Title 2. P1C2. Vector__With Solve.pdf ✅

†f±i  Mastery Practice Sheet 1 wØZxq Aa ̈vq †f±i Vector weMZ eQ‡i BwÄwbqvwis fvwm©wU‡Z Avmv wjwLZ cÖkœmg~n 1| C B A wÎfz‡Ri †KvY ̧wj KZ? [BUET 22-23] A  = 2i  + j  + 3k  , B  = 3i  – 3j  – k  mgvavb: C  = B  – A  = i  – 4j  – 4k  A B = cos–1      A   . B  AB = cos–1 0 = 90 B C = cos–1      B   . C  BC = cos–1     3 + 12 + 4 19 × 33 = 40.64 C A = 90 – 40.64 = 49.35 2| B C 37 36 m 37 vb = 10 ms–1 vr = 3 ms–1 A K. BC Gi •`N© ̈ wbY©q Ki| (L) mivmwi B †Z †h‡Z KZ †Kv‡Y iIbv w`‡Z n‡e? [BUET 21-22] mgvavb: (K) BC = (vr + vb cos ) t = (vr + vb cos ). AB vb sin  = (3 + 10 cos 37). 36 10 sin 37 = 65.72 m (Ans) (L) mivmwi B †Z †h‡Z, vr + vb cos  = 0   = cos–1     – 3 10 = 107.46 (Ans.) 3| i  + j  †f±‡ii w`‡K A  = 2i  + 3j  †f±‡ii Dcvsk wbY©q Ki| [BUET 19-20] mgvavb: awi, B  = i  + j   B  Gi w`‡K A  Gi Dcvsk = A  . B  |B|  . B  |B|  = A  . B  |B|  2 B  = 2 + 3 1 2 + 12 . (i )  + j  = 5 2 (i )  + j  (Ans.) 4| †Kvb GKw`b 30 ms–1 MwZ‡Z Dj¤^fv‡e e„wó cowQj| hw` evqy 10 ms–1 MwZ‡Z DËi †_‡K `wÿ‡Y eB‡Z ïiæ K‡i Zvn‡j e„wó †_‡K iÿv †c‡Z †Zvgvi QvZv †Kvb w`‡K †g‡j ai‡Z n‡e †ei Ki| [BUET 06-07] mgvavb: wPÎ n‡Z,  = tan–1     10 30 = 18.43 (c~e© w`‡Ki mv‡_ `wÿY eivei)  10 ms–1 30 ms–1 `: D: c~: 5| cÖwZ NÈvq 1800 m †e‡M 240 m cÖk ̄Í GKwU b`x wb‡Pi w`‡K cÖevwnZ n‡”Q Ges cÖwZ NÈvq 3600 m †e‡M muvZv‡i mÿg GKRb muvZviæ GKwU wecixZ we›`y‡Z †h‡Z B”QzK| †m †Kvb w`K eivei muvZvi †`‡e Ges †mB we›`y‡Z †h‡Z KZ mgq †b‡e? [BUET 03-04] mgvavb: †b.Kvi †eM, v = 3.6 km/h † ̄av‡Zi †eM, u = 1.8 km/h  wecixZ cÖv‡šÍ †h‡Z n‡j † ̄av‡Zi mv‡_ †b.Kv Pvjbv Ki‡Z n‡e,  = cos–1     – u v   = cos–1     – 1.8 3.6 = 120 (Ans.) cÖ‡qvRbxq mgq, t = d vsin = 0.240 3.6 sin120  t = 0.077 hr (Ans.) 6| ci ̄ú‡ii mv‡_ j¤^fv‡e wμqvkxj `yBwU e‡ji jwä 80 N| hw` jwä GKwU e‡ji m‡1⁄2 60 †Kv‡Y AvbZ _v‡K, Z‡e ej `yBwUi gvb wbY©q Ki| [RUET 17-18] mgvavb: P = Rcos60  P = 80 × 1 2 = 40 N Q = Rsin60  Q = 80 × 3 2 = 40 3 N (Ans.) R = 80N 60 Q P 7| p Gi gvb KZ n‡j †f±i v  = (5x + 2y)i  + (2py – z)j  + (x – 2z)k  mwjbqWvj n‡e? [RUET 15-16] mgvavb: v  = (5x + 2y)i  + (2py – z)j  + (x – 2z)k 
2  Physics 1st Paper Chapter-2   .v  = (5x + 2y) x + (2py – z) y + (x – 2z) z = 0  5 + 2p – 2 = 0  p = – 3 2 (Ans.) 8| †Kvb b`x‡Z GKwU †b.Kvi †eM † ̄av‡Zi AbyK~‡j I cÖwZK~‡j h_vμ‡g 18 Ges 6 kmh–1 | †b.KvwU KZ †e‡M †Kvb w`‡K Pvjbv Ki‡j †mvRv Aci cv‡o †cu.Qv‡e? [RUET 10-11] mgvavb: †b.Kvi †eM, u Ges † ̄av‡Zi †eM v  u + v = 18 u – v = 6  u = 12 km/h Ges v = 6 km/h v  u GLb,  = cos–1     – u v = cos–1     – 12 6   = 120 (Ans.) 9| 10 wK‡jvwgUvi/NÈv †e‡M e„wó co‡Q Ges 60 wK‡jvwgUvi/NÈv †e‡M c~e© n‡Z cwð‡g evZvm eB‡Q| c~e© n‡Z cwðg AwfgyLx PjšÍ Mvwoi MwZ‡eM wbY©q Ki hv‡ZÑ (a) Mvoxi mvg‡bi I wcQ‡bi KuvP wf‡R, (b) ïaygvÎ wcQ‡bi KuvP wf‡R| [RUET 04-05] mgvavb: – vc  va  = 60 km/h vr  = 10 km/h c~: c: (a) mvg‡bi I wcQ‡bi Dfq KuvP wfR‡Z n‡j Mvwoi mv‡c‡ÿ e„wói †eM n‡e Lvov wb¤œgyLx|  –vc  = va   vc = 60 km/h (Ans.) (b) – vc  va  vr  vrc  ïaygvÎ wcQ‡bi KuvP wfR‡j, wPÎvbymv‡i jwä n‡e vrc  A_©vr G‡ÿ‡Î, va > vc  vc < 60 km/h n‡e| (Ans.) 10| (a) †Kvb we›`y P Gi ̄’vbv1⁄4 P(2, –3, 4) n‡j we›`ywUi Ae ̄’vb †f±i wbY©q Ki| (b) A(2, –1, 3) Ges B(–1, 2, –3) we›`y؇qi ms‡hvMKvix w`K ivwkwU wbY©q Ki| [CUET 05-06] mgvavb: (a) OP  = 2i  – 3j  + 4k  (Ans.) (b) OA  = 2i  – j  + 3k  , OB  = – i  + 2j  – 3k   AB  = OB  – OA  = –3i  + 3j  – 6k  (Ans.) 11| GKwU BwÄb PvwjZ †b.Kvi †eM NÈvq 14 wK‡jvwgUvi| GKwU b`x AvovAvwo cvi n‡Z n‡j †b.KvwU‡K †Kvb w`‡K Pvjv‡Z n‡e? b`xi cÖ ̄’ 12.125 km n‡j Zv cvwo w`‡Z KZ mgq jvM‡e? † ̄av‡Zi †eM NÈvq 7 km| [CUET 04-05] mgvavb: AvovAvwo cvi n‡Z n‡j †b.Kv‡K † ̄av‡Zi mv‡_,  = cos–1     – u v = cos–1     – 7 14 = cos–1     – 1 2   = 120 †Kv‡Y Pvjbv Ki‡Z n‡e| (Ans.) Avevi, t = d vsin = 12.125 14 sin(120)  t = 1 hr (Ans.) 12| GKwU b`xi † ̄av‡Zi †eM 5 ms–1 | 10 ms–1 †e‡Mi GKwU †b.Kvi †mvRvmywRfv‡e b`x cvwo w`‡Z 1 min 40 second mgq jv‡M| b`xi cÖ ̄’ KZ? [CUET 03-04] mgvavb: †mvRvmywRfv‡e cvwo †`qvi †ÿ‡Î, t = d v 2 – u 2  d = 100 × 102 – 5 2 = 866.025 m (Ans.) u v 2 – u v 2 d 13| GKwU Mvwo 20 kmh–1 †e‡M c~e©w`‡K Pjgvb| evZvmI 4 kmh–1 †e‡M GKB w`‡K Pjgvb| G mgq e„wó Lvov wb‡Pi w`‡K 6 kmh–1 †e‡M co‡Z ïiæ K‡i| e„wó Mvwo‡Z Djø‡¤^i mv‡_ KZ †Kv‡Y AvNvZ Ki‡e? [BUTex 23-24] mgvavb: vw vc vr DËi (j) `wÿY (– j) cwðg (– i) c~e© (i) e„wói †eM, vr = – 6j  km/h; evZv‡mi †eM, vw = 4i  km/h Mvwoi †eM, vc = 20i  km/h evZvm I e„wói jwä †eM, vrw  = 4i  – 6j  km/h Mvwoi mv‡c‡ÿ, vrc  = vrw  – vc  = 4i  – 6j  – 20i  = – 16i  – 6j  Dj‡¤^i mv‡_ †KvY,  = tan–1     16 6 = 69.44 (Ans.) 14| hw` r  = bt2 i  + ct3 j  nq, †hLv‡b b I c abvZ¥K aaæeK| KLb †eM †f±i x I y A‡ÿi mv‡_ 45 †Kv‡Y _vK‡e? [BUTex 22-23]
†f±i  Mastery Practice Sheet 3 mgvavb: Y X 45 v †eM, v  = d r  dt = 2bti  + 3ct2 j  kZ©g‡Z, vx = vcos45  2bt = v 2 ..... (i) vy = vsin45  3ct2 = v 2  3ct2 = 2bt  t = 2b 3c t = 2b 3c n‡j †eM †f±i x I y A‡ÿi mv‡_ 45 †Kv‡Y _vK‡e| (Ans.) 15| GKwU Mvwo 40 kmh–1 †e‡M DËi w`‡K PjwQj| Gici MwZ cwieZ©b K‡i DËi w`‡Ki mv‡_ 60 †Kv‡Y 42 kmh–1 †e‡M Pj‡Z _vK‡jv| G‡ÿ‡Î †e‡Mi cwieZ©b KZ? [BUTex 21-22] mgvavb: 60 vi  vf  Avw` †eM, vi  = 40 j  km/h †kl †eM, vf  = 42 cos60 j  + 42 sin60 i   vf  = 21 3 i  + 21j   †e‡Mi cwieZ©b, vf  – vi  = 21 3 i  + (21 – 40)j  = 21 3 i  – 19j  km/h  |v | f  – vi  = (21 3) 2 + 192 = 41.036 km/h   = tan–1    21 3 19 = 62.42 (`wÿY w`‡Ki mv‡_ c~e© eivei) (Ans.)  21 3 19 16| Ggb GKwU GKK †f±i wbY©q Ki, hv xy Z‡ji mgvšÍivj Ges i  – j  + k  †f±‡ii mg‡Kv‡Y Aew ̄’Z| [BUTex 20-21] mgvavb: xy Z‡ji mgvšÍivj GKwU †f±i, A  = ai  + bj  cÖkœg‡Z, (ai )  + bj  . (i )  – j  + k  = 0  a – b = 0  a = b  wb‡Y©q GKK †f±i, A  |A |  =  ai  + aj  a 2 + a2 =  1 2 (i )  + j  17| A  . A  Ges A  × A  Gi gvb wbY©q Ki| [BUTex 18-19] mgvavb: A  . A  = A.A.cos(0) = A2 |A |  × A  = A.A.sin(0) = 0 18| A  = 2 i  + j  – k  , B  = 3 i  – 2j  + 4k  Ges C  = i  – 3j  + 5k  wZbwU †f±i| †`LvI †h, †f±i wZbwU GKB mgZ‡j Aew ̄’Z| [BUTex 04-05] mgvavb: A  , B  , C  GKB mgZ‡j Aew ̄’Z n‡j, A  .(B )  × C  = 0  A  .(B )  × C  =       2 3 1 1 –2 –3 –1 4 5 = 2(–10 + 12) + 1(4 – 15) – 1(–9 + 2) = 4 – 11 + 7 = 0  A  .(B )  × C  = 0 19| mggv‡bi `ywU †f±i a  Ges b  ci ̄úi mgvšÍivj n‡j, †f±i `ywU KZ n‡e? [BUTex 04-05] mgvavb: awi, a  = ax i  + ay j  + azk  ,  b  = pa  = pax i  + pay j  + pazk  [a  Ges b  mgvšÍivj] cÖkœg‡Z, | a|  = |b|   | a|  2 = |b|  2  ax 2 + ay 2 + az 2 = p2 (ax ) 2 + ay 2 + az 2  p =  1  a  = b  A_ev, a  = –b  (Ans.) 20| A  I B  †f±i `ywU Ggb †h, |A |  + B  = |A |  – B  | †f±i؇qi ga ̈Kvi †KvY wbY©q Ki| [BUTex 02-03] mgvavb: |A |  + B  = |A |  – B   A 2 + B2 + 2ABcos = A 2 + B2 – 2ABcos  4ABcos = 0   =  2 [∵ A  0 Ges B  0] (Ans.)
4  Physics 1st Paper Chapter-2 weMZ eQ‡i BwÄwbqvwis fvwm©wU‡Z Avmv eûwbe©vPbx cÖkœmg~n 1. x-Aÿ eivei †_‡K 30 GKwU †f±i | | A = 2; y Aÿ eivei n‡Z | |  B = 3 Z‡e | | A –  B = ? [BUET Preli. 22-23] 2.628 2.646 3.6148 1.5168 DËi: 2.646 e ̈vL ̈v: 30 – B A  B | |  A –  B = A 2 + B2 – 2AB cos = 2 2 + 32 – 2  2  3  cos 60 = 2.646 2. 4 kmh–1 †e‡M cÖevngvb † ̄avZhy3 b`x‡Z KZ †Kv‡Y †b.Kv Pvjv‡j 5 kmh–1 †e‡M †b.Kv Pvjv‡Z mÿg GKRb gvwS wVK Aci cv‡o †cu.Qv‡Z cvi‡e? [BUET Preli 22-23] 120 90 136.37 143.1301 DËi: 143.1301 e ̈vL ̈v:  = cos–1     – v u = cos–1     – 4 5   = 143.130 3. 5 ms –1 †e‡M cÖevngvb † ̄avZ wewkó GKwU b`x‡Z 10 ms–1 †e‡M Pjgvb †b.Kv 1 hour hvÎv K‡i †mvRv Aci cv‡o wM‡q †cu.Qvq| Zvn‡j b`xi cÖ ̄’ KZ n‡e? [BUET Preli 22-23] 30 km 31 km 32 km 29 km DËi: 31 km e ̈vL ̈v: t = d u 2 – v 2  d = 3600 102 – 5 2  31 km 35. GKwU †f±i A  , 30 †Kv‡Y abvZ¥K x-A‡ÿi w`‡K wμqviZ| Aci GKwU †f±i B  FYvZ¥K y-A‡ÿi w`‡K wμqviZ| |A| = 4, |B| = 8 n‡j, jwä †f±i †KvbwU? [BUET Preli 21-22] 2 3i  + 6j  7i  – 3j  2 3i  – 6j  None DËi: 2 3i  – 6j  e ̈vL ̈v: 30 B  X A  A  + B  = (4 cos30)i  + (4 sin30 – 8)j   A  + B  = 2 3i  – 6j  36. a  I b  `ywU mgvb †f±i mg‡Kv‡Y wμqv Ki‡j (a )  + b  I (a )  – b  Gi WU ̧Ydj KZ? [BUET Preli 21-22] 0 4a 2a None DËi: 0 e ̈vL ̈v: b  a  – b  a 45  45 b  a  + b  a  + b  , a  – b  Gi ga ̈eZ©x †KvY 90|  WU ̧Ydj = 0| 37. `ywU mgvb †f±‡ii jwä bvj †f±i n‡j †f±i؇qi ga ̈eZ©x †KvY KZ? [BUET Preli 21-22] 90 180 0 270 DËi: 40 38. xy mgZ‡j GKwU †f±i x-A‡ÿi mv‡_ 130 †KvY •Zwi K‡i| †f±iwU FYvZ¥K y-A‡ÿi mv‡_ KZ †KvY •Zwi Ki‡e? [BUET Preli 21-22] – 130 40 140 – 40 DËi: 140 e ̈vL ̈v: 50 130   = 50 + 90 = 140 4. GKwU KYvi Dci F  = (5i )  – 6j  + 3k  N ej cÖ‡qvM Kivi d‡j KYvwUi d  = (3i )  + dyj  + 5k  m miY nq| dy Gi gvb KZ n‡j m¤úvw`Z Kv‡Ri cwigvY k~b ̈ n‡e? [BUET 12-13] 0 5