Hsn Higher Mathematics

234567891011121314Vectors and ScalarsComponentsMagnitudeEqual VectorsAddition and Subtraction of VectorsMultiplication by a ScalarPosition VectorsBasis VectorsCollinearityDividing Lines in a RatioThe Scalar ProductThe Angle Between VectorsPerpendicular VectorsProperties of the Scalar Product1EFEFEFEFEFEFEFEFEFEFEFEFEFEFCfE EditionThis document was produced specially for the HSN.uk.net website, and we require that anycopies or derivative works attribute the work to Higher Still Notes.For more details about the copyright on these notes, please 5/scotland/11345789101114172021

Higher MathematicsVectorsVectors1Vectors and ScalarsEFA scalar is a quantity with magnitude (size) only – for example, an amountof money or a length of time.Sometimes size alone is not enough to describe a quantity – for example,directions to the nearest shop. For this we need to know a magnitude (i.e.how far), and a direction.Quantities with magnitude and direction are called vectors.A vector is named either by using the letters at the end of a directed line segment (e.g. AB represents a vector starting at point A and ending at pointB) or by using a bold letter (e.g. u). You will see bold letters used in printedtext, but in handwriting you should just underline the letter.BuAThroughout these notes, we will show vectors in bold as well us underliningthem (e.g. u).2ComponentsEFA vector may be represented by its components, which we write in a column.For example, 2 3 is a vector in two dimensions. In this case, the first component is 2 and this tells us to move 2 units in thex-direction. The second component tells us to move 3 units in the y-direction.So if the vector starts at the origin, it will look like:y32 xOhsn.uk.netPage 1CfE Edition

Higher MathematicsVectorsNote that we write the components in a column to avoid confusing them with 2 coordinates. The following diagram also shows the vector 3 , but in this case it does not start at the origin.y 1,2 1, 1 OxVectors in Three DimensionsIn a vector with three components, the first two tell us ho many units to movein the x- and y-directions, as before. The third component specifies how farto move in the z-direction.zWhen looking at a pair of x , y -axes, thez-axis points out of the page from theorigin.A set of 3D axes can be drawn on a page asshown to the right.zyOxFor example,z 4 3 1 yOis a vector in three dimensions. This vectoris shown in the diagram, starting from theorigin.413xZero VectorsAny vector with all components zero is called a zero vector and can be written 0 as 0 , e.g. 0 0 . 0 hsn.uk.netPage 2CfE Edition

Higher Mathematics3VectorsMagnitudeEFThe magnitude (or length) of a vector u is written as u . It can be calculatedas follows. a If u b then a If u b then c u a2 b2u a2 b2 c 2EXAMPLES 5 1. Given u 12 , find u . u 5 2 12 2 169 13 units. 5 2. Find the length of a 6 . 3 a 5 2 6 2 32 50 5 2 units.Unit VectorsAny vector with a magnitude of one is called a unit vector. For example: if u 1 2 230 then u 12 0 23 2 44 1 unit.So u is a unit vector.hsn.uk.net2Page 32CfE Edition

Higher MathematicsVectorsDistance in Three Dimensions The distance between the points A and B is d AB AB units. 1 For example, given AB 2 , we find d AB 5 2 1 22 52 30 .In fact, there is a three-dimensional version of the distance formula.The distance d between the points x1 , y1 , z1 and x1 , y1 , z1 isd x2 x1 2 y2 y1 z2 z1 units.22EXAMPLEFind the distance between the points 1,4,1 and 0,5, 7 . x2 x1 The distance is 2 y2 y1 z2 z1 222 0 1 5 4 2 7 1 2 12 12 8 2 1 1 64 66 units.4Equal VectorsEFVectors with the same magnitude and direction are equal.For example, all the vectors shown tothe right are equal.If vectors are equal to each other, thenall of their components are equal, i.e. a dif b e c f qpstr then a d , b e and c f . Conversely, two vectors are only equal if all of their components are equal.hsn.uk.netPage 4CfE Edition

Higher Mathematics5VectorsAddition and Subtraction of VectorsEFConsider the following vectors:abcAdditionWe can construct a b as follows:baa b means a followed by b .a bSimilarly, we can construct a b c as follows:bcaa b ca b c means a followed by bfollowed by c .To add vectors, we position them nose-to-tail. Then the sum of the vectors isthe vector between the first tail and the last nose.SubtractionNow consider a b . This can be written as a b , so if we first find b ,we can use vector addition to obtain a b .b b is just b but in the opposite direction. b and b have the same magnitude, i.e. b b . bTherefore we can construct a b as follows: ba bhsn.uk.netaa b means a followed by b .Page 5CfE Edition

Higher MathematicsVectorsUsing ComponentsIf we have the components of vectors, then things become much simpler.The following rules can be used for addition and subtraction. a d b e c f a d b e c f a d b e c f add the components a d b e c f subtract the componentsEXAMPLES 1 1 1. Given u 5 and v 2 , calculate u v and u v . 2 0 1 1 u v 5 2 2 0 0 7 2 1 1 u v 5 2 2 0 2 3 . 2 1 4 2. Given p 32 and q 3 , calculate p q and q p . 6 3 5 4 1 p q 32 3 6 3 5 5 3 2 21 5 hsn.uk.net 1 4 q p 3 32 6 3 5 3 9 2 . 9 5 Page 6CfE Edition

Higher Mathematics6VectorsMultiplication by a ScalarEFA vector u which is multiplied by a scalar k 0 will give the result ku . Thisvector will be k times as long, i.e. the magnitude will be k u .Note that if k 0 this means that the vector ku will be in the oppositedirection to u .For example:u 2u3uIf a u b c then1u2 ka k u kb . kc Each component is multiplied by the scalar.EXAMPLES 1 1. Given v 5 , find 3v . 3 1 3 3v 3 5 15 . 3 9 6 2. Given r 3 , find 4r . 1 6 24 4r 4 3 12 . 1 4 hsn.uk.netPage 7CfE Edition

Higher MathematicsVectorsNegative VectorsThe negative of a vector is the vector multiplied by 1 . If we write a vector as a directed line segment AB , then AB BA :BB AB AB BAA7APosition VectorsEF OA is called a position vector of point A relative to the origin O, and iswritten as a. OB is called the position vector of point B, written b.z yGiven P x , y , z , the position vector OP orP x p has components y . z O xyTo move from point A to point B we can moveback along the vector a to the origin, and alongvector b to point B. AB AO OB OA OBAaBbxO a b b a For the vector joining any two points P and Q, PQ q p .hsn.uk.netPage 8CfE Edition

Higher MathematicsVectorsEXAMPLE R is the point 2, 2, 3 and S is the point 4, 6, 1 . Find RS . 2 4 From the coordinates, r 2 and s 6 . 3 1 RS s r 4 2 6 2 1 3 2 8 . 4 8Basis VectorsEFA vector may also be defined in terms of the basis vectors i , j andkjk. These are three mutually perpendicular unit vectors (i.e. theyare perpendicular to each other).iThese basis vectors can be written in component form as 1 0 0 i 0 , j 1 and k 0 . 0 0 1 Any vector can be written in basis form using i , j and k . For example: 2 1 0 0 3 2 0 3 1 6 0 2i 3 j 6k . 6 0 0 1 There is no need for the working above if the following is used: a ai b j c k b . c hsn.uk.netPage 9CfE Edition

Higher Mathematics9VectorsCollinearityEFIn Straight Lines, we learned that points are collinear if they lie on the samestraight line. The points A, B and C in 3D space are collinear if AB is parallel to BC ,with B a common point.Note that we cannot find gradients in three dimensions – instead we use thefollowing.Non-zero vectors are parallel if they are scalar multiples of the same vector.For example: 2 6 2 If u 1 and v 3 3 1 3u then u and v are parallel. 4 12 4 15 5 20 5 If p 9 3 3 and q 12 4 3 then p and q are parallel. 6 2 8 2 EXAMPLEA is the point 1, 2, 5 , B 8, 5, 9 and C 22, 11,17 .Show that A, B and C are collinear. AB b aBC c b 8 1 5 2 9 5 22 8 11 5 17 9 7 3 4 14 6 8 7 2 3 . 4 BC 2AB , so AB and BC are parallel – and since B is a common point, A,B and C are collinear.hsn.uk.netPage 10CfE Edition

Higher MathematicsVectors10 Dividing Lines in a RatioEFThere is a simple process for finding the coordinates of a point which dividesa line segment in a given ratio.EXAMPLE1. P is the point 2, 4, 1 and R is the point 8, 1,19 .The point T divides PR in the ratio 2 : 3 . Find the coordinates of T.Step 1Make a sketch of the line, showing theratio in which the point divides theline segment.PT3RStep 2Using the sketch, equate the ratio ofthe two lines with the given ratio.Step 3Cross multiply, then change directedline segments to position vectors.Step 4Rearrange to give the position vectorof the unknown point.2PT 2 TR 3 3PT 2TR 3 t p 2 r t 3t 3 p 2r 2t3t 2t 2r 3 p 8 2 5t 2 1 3 4 19 1 16 6 5t 2 12 38 3 10 5t 10 35 2 t 2 7 Step 5From the position vector, state thecoordinates of the unknown point.hsn.uk.netPage 11So T is the point 2, 2, 7 .CfE Edition

Higher MathematicsVectorsUsing the Section FormulaThe previous method can be condensed into a formula as shown below.If the point P divides the line AB in the ratio m : n , then:p n a mb,n mwhere is a , b and p are the position vectors of A, B and P respectively.It is not necessary to know this, since the approach explained above willalways work.EXAMPLE2. P is the point 2, 4, 1 and R is the point 8, 1,19 .The point T divides PR in the ratio 2 : 3 . Find the coordinates of T.The ratio is 2 : 3 , so let m 2 and n 3 , then:n p mrt n m3 p 2r 5 1 3 2 2 8 15 5 3 4 2 1 1 3 1 2 19 5 NoteIf you are confident witharithmetic, this step canbe done mentally. 2 2 7 So T is the point 2, 2, 7 .hsn.uk.netPage 12CfE Edition

Higher MathematicsVectorsFurther ExamplesEXAMPLES3. The cuboid OABCDEFG is shown in the diagram.EFHDAGBOCThe point A has coordinates 0,0,5 , C 8,0,0 and G 8,12,0 . Thepoint H divides BF in the ratio 4 :1 . Find the coordinates of H.From the diagram: OH OA AB 54 BF OA OC 54 CG h a c 54 g cNote BH 4 , so BH 54 BF .BF 5 a c 54 g 54 c a 15 c 54 g 0 8 8 14 0 5 0 5 12 5 0 5 8 . 485 5 So H has coordinates 8, 485 ,5 .4. The points P 6,1, 3 , Q 8, 3,1 and R 9, 5,3 are collinear. Findthe ratio in which Q divides PR. Since the points are collinear PQ k QR for some k. Working with thefirst components:8 6 k 9 8 k 2. Therefore PQ 2QR so Q divides PR in the ratio 2 :1 .NoteThe ratio is 2 : 1 sincePQ 2 .QR 1hsn.uk.netCfE EditionPage 13

Higher MathematicsVectors5. The points A 7, 4, 4 , B 13,5, 7 and C are collinear. Given that Bdivides AC in the ratio 3 : 2 , find the coordinates of C. 3 AB 5 ACb ab a35ccNoteA sketch may help youto see this:2C3B 35 c a 35 c 35 a b 25 a 53 b 23 aA 13 7 53 5 23 4 7 4 17 11 . 9 So C has coordinates 17,11, 9 .11 The Scalar ProductEFSo far we have added and subtracted vectors and multiplied a vector by ascalar. Now we will consider the scalar product, which is a form of vectormultiplication.The scalar product is denoted by a .b (sometimes it is called the dotproduct) and can be calculated using the formula:a .b a b cos θ ,where θ is the angle between the two vectors a and b .This formula is given in the exam.hsn.uk.netPage 14CfE Edition

Higher MathematicsVectorsThe definition above assumes that the vectors a and b are positioned so thatthey both point away from the angle, or both point into the angle.aθbaθbHowever, if one vector is pointing away from the angle, while the other pointsinto the angle,aθbaθbwe find that a .b a b cos θ .EXAMPLES1. Two vectors, a and b have magnitudes 7 and 3 units respectively andare at an angle of 60 to each other as shown below.b60 aWhat is the value of a .b ?a .b a b cos θ 7 3 cos 60 21 12 212.2. The vector u has magnitude k and v is twice as long as u . The anglebetween u and v is 30 , as shown below.v30 uFind an expression for u .v in terms of k.u .v u v cos θRememberWhen one vector pointsin and one points out,u . v u v cos θ . k 2k cos30 2k 2 23 3k 2 .hsn.uk.netPage 15CfE Edition