Tiêu đề 2. P1C2. Vector (ভেক্টর)_With Solve_Ridoy 11.09.25.pdf ✅

†f±i  Engineering Practice Sheet ................................................................................................................................. 1 WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. (K) BC Gi ˆ`N© ̈ wbY©q Ki| (L) mivmwi B †Z †h‡Z KZ †Kv‡Y iIbv w`‡Z n‡e? [BUET 21-22] B C 37 36 m 37 vb = 10 ms–1 vr = 3 ms–1 A mgvavb: (K) BC = (vr + vb cos) t = (vr + vb cos) AB vb sin = (3 + 10 cos37) 36 10 sin37 = 65.72 m (Ans) (L) mivmwi B †Z †h‡Z, vr + vb cos  = 0   = cos–1     – 3 10   = 107.46 (Ans.) 2. 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.) 3. †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) (Ans.)  10 ms–1 30 ms–1 `: D: 4. 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: †bŠKvi †eM, v = 3.6 km/h † ̄av‡Zi †eM, u = 1.8 km/h  wecixZ cÖv‡šÍ †h‡Z n‡j † ̄av‡Zi mv‡_ †bŠKv Pvjbv Ki‡Z n‡e,  = cos–1     – u v   = cos–1     – 1.8 3.6 = 120 (Ans.) cÖ‡qvRbxq mgq, t = d v sin = 0.240 3.6 sin120  t = 0.077 hr (Ans.) 5. (a) †Kvb †Kvb kZ©vax‡b A  I B  †f±iØq ci ̄úi j¤^ Ges mgvšÍivj nq? (b) B  = 5 i  + 6 j  + 9k  Gi w`‡K A  = 10 i  + 8 j  – 8k  Gi j¤^ Awf‡ÿc †ei Ki| [BUET 00-01] mgvavb: (a) A  I B  ci ̄úi j¤^ n‡j Zv‡`i WU ̧Y Gi gvb k~b ̈ n‡e| A_©vr, A  .B  = 0 [A  0, B  0] Avevi, A  I B  ci ̄úi mgvšÍivj n‡j Zv‡`i μm ̧Y Gi gvb k~b ̈ n‡e| A_©vr, A   B  = 0 [A  0, B  0] (b) B  Gi w`‡K A  Gi j¤^ Awf‡ÿc, Acos = A  . B  |B|  = 50 + 48 – 72 5 2 + 62 + 92 = 13 142 71 GKK| (Ans.)