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Vectors (_ Fastracl« Revision ) > The sum of two vectors AB and BC Is AB + BC = AC. >» AB = Position vector of point 8 — Position vector of point A. ~ 2 > IF position vector of points A and B are a and b, then ar) position vector of mid-point of AB = a > Position vector of point P(a, b) = OP cai+ bj and modulus of vector OP = OP =Va" +b? Vector > Unit vector = —————___ Modulus of vector >» Suppose e Isa Position \ vector of polnt P(x, y, Z), then (i) reaxi+ y ie zk; x iy jand zkare component vectors and x, yand zare components of vector i along X, Y and Zaxes. a AndItlelxt4y]4zklayxr ay? +z > (il) If the vector r makes the angles a,fand y with x, Yand Z-axes, then it makes direction cosines cos a, cospand b a a cosy as cosa = 1 COS dae +b42 c OT Ee where a, band care direction ratios. (iil) If £ mand nare the direction casines of a vector, then we always have Pamsnt ot, & IFA(%4,4,%) and B(x, ,¥2,2,) are any two points in space, then direction ratios of AB are (0 = 4 )(Y2 - Wy - 4). > Three points A, 8, Cwhose position vectors are given, will be collinear if AC =m AB (where, mis any scalar] > Section Formulae: Let Aand 8 be two points with position vectors aand brespectively and let Pbe apoint dividing AB Internally in the ratio m:n. Let OP =r, then ea (mb+na) a (m+n) (internally) (externally) ee) > Three vectors a,b,c will be coplanar, iF os Sf asrb+pc (where, Xand p: are scalars) > The scalar product of two vectors 3 and b, having angle 0 between them, Is o a-b=|a||b|cos@=abcos - a > ~ > Projection of b in the direction of a = > Projection of a inthe direction of b » Vector component of a vector a on b 360 4 abe ab b (a:b) bl [BI (bl IB ( > Vector component of a vector b on a = , laP AAN > Scalar product of unit vectors |, J,k is » If the angle between two vectors aand bis @ then ae lallb| >» Two vectors a and b will be perpendicular, if a-b=0. > The vector product of two vectors 3 and b having angle 8 between them, is axb =( alib|sino) n=(oo singyn where, n is aunit vector perpendicutar,to a and b. Pifasa ita j+ak and bab, 14d, J+b5k fi ik Then, axbsla; d, a3 by bz by > The area of a parallelogram whose adjacent sides are 3 and bela x bl > Area of parallelogram = \d, x dl where, 4; and ad are the vectors of diagonals. — =» > The area of a quadrilateral ABCD Is esac xBD| > Area of A ABC= 41 AB x ACI where, AB and a are adjacent sides. > Condition for collinearity of three points, whose position vectors are a,b,c is (ax b) + (bx ¢) + (cx a)= 0. 23 axb jaxbl > Unit vector perpendicular ta vectors a and b isn= AK » Vector product of unit vectors i, j,k i a ixico, jxjeg, k xkeo, ixjek, Jxkei, kxtaj > IFthe angle between two vectors a and bisa, then _laxb] sin@= BE ha Practice Exercise j a) Multiple Choice Questions Q1. If a and bare two collinear vectors, then which of the following statement is not true? (NCERT EXERCISE) a. B =A. for some scalar based c. the consecutive components of 2 and b are proportional d. the direction of bath vectors a and b are same but magnitudes are different Q 2. ABCD is arhombus whose diagonals intersect at E. Then EA + EB + EC + ED equals: (CBSE SQP 2023.24, CBSE 2020) c2B0 86d2ad = ad b. AD Qa. The magnitude of the vector 6 in 2 j+3k (CBSE 2023) al b.5 a7 d.12 Q4. Aunit vector along the vector 4)-3kis: (CBSE 2023) ad (ai-ak) v.d(aiak) a 5 1, A oA 1,4 4 : 4i-3k d. 4|-3k (s A ) i! 4 ) 06. If |al=3 and -1
ALA A a ae Q8. The vectors a=x i-2j+Skand b=i+yj-zk are collinear, if: axeslys-2z=-5 b.x=V2y=-4z=-10 ax=s-V/2y=4.z=10 d.Allafthese > A A Q9. Two vectors a=a, i+a, j+az kand > A AA b =b, i+ bz j+b3 kare collinear, if: —(cgse2023) a. a,b, + 03b; +a3b, =0 Ob as as C2 2D, bz by C.a)=b) 03 = bz, a3 =b3 d.a)}+03 +03 =b)+b,+b3 Q10. The position vectors of the points A, B, C are AAR A RA a A A (2i+ j-k),@i-2j+k)and (i+4j-3k) respectively. These points: a. form an Isosceles triangle b. form a right angled triangle c. are collnear d. form a scalene triangle Q11. Consider the points, A,B,C and D with position vectors 7i-4j + 7k, j-6] + 10k, i -3] 44k and5i-j+5 k respectively. Then ABCD is a: b. rhombus d, None of these a, square c. rectangle nN Q12. The figure formed by the four points i+ j -k, A ALA A An 2143 j,5 j-2k and k= jis: a. square c. parallelogram Q13. If Ois origin and C is the mid-point of A (2, -— 1) and — B (-4,3), then the value of OCis: a4 b= b. rectangle d. None of these nA aA nN ait] c-i+| | del = De Q34. Q a6. Q 36. Q 37. The area of a triangle with vertices A, B, Cis given by: (CBSE SQP2022-23) al AB x AC! b.tTAB «ACI eee. Tee c.—| AC x AB] d.—| AC x AB] 4 8 flal=8, | bl=3and] ax b|=12, then the value of a bis: a.6V3 c.12¥3 Let vectors a and b be such that lal=3 and (NCERT EXEMPLAR) b.8v3 d. None of these Ibl= # then a x bi is a unit vector if the angle > > between a and b is: (NCERT EXERCISE) b.= ae 4 2 wia n a= 6 > 3 If 0 is the angle between two vectors a and b and 27> > : |a-b]=| ax b], then 0is equal to: iycear exercise) ao b= dr Nia -@ Assertion & Reason Type Questions Directions (Q. Nos. 38-46): /n the following questions, each question contains Assertion (A) and Reason (R). Each question has 4 choices (a), (b), (c) and (d) out of which only one is correct. The choices are: ae. Qa9. a. Both Assertion (A) and Reason (R) are true and Reason (R) Is the correct explanation of Assertion (A) b. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A) c. Assertion (A) Is true but Reason (R) Is false d. Assertion (A) Is false but Reason (R) Is true Assertion (A): The magnitude of the resultant of + AAW > A A A vectors a=2i+j+kand bait 2j+3kis-\34. Reason (R): The magnitude of a vector can never be negative. Assertion (A): The unit vector in the direction of ARN KA AK AK AK sum of the vectors i+ j+k, 2i-j-kand2j+6k 104 oA A is -=(3 i +2 j+6k). T = z Reason (R):Let a be a non-zero vector, then is lal > aunit vector parallel to a. Qa. gai. Q 42. 943. Q 4a, 045. Q 46. > +> Assertion (A): If the points P=(a+ b-c), > > = > > > Q=(2a+ b)and R=(b+ tc) are collinear, where > a,b, © are three non-coplanar vectors, then the value oft is -2 — Reason (R): If P,Q,R are collinear, then PQ || PR or PQ =2 PR, 2ER. Assertion (A): The sides of a > A UA asi+2j and adjacent parallelogram are along b =2 ji + i The angle between the diagonals is 150°. Reason (R): Two vectors are perpendicular to each other if their dot product is zero. 2 a] Sh > Assertion (A): If a+ b+c=0, |al=3, |b|=4, > > >> | ¢ |=5, then abt bec+ s ais equal to -25. 2734 Reason (R): If a+b+c=0, then the angle 6 obec — 2bec Assertion (A): The length of projection of the is a a between b and c is given by cos0 = vector3 j -j -2konthe vector i +2]-3kis—Z, Tia Reason (R): The Projection of a vector a on > another vector b is @ 5) |b] Let a and rf be proper vectors and 0 be the angle between them. | > hy oy mr re Assertion (A): (ax b)” + (a+ b)* # (a)” (b) Reason (R): sin’ 64+cos’ O=1 > Assertion (A): If asi-jek, a:‘b=0 and axbeq where c=-2i-j+k then b=(0,1,1). + A A “ Reason (R): lf a=x, i+y, j+z, kand = A A " bax, ity, j+z, k, then nA” ij k 7 3 axbelx, yz Zz mG MQ 24 > >, >>) Assertion (A): If (ax b)* +(a+b)* =400 and 13 [=4 then ins Reason (R): If a and b are pny two vectors, then (ax b)? = (a)? (by -(a- bY. De Answers 1. (9) 2. () 3. (c) 4. (b) 5. (a) 6. () 7. (d) 8. (d) 9. (b) 10. (a) 11. (0) 12. (d) 13. (c) 14. (a) 15. (a) 16. (0) 17. (d) 18. (d) 19. (a) 20. (6) 21. (0) 22. (c) 23. (b) 24. (b) 25. (d) 26. (6) 27. (d) 28. (a) 29. (b) 30. (a) 31.4) 32.) 33. (a) 34. (6) 35. (c) 36. (6) 37. (b) ~—-38. (b) 39. (¢) 40. @) 41.() 42.0) 43. a) 44. (d) 45. (c) 46. (d) wae and @=P5=05-0P @ Case Study Based Questions “Gf ita aff) Case Study 1 =-314]-2k PORS is a parallelogram Now, p4q=(-1+]-k)+(-314]-2k) whose adjacent sides are A LA A =-4/4+2)-3k ae represented by the vectors p So, option (a) is correct. >. > A A LA and q. Three of its vertices 2. From part (1). p+q=-4i+2j-3k P B Q are P (4, —2, 1), O(3,-1, 0) N 2 2 aap tat and S (1,-1,—D. low, Ip+ql=l * i ; | . Based on the above information, salve the following =y(-4)? +2)? +(-3) questions: =e 459 = 29 + > — ce". Q1. The vector p+ q is: - Aunit vector along the vector (p+ q)= 244) A A A A LA lA fe, dy _A Ip+ql a.-414+2)|-3k b.41-2j-3k -41+2j-3k a 5% af a An a a c-41+2j)+3k d.-i+j+k a So. option (b) is correct. Q2 Auunit vector slang the vector (hea)is: 3 Diagonal ¢2p-qe(-t+}-@)-(-3l+}-2h) 2l-i+k -414+2]-3k AA b. =2i+k ‘ V6 Po) WOW KR a A So, option (c) Is correct. cl+ick g irk be Now. oeeetta® fya(-atad ot se ar . Now, p xq=(-i +] -k)x(-31 4 | -2k) nA . 2 i 7k Q.3. The diagonal s is: AOA “ a“. AA AA =]-1 1 -1 a(2i-{) b()/+2k) c(2i+k) (ik) 4a 1-2 Q4, Area of PORS, whose adjacent sides are pand a A a a is: = | (-241)-| (2-3)+k (-1+3) a. /2 59. units b. /3 sq. units nit t , t 2k c. 15 sq, me an 4. ¥6 sq, units -. Area of parallelogram PORS=|p xql Q5. The value of =| rx sis: al-14f42hl= f(y? + (02 +2? ave b.2/2 av a5 3 [4144 = V6 sq. units - So, option (d) is correct. e——_—_————__ Solutions > > > A A A 5. Dilagonalr =p+q=-41+2|-3k 1. Position vector of the points P,Q and S are (from part (1)) ee ee re ees > aA # OP =41-2j+k,00 531 -] and diagonals =21 +k {from part (3)) —m K A KA and OS =i eo | k LS > —> => => Now, rxSml-4 2 -3 p=PQ=0Q-OP 201 (ai -f)-(4i -27 +k) =(3i-j|)-(41-2] + nN a a nk = i (2-0)-j| (-4+6)+k(0-4) e-l+j-k , A Case Study 3 A building is to be constructed in the form of a triangular pyramid ABCD as shown in the figure. Let its angular points be A (0, 1, 2), B (3, 0, 1), C (4, 3, 6) and D (2, 3, 2) and G be the point of intersection of the medians of ABCD. Based on the above information, solve the following questions: Q1. The coordinates of point G are: a.(2.3,3) b.(33,2) (32.3) Q 2. The length of vector AG is: a AAT units b. /Munits c.V13 units d.J19 units Q3. Area of AABC (in sq. units) is: a Vid b2vi0 c.avI0 Q 4. The sum of lengths of AB and ac is: a.5units b.932units c. units d. 11 units Q5. The length of the perpendicular from the vertex D on the opposite face is: d (0,2, 3) d.5vid a 6 units b. 2 units Via “io OF units d.8V10 units Solutions 1. as centroid. TRICK [ Intersection point of medians in a triangle is foe Clearly, G is the centrold of ABCD, therefore coordinate of G are e 04343 14642 30° 3 ' 3 So, option (c) Is correct. 2. Since, As(0,1, 2) and Gs (3, 2,3) —> A a iN AG = (3-0) i +(2-1) | +(3-2)k Aww =31+|+k => [AGP 232412412 29414101 => | AGle= Vilunits So, option (b) Is correct. 3. Clearly, area of AABC = 3108 x ACI )- (3.2.3) AO OB ys ew [Lt Here, ABxAC=/3-G O-1 1-2 -O 3-1 6-2 i jk aS 1 of 424 =i (-442)-](12+4)+k(6+4) =-21-16j 410k VB x AC | =f(-2)? + (16)? + 107 = 472567100 = 360 =6V10 Hence, area of AABC =5xeVi0 = V0 sq, units Sa, option (c) Is correct 4. Here, AB =31-j-k > |AB|=/9a15 =v we. ae Also, AC =4742744k = | AC|= 6+ 4516 = 56 =6 Now, | AB 1+] AC |= 11+ 6 =332 +6 =9.32 units So. option (b) is correct. 5. The length of the perpendicular from the vertex Don the opposite face =IProjection of AD on AB x ACI A LA A A A (27424)-(-27-16] 410k) (-2)* + (-16)* + 107 -4-32 36 6 eo =5 ia = vio "s So, option (a) Is correct. Case Study 4 Three slogans on chart papers are to be placed on a school bulletin board at the points 4,B and C displaying A (Hub of Learning), B (Creating a better world for tomorrow) and C (Education comes first). The coordinates of these points are (1, 4, 2), (3, —3, —2) and (—2, 2, 6) respectively. FA kkk KY 4 [Hub of Loaring} 5 » — Educution comas first] tape Creating a batter world for tomorrow KERRIER TRE dick ddddiddik ddd De Based on the given information, solve the following questions: Qi. Let 2b and e be the position vectors of points A, BandC respectively, then at b+ C is equal to: a2zis3j+46k b.21-3)-6k c.214+8)43k d.2(71+8] 43k) Q 2. Which of the following is not true? Qa. a.AB+BC+CA=0 b.AB+ BC-AC=0 c. AB + BC - CA = O d. AB - (eB 4CA= Oo Area of AABC is: a. 19 sq. units b. 11937 sq. units cA 8I7 54 units d. (1837 sq. units Q 4. Suppose, if the given slogans are to be placed ona g6. straight line, then the value of a lax b+bx c+cx alwill be equal to: a.-1 b-2 a2 do if 2 =2i1+3 j + 6k, then unit vector in the 4 direction of vector a is: Ze Be Bik bil+s)+=sk 7 7 7 d. None of these Solutions ————————_ acis4[s2kb= aia} 2k -2isa}ssk ~ and ec » asbece2t 13 16k So, aption (a) is correct. . Using triangle law of addition in AABC, we get ‘AB 4 BC 4 CA = 0. which can be rewritten as AB 4 BC AC = i or AB - a + We a So, option (c) Is correct. . We have, A (1, 4, 2), B (3, -3, -2) and C (-2, 2, 6) — > 2 A An Now, AB =b-a=21-7|-4k — > » a A On and AC =c-a=-31-2)4+4k A AOA sap cay |! oh ABxACs|2 -7 -4 -3 -2 4 a = 1(-28-8)—] (B-12)4 k (4-21) ra GLP =-36 144] -25k Now, |AB x AC |= (-36)? 4 47 + (-25) = fi05962 162625 = 1537 . Area of AABC = 3I%B «AC l= + Ji537 ‘sq. units So, option (c) is correct. 4. If the given paints le on the straight line, then the points will be collinear and so area of AABC =A. > laxb+bxc+cxal=0 (lFa a. hc are the position vectors of the three vertices A,B and C of AABC, then area of triangle 1 = So = So. =5lla xb+bxc+cxal] So, option (d) is correct. 5. Here,|al= 27432 +6" =V4494356 =V49 =7 «. Unit vector in the direction of vector a Is a 2t43;4+6k 2% 3% 6s 3 a esis tk So, optian (b) Is correct. Case Study 5 Rakesh purchased an air plant holder which is in the shape ofa tetrahedron. Let P,Q, R and S be the coordinates of the air plant holder where P =(3, 3, 4), = Gy1,:2),. R= (2; 1, 3) and 8(1, 1,1). Based on the above information, solve the following questions. Q1. Find the position vector of Ps. Q 2. Find the area of APOR. Qa. Find the unit vector along Ps. Or Find the projection of PQ on PR. ° Solutions ° ee ee a er 1. Here, OP =31+3)+4kand 05 si+) +k — = -. Position vector of PS = OS - OP A AA A AR ogc A lA aif =(14+]+k)-(3143]4+4k)=-21-2|-3k 3 A AR me 2. Here, OQ =31+|+2kand OR=214j)+43k 5 Now, position vector of BQ = =(at4}42k)-@i +3) 44k)=-2 7-2 and position vector of PR =OR-OP iN 2(214]43k)-(3143] +44) A AON wel-2)~k i % A = 1(2-4)-](0-2)+k (0-2) aD iz] <2k De o——————_ Solutions > 1. Position vector of Q= 5i4 aj and position vector of R=6 145} T42] OR =(6-5)1+(5-3)j 2. Position vector of P =274 2j A A and position vector of S=9 i +8 j PS =(9-2)i+(8-2)[=71 46] Now, | PSP =(7)2+(6)2=49+36=85 > IPS | = VO5 units 3. PR =OR-OP =(6145})-(21+2)) =(4743]) Unit vector of PR or BR = au IPR 08 = 05 -00=(97 +8) -(51 +3]) = 1 5} — - & and PR =41 43] =(0-0)1-(0-0) f+ (20 -12)k 28k Case Study 8 If two vectors are represented by the two sides of a triangle taken in order, then their sum is represented by the third side of the triangle taken in opposite order and this is known as triangle law of vector addition Based on the above information, solve the following questions: Q1. If ABCD is a parallelogram and AC and BD are its diagonals, then find the value of AC + BD. — Q 2. If ABCD is a parallelogram, where AB sla and BC a2 b, then find the value of AC - BD. 2a" D c — > 2b 2b Or If ABCD isa a quadrilateral, whose diagonals are ra and BD, then find the value of BA + O. c A iB Q3. If Tis the mid-point of side YZ of AXYZ, then find the value of XY + xz. x Y T Zz Solutions ° 1. From triangle law of vector p, c addition, AC + BD = AB + BC 4 BC + wo A =AB+2 BC +CD — =< wy =AB+2 BC -AB = = = =2 BC [- AB =- CD) 2. INAABC, AC=2a+2b ) (by triangle law of addition) = b 2a e 2 Bz Al 8 a ie =e andIndAABD, 2b=2a+ BD saf2) (by triangle law of addition) Adding eqs. (I) and (2). we have AC +2b=4a48D42b = AC -BO=4 a Or InAABC, BA+ ACs BC (by triangle law) ...(1) (by triangle law) ...(2) From eas. (1) and (2), BA +AC=BD-CD = BA4CD=BD-AC=8D+CA InaBCD, BC+ CD=B8D 3. Since, T Is the mnldepoint of YZ. So, Wei Now, XW 4X2 = (XT 4 TY)+ (XT + TZ) [by triangle law) Very Short Answer type Questions Q1. Write the associative law of vector addition. — >> > o> Q2. InAABC, prove AB + BC + CA = 0. (NCERT EXERCISE) — = = — Q 3. Find the sum of vectors BA, BC, CD and DA, where ABCD is a quadrilateral. Q4. In the figure, ABCD is a D Cc parallelogram in which E and F are the mid-points of F AB and GC respectively. — + A B lf AB=a,AD=b, then E find vector TF. Q5. If the position vector of P is je] and position vector ofQis4j 5 i, then find PO. (NCERTEXERCISE) Q6. Three forces 2i 43] +4R,- 4j ai and 5 ~4k-37 act on a particle. Prove that the particle is equilibrium. Q7. Find for what value of a, the vector ( 21-3 j +4) and fai +6 j -8 ky are collinear? Q8. Find the direction cosines of the vector i +2 j 3k. (NCERTEXERCISE) Q9. Find a unit vector along the vector 2 i =i) 42k AKA Q10. Prove that (i-j)k-+ (j-k) i+ (k- AA > i) j=0. Qi. if vectors ai+2j43k and 3i+b j are perpendicular, then prove that 3a +26 =0. (CBSE 2020, 19, 18) Ah x A AA Q12. If the vectors 2i-j+pk and i+j-k are perpendicular to each other, find the value of p. > > Q1a. The magnitudes of two vectors a and b are 1 and 2 respectively and abs 1, find the angle between these vectors. (NCERT EXERCISE) Q14. Find the magnitude of each of the two vectors ‘aand b, having the same magnitude such that the angle between them is 60° and their scalar product iso (COSE 2018) Q 16. Find a vector perpendicular to vectors a=i ja k and b=2i+3j-k ROA KR A NK Q16. If (2 i+ 6 j+27k)x (i+Aj+uk)=0, then find Aand [. Q17. Find the area of that triangle whose two adjacent sides are represented by a=3i+4j and Becsia7]. Q18. (i) Find i-(jxk)+(ixk): j Gi) Find 3-(jxR)+ (Rx i) heh fs (NCERT EXERCISE) y Short Answer type-i Questions Q1. If E is the mid-point of side AC of AABC, then — iwor om prove that BE =5( BA + BC). Q2. Ina pentagon ABCDE, prove that => aor as a ss a eae a oe rao AB + AE+ BC + DC + ED + AC =3 AC. Q3. ABCDEF is a regular hexagon in which the forces Sy ay ee ies = AB, AC, AD, AE and AF are acting at A, prove that their resultant is 3 AD. Q 4. Show that the line passing through the points (4,7, 8) and (2, 3, 4) is parallel to the line passing through the points (—1, -2, 1) and (1, 2, 5). Q6. Find the coordinates of the points which divides the line joining two points (2, -5, 1) and (1, 4, - 6) in the ratio 2 : 3 internally. Q6. Find aunit vector, along the sum and difference of > AR A OR > oA a vectors a=2i+2j-Skand b=2i+j+3k. (NCERT EXERCISE) 2 Ay ok * i K Q7. lf a=i-j+7k and b =5 i-j+/k, then find the as >> value of 4 so that the vectors a+ b and a—b are orthogonal. (CBSE SQP 2022-23) QB. If a=5i-j-3k and b= i+3j-5k, then show > => > > that (a+ b) and (a —b) are perpendicular. (NCERT EXERCISE) Q9. If the projection of the vector i j i k on the AK 4 vector p i+ j-2k is ? then find the value(s) of p. (CBSE 2023) Q 10. Write the projection of the vector (b+ ©) on the > ee ee vector a, where a=2i-2j+k, b=i+2j-2k > LAA and¢e=2i-j +4h. (CBSE 2022 Term-2) = > A A a Qi1. Find |x|, if (x—a).(x+a)=12, where ais a unit vector. (CBSE SQP 2022-23) > A A AD A An A > lA A 012. If a=2i+2j+3k, b=-i4+2j+kand c=3i+j such that (a +A, b) is perpendicular to rs then find the value of i. (NCERY EXERCISE, CBSE 2022) De Q15. Q16. qu. Q18. Qg19. Q20. Q21. Q 22. Q 2a. Q24. Q265. 026. For any two vectors a and 6, show that 27> > la-bl<|allbb (NCERT EXERCISE) > = For any two vectors a and b, show that a [a+ bls] al+]bl- (NCERT EXERCISE) > oR BW oR KR 5 If a=2i+j-k, b=4i-7j+k, find a vector c Sey ey suchthat ax c =band a.c =6. (CBSE 2017) Se A A ow: nA UA Let a=4i+5j-k, -4j+5k and casit— Find a vector d which is 5 oF et perpendicular to both c and b and d-a=21. (CBSE 2018) Find the area of a parallelogram ABCD whose side AB and the diagonal DB are given by the vectors si +7 and 2j 42]43k respectively. (cgse2017) Find the area cid parallelogram, whose diagonals are dy =3i+2j- kand dy = 1-3 j+2k. if ac3i-j+Skand b=i+2j—k, then find the area of that triangle whose two sides are 2 > represented by a and b. Using vectors, find the area of AABC, with vertices A(1, 2, 3), B(2, -1, 4) andc (4, 5, -1). (CBSE 2020, 17) Show that the points A, 8,C with position vectors 2i-j+k,i-3j-5k 3i-4j-4k respectively, are the vertices of a right angled triangle. Hence, find the area of the triangle. (CBSE 2017) and 2 oo Given that vectors a, b, c form a triangle such >> > that a=b+c. Find p,g,r,s such that area of triangle is 5/6, where na A On a > AA A b=si+3j+4kandc=3i+j-2k. Prove that the area of triangle, the position vector 3 “i whose vertices are a, b, c, is: ee) Flbx vex asad. > A nA #W aspitgq j+rk, (CBSE 2016) i eee A OK IFazi+ j+kand bei 142] 43k, then find aunit vector perpendicular to both a+ band a-b. (COSE 2023) Or Find the unit normal vector to each of the vectors (a + b) and (ab), where asia jek and idk oh ai b=i+2j+3k. (NCERY EXERCISE) x , then show that (CBSE SQP 2022 Term-2) -Q Long Answer Type Questions Qu. Qg2. Q3. Qa QE. 06. Q7. Qe. Qo. The two adjacent sides of a parallelogram are AOR OOK represented by vectors 2i-4j+5k and j-2 j 3k. Find the unit vector parallel to one of Also, find the area of the (CBSE 2022 Term-2) its diagonals. parallelogram. Ifaci+ j+kbo4i+3]+4k and csi + aj+pk are linearly dependent vectors and| c|=.J3, then find the values of o and B. Let 3, band c be three non-zero vectors such that no two of these are collinear. Ifthe vectors a +2 b is collinear with t and b + ae is collinear with a then find the value of a+2b+ 6c. Tha scalar, product of vector T from the vectors i i Sik, zi i+7 jandi i + i + icare respectively — 1,6 and 5. Find vector r. Find the length of parallelogram constructed on 5 a+2b and 2-36, given that lal =2v2, Ibl=3 and angle = eH between a and b ise longer diagonal of eed If a,b,c are unit vectors, then show that > 9 9 > > Ja-bl° +1b-cl° +1 ¢-al° doesnot exceed by 9. +35 > > > Let u, v,w be such that} u|=1,|v|=2,|wl=3. If the projection v along Uis equal to that of w along wand vw are perpendicular to each other, > Ifa a 42b43c= , then prove that the value of = eb ied tet is 6(bxc) or 2(ax b) or > > 3(cx a). . Using vector method in AABC, prove that: sinA_sinB_ sinc (NCERT EXEMPLAR) De Very Short Answer Type Questions 1. The associative law of vector addition Is as follows: 24(b4 c)=(a4 b)+ c 2. Accarding to the triangle law of vector addition. AB + BC = AC => AB4EC=-CA fs AC =-CA) = ABLBCLCA =0 Hence proved. 3. Required sum = BA BC+ CO+DA =BA+BD+4DA (Be SRT UB =2BA 4. InAEBF, EF = EB + BF =p h8 +p Rai (A+ BC) (by triangle law] — (AB + Ab) = +@ + b) (- Inilgram, BC = ru) i “2 5. Po =Position vector of Q— Position vector of P A Ay A om he, ie he =(4|-5k)-(i + j)=-143]-5k 6. Resultant of these forces: a tae AA OR OKA =(214+3/+4k)+(-4])+i)+()-4k-31) A A A > =(2+1-3)1+(3-441)|+(4-4)k=0 Since, the resultant of these forces Is zero, so the particle Is in equilibrium. Hence proved, 7. Since, given vectors are collinear. 2 4 = a=-4 SeoR kK a=i+2|+3k laleq()?4(@2)? 4) =¥ie409 = V5 1 2 3 <. Directl I = ‘ F rection cosines Wye ye 9. Required unit vector = Z er I21-) +2k| TiP Read the question carefullyand practice more problems on unitvectors. @. Let AR eb Aa aA 2t-fa2k — 2t-fs2k Bryn “Vari s Rie att wth Oi oP 2i-j4+2k 27-j+2k AA 10. Defame ke Hence proved. Tl. IF vectors (ai +2] 43k) and (31 +bj) are perpendicular, then (ai 42] 43k)-(i4b])=0 = )(0) + (6) (2) =0 => 3a+2b=0 Hence proved. 12. Let a=2i-japk and beisj-k Let a and b be perpendicular to each other then a-b=0. le. (21-j4pk)-(4]-k)=-0 = @N)+-)+(@))-0 > 2-1-p=0 => p=! 13. Given,a-b=1Jal=land|b|=2 Let Abe the angle between aandb. cos = EE 8 a cas 2S e = S=cass jalio) S24 @== 3 14. Given that, the two vectors a and b having the same ss lal=lbl (y TiP uA @ is the angle between two nan-zero vectors a and b, then the scalar product is given by a a:b =|a||b|cos@where OS Osx Angle between a and bis 60° and their scalar product 9 Iss. le, 3 2: [all Bicos@ =2 2) » 2 (Here ‘@ Is the angle between a and b) = laliblcos6o° => (- @=605) 9 > = Gy =lb = 5 § lalslbQ = =9 => lal=4 Thus, 23 a a 15. Glven,a 3|- ot (-1-3)-] (-1-2) +h (3-2) =-4laaj+k ” > Therefore, a vector perpendicular to a and b Is -41 43] +k. -, Unit vector along the sum = oa0 la+b = Fg lti 3] -2h) a KR hy pA ie he OB Again a —b =(21+2|-5k)-(21+ | +3k)=|-Ok and la -bl=,i? + (8)? = VE5 *. Unit vector along the difference = ERROR Some students find any vector instead of unit vector and others find the unit vector in the same direction. > AK LA > AA A 7. Given, a=i-j+7k and b=5/-j+ak + a AA A ee, Now, (a+b)=(1-j|+7k)+(51-j+ak) SW a =61-2)4+(7+a)k and (3-b)=(i ~j=7k)-(61-j 42k) i =-474(7-a)k (a+ b) and (a-b) will be orthogonal if (a46).(a-b)=0 = (i-2]4(74+A)k)-(4 7+ (7-2)k}=0 => (6)(-4)+(7+4)(7-a)=0 > -24+49- =0 => Y=25 > etd 8. Given,a=5i-|-3k andb=143]-5k aot kA A ww OD Now, a+b=(51-j-3k)+(1+3]-Sk) =6142]-ak and a-b=(51-]-3k)-(i 3] -5k) Ah =(4/1-4) 42k) -.(a4b)-(a—b)=(6142 | -8k)-(41-4} 42k) = (6) (4) + (2)(-4) + (-8) (2) =24-B-16= 24-2450 Therefore, vectors (2 + b) and (a- b) are perpendicular. 9. TRICK > > ab The projection of a on b Is given by —- \b| Hence proved. >> 10. M. > AAA > AAA Leta =i+j+kandb=pi+j-2k 1b k) -. The projection of aonb= (p)? + (1)? +(-2)* 1 p+l-2 z => see = p*?+5=3(p-1) -() a p?+5 P ss Squaring both sides, we get p*+5=9(p*+1-2p) > 8p*-1ep+4=0 = 4p*-9p+2=0 = (4p-1)(p-2)=0 1 =2— => p A But p = ae does not satisfy the eq. (1). Hence, p =2. a Se Given, a=2i-2])+k,b=i+2]-2k > AAA and c=2i-j+4k a ee ee AAA Now, b+c=(i+2j-2k)+(2i-j+4k) A A LA =3i1+j+2k (2)? + (-2)? +(1)? (3)(2)+ ()(-2) + (2)() a Glven.a Is a unit vector. a lalel sae e5e3 lal BS hye eK we have;(x-a).(x4a)=12 => (x-a)-(x4+a)=12 > , oo oo» = X'X-ax+xX-a-a-aal2 = (xP-x-a+x-a-(a)? =12 (-a-beb-a) _ IxP-laPai2 (-a? @lP) ‘ > |x? (1)? =12 oi => [xP 212+1813 = Ixl=vi ok NK . Given, ae2i42[4+3k i & x * be-i+2]+k A o and co3l+{ a+aba(2i42}43k)+a(-t42, +k) =(2-a)i+(2+2a) [4 (B+a)k . a+abis perpendicular to c (a4ab)-c=0 = ((2-a) i +(2+2a)j +(+ak)-Bi+jJ=0 => 3(2-a)+(I)(24+2a)=0 > 6-30+2+21=0 -2+8=0 > A=8 TRICK (a, i+ by J+ k)-(0, i +b, j+e k) = (aya) +bybz +¢,C2) ee 13. Given,|a+bl=la—-bl Squaring bath sides. we get 2 Sa oe Sos la+b[? s|a-b/* or (a+b)? =(@-b)? (- square of vector = square of its modulus) or (a)? +(b)?4 2(a-b)=(a) +(b)? -2(a-b) or 4(a b)=0 or 3-b=0 Therefore, vectors a and bare perpendicular to each other. Hence proved. 14. Let a and b be unit vectors. Then, according to question, > > Here, Cis also a unit vector. > ee lal=lbl=|cl=1 (2) 2 ee From eq. (1), la+bl=Icl = la+ble-1 2 ig ww a We know that, la+bl?+/a-bP=2(lal? +b?) = 4la-bl? =2(1+1) a ja-b? 24-163 = |a-bl=.3 > A A A 15. Given,a=i-2)+k Let the angle formed by vector 3 from X-axis = 8 : Unit vector alang X-axls = | .. Angle formed by vector 3 from X-axis ae = Angle between the vectors a and | = flied ees Wreeare(y WaT => O=cos"! (=) which Is the required angle. 16. Let pools] +3kandg=3i-2]+k -- The angle between the vectors Is 60°. PF Lcosgoeet Ipliql ois j+ak) af 274k qo? +12 +3? ya? + (-27 +1? x3+41x(-2)+3x1 Jor +10 ia da+1 yo? +10 Via or (2(30+1)}? =(a7 +10)-14 (squaring on bath sides) 490? +60 +1) =1407 +140 2207 + 240-136 =0 Na? 4120-68 No? + 340-220 -68 =O or a(lla +34)-2(1la+34) =0 (lla + 34)(a-2) =0 ao=2 or g=-— an or or NI- NI- NIA= Youd J aT 7. (a- b) =a*b? (abcosé)? =0%b* where, lal=a and |bl=b ab? cos 0=07b* cos*0=1 = cosd=+l 0=0° or 180° Therefore, the given relation will be true when the angle between two vectors Is O° or 180° 2, when two vectors are parallel. => = = = 1B. ~ axb=absinén LHS =(a xb)? =(absinon)2 =a7b? sin* a(n-n) =07b? sin? 9-1 = a7b? (1-cos* 0) =a7b? —(obcos 6)? =07b? -(3.b)?=RHS (sin? 0+ cos? 0=1) Hence proved. > AA LA 2 A LA A 19. Glven,asi-j|+3kandb=2/-7|+k The area of parallelogram having adjacent sides a and bls given by A KOA sw |! | kK laxDl=|] 1 4 2-71 al 1(-14-21)-](1-6) + k(-7+2)I a A LA AKA 3120145 |-5kl=514 i+ j-k! = 5 (4)? + (1) + (A)? = Vibe Ts = 5Vi8 = 15,2 sq. units De xd-4b xd+4bxd-2exd (from eqs. (1) and (2)) Hence, (a —2 d) is parallel to (2 b-c) Hence proved. Short Answer Type-ll Questions 1. -Dand E are the mid-points of sides AB and AC respectively of AABC. oho -. DE and BC are parallel [Oey 1 B and DE =—B8C 2 In ABLE, and in ADCE, DC = DE + EC. BE + DC =(BC + CE) + (DE + EC) =BC+ De +(Ce + EC) =BC + DE+(-EC+ EC) [-Ce=-EC) =BC4DEe+ 0 =BC+DE= BC+ I BC (from eq. (1)) Hence proved. 2. The diagonal of a parallelogram blsect each other. *. Point of intersection G will be the mid-point of the dlaganals AC and BD both. Then, OA+0C=206 and 08+ Of Adding, we get OA +08 +0C+ 00-406 Hence proved. = =20G —> Ak a 3. Given, PQ =31-2|+2k s R Pp Q and PS s- ; -2 k Now, diagonal R= Po + OR (by triangle law) =PO+PS (“OR = PS) =(3l-2742k)4(-1-2k) « * =21-2] [PR I=)? + 27 lai 22 units + RS — = = PS -PO [. in Ilgm. opposite sides are equal] i-2k)-(31-2]+2k) A LA LALA e+ |~2k-3142 |-2k A nN Ow =-4142]-4k 105 = (4) + (22+ (4 = {16+ 4416 = 36 =6 units 4, Given two adjacent sides of parallelogram ABCD are a=t+2],43k and b= 214475. TiP Practice problem based on parallel and perpendicular’ vectors. and diagonal Os 0 a c e ry A oj B a Here, diagonal of parallelogram are AC and BD. AC a+b (by triangle law) and -3 (by triangle law] a A A oA Now, a(t427 43k)4 (27447 -5k) Kk ck =314+6)-2k ae ee ee and BD =(21+4j -5k)-(1 +2) +3k) A A UA =i+2]-8k * Unit vector arallel te the dlaganal rd Ke - 3h+6]-2k “Vacl Warr @F +2) A A A A 3las;-2k 3ts6)-2k “pr3e+4 7 — and unit vector parallel to the diagonal BD= BD 1501 fazt-ak — 742}-8f (Pee P ear Weare U(i+27-a) "oes +2] - ERR@R Instead of finding the parallel vectors, some students take the cross product to find the perpendicular vector. De 5. Given two adjacent sides of parallelogram ASCD are 3=21-4) 5k and be2i+2)+3k 4 a Cc = > b b Al 8 3 a Here, dlagonal of parallelogram are “AC and BD AC=2+b (by triangle law) and BOeb-a (by triangle law) Now, ‘AC =(27-4] -5k)+(21 +2] +3k) A A LA =4)-2]-2k and BD =(21+2[+3k)- (27-4) -5k) =6]+8k . Unit vector parallel to diagonal oe 1ACI qi-2[-2e —_ Wi=2fuvk 21 =[-k “Terre eap Tera Fe BO BO _ Sivek sirak aj+ak “Ter+er YBe+64——~COS Now, area of parallelogram = 3 ‘AC x BDI AA A on - | -k) «2(3] + 4k)I A A A -1 -1)=2|-i-8)+6k1 =24(-1)7 + (8) + (6) =2,/1+ 64436 =2 VIO) sq. units [COMMON] ERR(R Some students get confused between the formula for areas of triangle and parallelogram. => A LA Amo nA 6. Here, OA =-214+3)4+5k. 0B a/+2|+3k — An and OC =71-k =— = = Now, AB = 08 -OA A LA LA aA a A =(1+2])+3k)-(-21 43] 45k) AA LA =3/1-]-2k =— AC =O0C-OA an a na =(71-k)-(-214+3]+5k) a ar. =91-3)-6k and ‘BC = OC OB =(71-k)-(i +2] +3k) =61-2]-4k = AB =| AB |= (3)? 4 (-1)2 + (-2)° > AC =| BC |= (6)? + (2) +4? = 36+ 4416 = V56 -2/14 and AC =| AC = of(9)? + (-3)? +(-6) = 8149436 = VI26 =3V14 Clearly, AB+8C = AC Hence A Band Care collinear Hence proved. . Given that,a= i+] +k, b=2144]-5k and Cant +2] 4+3k Now, bac=(2'+4]-Sk)+(,742] 43k) A Lh A =(A+2)i +6] -2k and [b+ cl=y(a+2)? + (6)? +(-2) =024+444145644 =f 440444 (b+c) i= Ib+cl _(a42)i46)-2K Yea ana 44 According to question, bic le #e iy (t ial: nace: . Unit vector along (b 1 c)= arrerras => (1)(A+2)+ (1)(6)+(I)(-2) = i444 44 => ae ah (Squaring on both sides) = 24364120527 + 424 44 = Bi=8 > rel sa LA 142) +6] -2k Thus, unit vector along (6 + c)= lt)? +4 «14.44 . Given that at band c are three unit vectors. 2 3 @ So, lal=|bl=lcl=1 > > > = la = [b= cl? =1 a > => a.aeb.beccel a and given,a+2b+3c=0 Taking the scalar product with a.Bande respectively. De
