product is given by

22

11

22

11

22

1121 yx

yxk

zxzx

jzyzy

ivv +−=× .

Notes i) Unlike the dot product which takes two vectors and returns a number, the vector product

returns another vector. ii) The resultant product vector will be perpendicular two both of the original vectors. See

in class for the “right-hand rule”. iii) The vector product is often called the cross product. iv) When writing a cross product on paper we can use ∧ instead of × ; this avoids

confusion with an x. It’s still referred to as a cross product! v) Applications include moments of forces (torques), shearing forces and rotational

dynamics.

16 Summary Examples in 3D (6) For the 3-dimensional vectors )3,1,2(1 −=v and )4,2,3(2 =v determine (i) 21 vv + ; (ii) 21 vv − ; (iii) 21 . vv ; (iv) ) :notation eAlternativ( 2121 vvvv ∧× ; (v) The angle of separation between the vectors; (vi) The component of 1v in the direction of 2v . (i) )7,1,5()43,2)1(,32(21 =++−+=+ vv (ii) )1,3,1()43,2)1(,32(21 −−−=−−−−=− vv (iii) 16432)1(3221 . =×+×−+×=vv

(iv) 42331221 −=×kji

vv

)7,1,10(

7110

]34[]98[]64[

]3)1(22[]3342[]234)1([

2312

4332

4231

−=

++−=

++−−−−=

×−−×+×−×−×−×−=

−+−

−=

kji

kji

kji

kji

(v) Over . . .

17 (v) 16432)1(3221 . =×+×−+×=vv 143)1(2 222

1 =+−+=v 29423 222

2 =++=v . From the definition of the dot product

79407.02914

16

cos21

21 .

=

=

=vv

vvθ

and so the angle of separation is °= 43.37θ . (vi) Magnitude of 2v : 29423 222

2 =++=v

Unit direction vector: )4,2,3(291

2

2 ==vv

u

Component of 1v in the direction of 2v :

9711.2

16291

)34)1(223(291

)3,1,2()4,2,3(291 .1.

=

×=

×+−×+×=

−=vu

18 (7) A triangle in 3D space has vertices , )4,2,3(A )5,4,1( −B and )1,0,2( −C .

Determine the angle between the sides AB and AC . Angle between sides AB and AC is the angle of separation of the relative position

vectors BA and CA .

A

B

C

θ

)1,6,2()45,24,31( −−=−−−−=BA 41=BA

)5,2,1()41,20,32( −−−=−−−−=CA 30=CA

9)5(1)2()6()1()2(. =−×+−×−+−×−=CABA

25662.0

30419

cos .

=

=

=CABA

CABAθ

Angle: °= 13.75θ

19 Tutorial Exercises Two-Dimensional Vectors (1) Sketch the following vectors (given in polar form) emanating from the specified points: (i) from (ii) from )30,2( ° )2,3(− )140,5( ° )4,1( (iii) from (iv) )200,25.0( ° )4,5( )30,6( °− from )2,2(− . (2) Sketch the following vectors (given in rectangular form) emanating from the specified

points: (i) from (ii) ))4,3( )1,2( 2,4(− from )4,2( − (iii) from )5.3,5.6( − )4,4( −− (iv) )5.4,5.2( −− from . )2,6(− (3) Convert the following vectors to rectangular (Cartesian) form, first using the conversion

formulae stated on p8, and then using the conversion functions on your calculator: (i) (ii) )30,2( ° )80,3( ° (iii) (iv) )120,1( ° )315,5( ° (v) (vi) )200,4( ° )75,2( °− . (4) Convert the following vectors to rectangular (Cartesian) form without the aid of

conversion formulae or calculator: (i) (ii) )0,4( ° )90,8( ° (iii) (iv) )180,6( ° )270,5( ° (v) (vi) )90,3( °− )180,10( °− . (5) Convert the following vectors to polar form, first using the conversion formulae stated

on p8, and then using the conversion functions on your calculator: (i) (ii) )2,3(− )2,3( − (iii) (iv) )3,4( −− )3,4( (v) (vi) )1,2(− )1,2( − .

20 (6) Convert the following vectors to polar form without the aid of conversion formulae or

calculator: (i) (ii) )0,5( )10,0( (iii) (iv) )0,8(− )2,0( − . (7) Given the vectors )3,4(1 −=v , )4,2(2 −=v and )1,5(3 −−=v , determine the

following, displaying the results graphically: (i) 21 vv + (ii) 321 vvv ++ (iii) )( 21 vv −+ (iv) 31 vv − (v) 12v (vi) 23v (vii) 21 32 vv + (viii) 321 524 vvv +− . (8) For each of the following pairs of vectors, determine the magnitudes 1v , 2v , the

scalar (dot) product 21 . vv and the angle of separation θ :

(i) ⎭⎬⎫

−==

)3,1()2,2(

2

1

vv

(ii) ⎭⎬⎫

−=−=

)8,5()6,4(

2

1

vv

(iii) ⎭⎬⎫

==

)0,1()4,3(

2

1

vv

(iv) ⎭⎬⎫

==

)1,0()4,3(

2

1

vv

.

(9) A triangle has vertices , and . )1,1(A )2,4(B )4,3(C (i) Draw this triangle accurately on an Oxy axes system. (ii) Determine the relative position vectors BA and CA . (iii) By calculation, determine the angle of the triangle at vertex A . (iv) Confirm your answer by the direct measurement of the angle on your diagram.

21 (10) A triangle has vertices , and . )1,2(−A )1,3(B )5,1(C (i) Draw this triangle accurately on an Oxy axes system. (ii) By calculation, determine the angle of the triangle at vertex B . (iii) Confirm your answer by the direct measurement of the angle on your diagram. (11) For each of the following pairs of vectors, determine the component of 1v in the

direction of 2v (see p13 and the 3D Example (6)(vi) on pp16-17):

(i) ⎭⎬⎫

−==

)3,1()2,2(

2

1

vv

(ii) ⎭⎬⎫

−=−=

)8,5()6,4(

2

1

vv

.

Three-Dimensional Vectors (12) Given the vectors )1,5,4(1 −=v , )3,6,5(2 −=v and )4,2,6(3 −=v ,

determine the following : (i) 321 vvv ++ (ii) 321 234 vvv −+ (iii) )( 31 vv −+ (iv) 321 52 vvv ++− (v) 21 . vv (vi) 32 . vv (vii) ) :notation eAlternativ( 2121 vvvv ∧× (viii) 2332 and vvvv ×× [Comment?] (ix) The angle of separation between 1v and 2v . (x) The angle of separation between 1v and 3v . (xi) The component of 1v in the direction of 2v . (xii) The component of 3v in the direction of 1v .

22 (13) A triangle in 3D space has vertices , and . )3,4,1(A )0,2,4(B )6,4,5(C (i) Determine the relative position vectors BA and CA . (ii) Determine the angle of the triangle at vertex A . (14) A triangle in 3D space has vertices )5,3,2( −−A , and .

Determine the angle of the triangle at vertex C . )4,0,2(B )1,5,1(C

(15) The unit vectors i , j and k are written in Cartesian form as follows: )0,0,1(=i )0,1,0(=j )1,0,0(=k . (i) Determine the scalar (dot) products ii . , jj . , kk . , ji . , ki . and kj . . (ii) Determine the vector (cross) products ii × , jj × , kk × , ji × , kj × and

ik × . (16) A force )5,2,3( −=F acts on a particle which moves from point )1,4,1( −P to

point . )1,3,2(−Q (i) Determine the displacement vector d of the particle. (ii) Determine the work done by the force, i.e. dF . . (17) A force F acts through a point P . The moment of force (torque) about a second point

Q is given by FrM ×= , where r is the position vector of Q relative to P . Determine M when kjiF 254 −+=

)5,3,2(is −P )3,2,1(is −Q .

23 Answers (1) (i)

y

x O

(ii) – (iv) Similar (2) (i)

y

x O

(ii) – (iv) Similar (3) Figures, where approximated, expressed to 3 decimal places. (i) (ii) )1,732.1( )954.2,521.0( (iii) (iv) )866.0,5.0(− )536.3,536.3( − (v) )368.1,759.3( −− (vi) )932.1,518.0( − (4) Hint: Sketch the vector first. (i) (ii) )0,4( )8,0( (iii) (iv) )0,6(− )5,0( − (v) (vi) )3,0( − )0,10(− (5) Figures, where approximated, expressed to 3 decimal places. (i) (ii) )310.146,606.3( ° )690.33,606.3( °− (iii) (iv) )130.143,5( °− )870.36,5( ° (v) (vi) )435.153,236.2( ° )565.26,236.2( °−

24 (6) Hint: Sketch the vector first. (i) (ii) )0,5( ° )90,10( ° (iii) (iv) )180,8( ° )90,2( °− or )270,2( ° (7) (i) )1,2(21 =+ vv (ii) )0,3(321 −=++ vvv (iii) )7,6()( 21 −=−+ vv (iv) )2,9(31 −=− vv (v) )6,8(2 1 −=v (vi) )12,6(3 2 −=v (vii) )6,2(32 21 =+ vv (viii) )25,5(524 321 −−=+− vvv (8) (i) 81 =v 102 =v 421 . −=vv °= 57.116θ (ii) 521 =v 892 =v 6821 . −=vv °= 32.178θ (iii) 51 =v 12 =v 321 . =vv °= 13.53θ (iv) 51 =v 12 =v 421 . =vv °= 87.36θ (9) (ii) )1,3(=BA )3,2(=CA (iii) °= 9.37θ (10) (ii) °= 4.63θ

(11) (i) 265.1104

−≈− (ii) 208.7

8968

−≈−

(12) (i) (ii) )6,3,15( − )3,6,19( − (iii) (iv) )5,7,2( −− )25,26,27( − (v) −13 (vi) 54 (vii) (viii) )49,17,9( )26,2,18( −− ; )26,2,18( − (ix) 103.87° (x) 78.1° (xi) −3.153 (xii) 1.543

25

(13) (ii) )3,2,3( −−=BA )3,0,4(=CA (iii) °= 65.82θ (14) 105.40° (15) (i) 1... === kkjjii 0... === kjkiji (ii) 0)0,0,0( ==×=×=× iiiiii (i.e. the zero vector) jikikjkji =×=×=× ;; (16) (i) )2,1,3( −−== QPd (ii) 3. =dF (17) )2,1,1( −−== QPr kjiFrM −+−=−−=×= 68)1,6,8( Note: A torque produces a rotation. The direction of the torque vector gives the

direction of the axis of that rotation.