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b 3.2.4 The Sine Rule: In. AABC, — — . = £ = 2R, where R is the circumradius of AABC. sinA sinB sinC Proof: Let AD be perpendicular to BC. a AD=bsinC 1 ¢ b A (AABC)= 5 BC x AD = i ax bsinC - B D a Cc A (AABC) = : ab sin C 2A (AABC) = ab sin C Similarly 2A (AABC) = ac sinB and 2A (AABC) = be sin A be sin A = ac sin B = ab sin C Divide by abc, besinA _ acsinB _ absinC abe abc abc sinA _ sinB _ sinC a b c a b =——=—— ..(1) sind sinB sinC Scanned with CamScanner Scanned with CamScanner To prove that cach ratio is equal to 2R. As the sum of three angles is 180", at least one of the angle of the triangle is not right angle. Suppose A is not right angle. Draw diameter through A. Let it meet circle in P. AP = 2R and AACP is a right angled triangle. 4 ABC and 2 APC arc inscribed in the same arc. m ZABC=m Z APC sin B= sin P= $+ A 2R b sinB= 5p 6b R > sinB @Q) From (1) and (2), we get qa _ b _¢ =2R sind sinB sinc Different forms of Sine rule : Following are the different forms of the Sinc rule, In AABC. a b c =—_= =2 sind sinB sinC aR (i) (ii) a=2Rsin A, b=2R sinB, c=2R sinC ... sind sinB_ sinC (iii) ~——= = = a k b c (iv) bsin A=asin B, csinB=bsinC,csinA=asinC a_sind b_sinB (Vv o> b sinB’c sinC Scanned with CamScanner Scanned with CamScanner 3.3.6 The projection Rule: In AABC, (i) a=bcosC+ccos B (ii) b=ccos A+acosC (iii) c=acosB+becosA Proof : Here we give proof of onc of these three statements, by considering all possible cases. To prove that a= b cos C +c cos B Let altitude drawn from A meets BC in D. A BD is called the peojection of AB on BC. DC is called the projection of AC on BC. c b Projection of AB on BC =c cos B And projection of AC on BC = DC = b cos C Case (i) B and C are acute angles. B D a Cc Projection of AB on BC = BD = c cos B . And projection of AC on BC = DC =b cos C Fig 3. From figure we have, a =BC=BD+DC =ccos B+bcosC =bcosC+ccosB “a =bcosC+ccosB Case (ii) B is obtuse angle, *. Projection AB on BC = BD =c cos (x - B)=-c cos B And projection of AC on BC = DC = b cos C From figure we have, a =BC=DC-BD = bcos C -(-c cos B) =bcosC+ccosB “a =bcosC+ccosB Case (iii) B is right angle. In this case D coincides with B. R.HLS. = b cos C+ ¢ cos B = BC+0 = a=LHS. a=bcosC+ccosB Similarly we can prove the cases where C is obtuse angle and C right angle. Therefore in all possible cases, a=bcosC+ccosB Similarly we can prove other statements. Scanned with CamScanner Scanned with CamScanner nol? Let's have a theorem. Theorem 3 : Homogenous equation of degree two in x and yy, ax? +2hxy + by? = 0 represents a pair of lines passing through the origin if i? - ah > 0. Proof : Consider the homogencous cquation of degree two inx and y, ae? +2hxr + by? = 0 (1) Consider two cases b = 0 and b # (). These two cases are exhaustive. Case 1: If b= 0 then equation (1) becomes ax? +2hxy = 0 x(ax +2hy ) =0, which is the combined cquation of lines Y ax + 2hy=0 y" x=0 Figure 4.3 Scanned with CamScanner Scanned with CamScanner Theorem 4.4 : The acute angle 0 between the lines represented by av + 2hxy + hy? = 0 is given by 2Vir — ab tan0 = ath Proof: Let m, and m, be slopes of lines represented by the equation ae + 2hxy + by? = 0. 2h a omy tm, =-— and mm,= 7 ° b 2b o (m,— mF = (m, + m,)? — 4mm, Fy aU) 4h? 4ab | an Scanned with CamScanner Scanned with CamScanner 4h? —4ab 2 2 my—m,=t 5 Be 4(h? - ab) bb vi? - ab b As 0 is the acute angle between the lincs, tan@ = m,—m, I+ myn, + 2Vh* — ab b 142 b 2Vir - ab ath TM yemx Figure 4.4 Remark : Lines represented by av? + 2hxy + by? = 0 are coincident if and only if m, =m, _ 2Vi -ab m,— m= 0 =() b “I? -ab=0 2. P=ab Lines represented by ax? + 2hxyv + by? = 0 are coincident if and only if 4° = ab, Scanned with CamScanner Scanned with CamScanner 2, (Midpoint formula) If R(F) is the mid-point of the line segment AB then m=nsom:n=1:1, _ Wb+la = a4 r= 141 that is " = 2 Theorem 6 ; (Scction formula for external division) Let A(@ ) and B(h ) be any two points in the space and R(7) be the third point on the line AB dividing the segment AB extcrally in the mb -na ratio m:n. Then = mn Proof: As the point R divides line segment AB exterally, we have cither A-B-R or R-A-B. Assume that A-B-R and AR : BR =™in AR Lm “BR on As n(AR) and m(BR) have same magnitude and direction, * n(AR) = m(BR) .n(F-@) =m(r—-5) “nF —na = mF -mb Fig 5.36 so n(AR) = m (BR) cmb — nd = mF -nF = (m-n)F —_ mb-na ~ Fs A(a) m-n Note : 1) Whenever the ratio in which point R divides the join of two points A and B is required, it is convenient to take the ratio as kK: 1. Se kb+a B(b) Did) (ae) Then, F = , if division is internal, Fig $37 kb —@ , if division is external. k-1 2, In AABC, centroid G divides the medians internally in ratio 2: | and is given by (sce fig. 5.37) F g= G+b+e (Verify). 3. In tetrahedron ABCD centroid G divides the line joining the vertex of tetrahedron to centroid of opposite triangle in the ratio 3: | and is given by a+h+ted g= 4 Scanned with CamScanner Scanned with CamScanner 10. Prove that medians of a triangle are concurrent. Solution: Consider A ABC. Let P, Q, R be the midpoints of the sides BC, CA, AB respectively. Let the medians BQ and CR intersect at G. Std. XII Sci.: Mathematics - 1 1 Chapter 05: Vectors Target Publications’ Pv. Ltd To prove that the third median AP also passes through G. Leta, b,c, P. q. tr. £ be the position vectors of the points A, B, C, P,Q, R, G respectively. Since P, Q, R are the mid-points of the sides BC, CA, AB respectively By midpoint formula, we get _ bte “20 ct+a 2 =_ a+b r=—— 2 From (i), (ii) and (iii), we get i) 5 ! «-(ii) ol I 1 1 ..iii) 2p=b+c=>2pta=at 2q=cta>2q+ +b= 2r=a+bo2r+c= 3 3 3 2pta _ 2q+b _ 2r+e 241 241 241 a+ at 2p+a_2q+b_2r+e_ a+b+te 3 at This shows that the point G whose position vector is £ lies on the three medians AP, BQ, CR dividing them internally in the ratio 2:1. Hence, the three medians are concurrent. Scanned with CamScanner Scanned with CamScanner Ex. 6. Prove that the angle biscctors of a triangle arc concurrent, Solution : Let A, B and C be vertices of a triangle. Let AD, BE and CF be the angle biscctors of the triangle ABD. Let @, b,é,d,@ and S be the position vectors of the points A, B, C, D, E and F respectively. Also AB =z BC = x AC=y. Now, the angle bisector AD mects the side BC at the point D. Therefore, the point D divides the line segment BC internatty in the ratio AB : AC, that is z : . Hence, by section formula for internal division, we have d == Similarly, we get —_ XO+20 yb+xa t= and f =——— Xz v+x 5 _st+yh As q@=*? x+y ‘ (ety)d =20 +yh A F B ie. (ety)d txd =x0 +yhb +27 “ wo = = - = Fig 5.40 similarly (xtz)@ tyb =xd +yh +26 ‘9 and (x +f +20 =x@ +yb +20 (zt+y)d +40 (x+z2)@+ yb (x+y) f+2e xO+yh+2e i = = = = =h (sa X+y+2 X+ytc Xtytez Xt yer (say) Then we have ie (ytz)dtaG _(xtz)e+yh _ (x+y) +20 (vtz)+x (xtz)+y (xty)+2 That is point H(i) divides AD in the ratio (+ z) : x, BE in the ratio (x +z) :y-and CF in the ratio (x +"): z. This shows that the point H is the point of concurrence of the angle bisectors AD, BE and CF of the triangle ABC, Thus, the angle biscctors of a triangle are concurrent and H is called incentre of the triangle ABC. Scanned with CamScanner Scanned with CamScanner 5. Prove that the line segments Joining . mm a | points of adjacent sides of a quadrilater form a parallelogram. Solution: Let ABCD be a quadrilateral, Let P,Q, R Sh the midpoints of sides AB, BC, CD, DA respectively. Let a,b, e, ds ps 45, 8 be the position vegta,. of A, B, C, D, P,Q, R, S respectively, A— 8 Ss Q D R Cc By using midpoint formula, - a+b -_bte -_ctd - Gay = =— ,r=— a = a Pp 7 q 2 2 ind s > li) Prove that OPQRS is a parallelogram. i.e., to prove that [eq] = [SR] and PO II SR, and QR || PS far]= Consider, PQ = q -p Il —m ol N+ a if je! rm] t ls 3 g i=3 PQ = SR = = <(¢+d-d-a) aR = 4 (6-2) ii) PQ = SR ...[From (ii) and (iii)] [P| = [sR| and PQ || SR Consider, QR = - q = ae ->4¢ [From @)] = 1(¢+d-b-<) — 1, = R= —(d- b) (iv) Scanned with CamScanner Scanned with CamScanner 7. Prove that a quadrilateral is a parallelogram if and only if its diagonals bisect each other. Solution: Let a, b, c and d be the position vectors of the vertices A, B, C and D respectively of the parallelogram ABCD. Scanned with CamScanner Scanned with CamScanner Then, (AB) = (DC) and AB || DC. : ‘ am ... [Opposite sides of a parallelogra | lol > o Il | Q J +t I Ql pt I Sloat + 1 ea] o] > | + ol co! + a] »| 2 wy: e The position vectors of the midpoints of th ate b+d and and diagonals AC and BD are they are equal. The midpoints of the diagonals AC and BD are the same. This shows that the diagonals AC and BD bisect each other. Conversely, suppose that the diagonals AC and BD of DABCD bisect each other. They have the same midpoint. The position vectors of the midpoints of AC and BD are equal. ate _ bed 2 2 atc=bt+d b-a=c-d AB= DC (AB) = (DC) and AB || DC. OABCD is a parallelogram. Scanned with CamScanner Scanned with CamScanner alan, lel ao bees de. F be the position Ve pointy ALB, CY DEE respectively, SCloty af Prove that Er in parallelto AL and ie Since side AB is parallelto side boo AL is parallel to De There exists a sealar (suet (hit AB ODE “ly Hee and Ff are the midpointy of AD En BC respeetively, rd a, bt oe LEE and Ps = 2 2 Now, consider EE = f = ¢ I oa] _ bre at 2 2 bea a 2 2 AB, DC -AB,DC «| 5 3 (ii) _t-DC , DC : . = | a [From (i)] — +- I}. Bh = aa DC, where [Sy is a scalar, 2 2 EF is a scalar multiple of DC. EF is parallel to DC EF is parallel to AB as AB and DC are parallel to each other. Also from (ii) we get, BP = ~ (AB+DC) =! 2 (EF) = = |AB+DC| [BF |AB+D¢| Scanned with CamScanner Scanned with CamScanner 9. Prove by vector method that the angle subtended on semicircle is a right angle. Solution, (>> TTT 1 Let r be the radius and O be the centre of the circle. A, B and C are three points on the circle such that, AB is the diameter. Let a, b and c be the position vectors of points A, B and C respectively. C is on the circle, lel =r Also, lal=lb] =r and b = -a Consider, CA- -CB =(a-c): (b-c) =(a-c)- (-a-c) =(a-c): (-1)(a+e) =(- (a ~e)-(a +e) (FF I i So Ag CA-CB =0 CA is perpendicular to to CB. The angle between CA and CB isa right angle. mZACB = 90° The angle subtended on a semicircle is a right angle. Scanned with CamScanner Scanned with CamScanner