Tiêu đề Vector Engineering Practice Sheet Solution (HSC 26).pdf ✅

†f±i  Engineering Practice Sheet Solution (HSC 26) 1 02 †f±i Vector WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1|  Gi †Kvb gv‡bi Rb ̈ wb‡¤œi †f±i wZbwU mgZjxq n‡e? cÖ`Ë †f±i·qi mgZ‡ji Ici j¤^ GKK †f±i wbY©q Ki| a  = i ^ + j ^ + k ^ , b  = 2i ^ – 4 k ^ , Ges c  = i ^ + j ^ + 3k ^ . [BUET 20-21] mgvavb: a  = i ^ + j ^ + k ^ , b  = 2i ^ – 4k ^ , c  = i ^ + j ^ + 3k ^ , †f±i wZbwU mgZjxq n‡e hw`,      1 2 1 1 0  1 – 4 3 = 0  1(0 + 4) – 1(6 + 4) + 1(2 – 0) = 0  4 – 10 + 2 = 0   = 5 3 (Ans.) cÖ`Ë †f±i·qi mgZ‡ji Dci j¤^ GKK †f±i = a  I b  Gi mgZ‡ji Dci j¤^ GKK †f±i [⸪a  , b  , c  mgZjxq]  j¤^ GKK †f±i,  ^ = a   b  |a |   b   a   b  =      i  ^ 1 2 j ^ 1 0 k ^ 1 – 4 = i ^ (– 4 – 0) – j ^ (– 4 – 2) + k ^ (0 – 2) = – 4i ^ + 6j ^ – 2k ^ Ges |a |   b  = (– 4) 2 + (6) 2 + (– 2) 2 = 2 14   ^ =  – 4i ^ + 6j ^ – 2k ^ 2 14 =  – 2(2i ^ – 3j ^ + k ^ ) 2 14 =  2i ^ – 3j ^ + k ^ 14 (Ans.) 2| a  = 2i ^ + j ^ – 3k ^ Ges b  = i ^ – 2j ^ + k ^ †f±i `ywUi Dci j¤^ GKwU †f±i wbY©q Ki hvi gvb 5 GKK| [BUET 19-20] mgvavb: a –  b – =      i  ^ 2 1 j ^ 1 – 2 k ^ – 3 1 = i ^ (1 – 6) – j ^ (2 + 3) + k ^ (– 4 – 1) = – 5i ^ – 5j ^ – 5k ^   ^ = a –  b – | | a –  b – =  – 5i ^ – 5j ^ – 5k ^ (– 5) 2 + (– 5) 2 + (– 5) 2 =  5i ^ + 5j ^ + 5k ^ 5 3 =  i ^ + j ^ + k ^ 3  wb‡Y©q †f±i = 5 ^ =  5 3 (i ) ^ + j ^ + k ^ (Ans.) 3| GKwU GKK †f±i wbY©q Ki hv a  = i ^ + j ^ + k ^ Ges b  = i ^ – j ^ – k ^ †f±i؇qi mgZjxq Ges a  †f±‡ii Dci j¤^| [BUET 18-19] mgvavb: wb‡Y©q †f±i, r  = a  + mb  = (i ^ + j ^ + k ^ ) + m (i ^ – j ^ – k ^ ) = (1 + m)i ^ + (1– m)j ^ + (1– m)k ^ Avevi, r  .a  = 0  1 + m + 1 – m + 1 – m = 0  3 – m = 0  m = 3  r  = 4i ^ – 2j ^ – 2k ^  wb‡Y©q GKK †f±i, r ^ =  r  | r|  = 4i ^ – 2j ^ – 2k ^ (4) 2 + (– 2) 2 + (– 2) 2 = 4i ^ – 2j ^ – 2k ^ 2 6 = 2i ^ – j ^ – k ^ 6 (Ans.)

†f±i  Engineering Practice Sheet Solution (HSC 26) 3 AB  I AC  †K mwbœwnZ evû a‡i Aw1⁄4Z mvgvšÍwi‡Ki †ÿÎdj, = |AB |   AC  = (5) 2 + (– 11) 2 + (– 18) 2 = 470 (Ans.) 10| P I Q we›`yi ̄’vbv1⁄4 h_vμ‡g (1, 1, 1) Ges (3, 2, – 1) n‡j, PQ  †f±i wbY©q Ki| [BUET 03-04] mgvavb: OP  = i ^ + j ^ + k ^ OQ  = 3i ^ + 2j ^ – k ^ PQ  = OQ  – OP  = (3 – 1)i ^ + (2 – 1)j ^ +(– 1 – 1)k ^ = 2i ^ + j ^ – 2k ^ (Ans.) 11| a – = i ^ + 2j ^ – k ^ Ges b – = j ^ – i ^ – 2k ^ †f±i؇qi ga ̈Kvi †KvY wbY©q Ki| [BUET 02-03] mgvavb: cos = a – .b – |a – |.|b – | = – 1 + 2 + 2 1 2 + 22 + (– 1) 2 . (– 1) 2 + (1) 2 + (– 2) 2 = 3 6. 6 = 3 6 = 1 2  cos = 1 2 = cos60   = 60 (Ans.) 12| hw` (ai ^ + bj ^ + k ^ )  (2i ^ + 2j ^  3k ^ ) = i ^ – j ^ nq, Z‡e a Ges b Gi gvb wbY©q Ki| [BUET 01-02] mgvavb:      i  ^ a 2 a ^ b 2 k ^ 1 3 = i ^ – j ^ = i ^ (3b – 2) – j ^ (3a – 2) + k ^ (2a – 2b) = i ^ – j ^ Dfq cÿ n‡Z i ^ I j ^ Gi mnM mgxK...Z K‡i cvB, 3b – 2 = 1 Ges 3a – 2 = 1  b = 1  a = 1  a = 1, b = 1 (Ans.) weMZ mv‡j KUET-G Avmv cÖkœvejx 13| 2i ^ – j ^ + 2k ^ †f±iwU x A‡ÿi mv‡_ †h †KvY Drcbœ K‡i Zv wbY©q Ki| [KUET 04-05] mgvavb: a  = 2i ^ – j ^ + 2k ^ ; |a  | = 3 x A‡ÿi eivei †f±i b  = i ^ ; |b  | = 1 a  . b  = ab cos  cos = a  . b  |a  | |b  |  cos = (2i ^ – j ^ + 2k ^ ) . i ^ 3   = cos–1     2 3 (Ans.) 14| hw` a  = i ^ + j ^ + k ^ , b  = 3i ^ + 3j ^ – 2k ^ nq Zvn‡j b  †f±‡ii Dci a  †f±‡ii Awf‡ÿc †ei Ki| [KUET 03-04] mgvavb: a  = i ^ + j ^ + k ^ , b  = 3i ^ + 3j ^ – 2k ^ a  . b  = ab cos  acos = a  . b  b = (i ^ + j ^ + k ^ ) ( 3 i ^ + 3j ^ – 2k ^ ) ( 3) 2 + 32 + 22 = 3 + 3 – 2 16 = 3 + 1 4 (Ans.)  a – b – |a – | cos weMZ mv‡j RUET-G Avmv cÖkœvejx 15| abvZ¥K x A‡ÿi m‡1⁄2 †f±i A  = – 3i + j †h †KvY Drcbœ K‡i Zv wbY©q Ki| [RUET 18-19] mgvavb:  = cos–1       Ax A 2 x + A 2 y = cos–1      – 3  3 + 1   = 150 Ae ̄’vb 2q PZzf©v‡M 1 150 3
4  Higher Math 1st Paper Chapter-2 16| AB  = 3i ^ + 2j ^ – k ^ Ges AC  = 5i ^ – j ^ + 2k ^ n‡j, AB I AC †K mwbœwnZ evû a‡i AswKZ mvgvšÍwi‡Ki †ÿÎdj wbY©q Ki| [RUET 13-14; BUET 09-10, 04-05] mgvavb: AB   AC  =      i  ^ 3 5 j ^ 2 – 1 k ^ – 1 2 = 3i ^ – 11j ^ – 13k ^  AB  I AC  †K mwbœwnZ evû a‡i AswKZ mvgvšÍwi‡Ki †ÿÎdj, |AB   AC  | = 3 2 + (11) 2 + (– 13) 2 = 299 (Ans.) weMZ mv‡j CUET-G Avmv cÖkœvejx 17| aaæeK a Gi gvb wbY©q Ki †hb 2i ^ + j ^ – k ^ , 3i ^ – 2j ^ + 4k ^ , i ^ – 3j ^ + ak ^ †f±i wZbwU GKB mgZ‡j _v‡K| [CUET 07-08] mgvavb:       2 3 1 1 –2 –3 –1 4 a = 0  – 4a + 24 – 3a + 4 + 7 = 0  a = 5 (Ans.) 18| †f±i c×wZ‡Z GKwU wÎfz‡Ri †ÿÎdj wbY©q Ki, hvi kxl©we›`yÎq h_vμ‡g A(1, 3, 2), B(2, – 1, 1) Ges C(– 1, 2, 3). [CUET 04-05] mgvavb: OA  = i ^ + 3j ^ + 2k ^ , OB  = 2i ^ – j ^ + k ^ , OC  = – i ^ + 2j ^ + 3k ^ GLb, AB  = OB  – OA  = i ^ – 4j ^ – k ^ , BC  = OC  – OB  = – 3i ^ + 3j ^ + 2k ^  ABC = 1 2 |AB   BC  | AB   BC  =      i  ^ 1 –3 j ^ –4 3 k ^ –1 2 = – 5i ^ + j ^ – 9k ^  ABC = 1 2 25 + 1 + 81 = 1 2 107 (Ans.) weMZ mv‡j BUTex-G Avmv cÖkœvejx 19| †`LvI †h, A  = 8i ^ + j ^ – 6k ^ Ges B  = 4i ^ – 2j ^ + 5k ^ †f±i `yBwU ci ̄úi j¤^| [BUTex 07-08] mgvavb: A  . B  = (8i ^ + j ^ – 6k ^ ).(4i ^ – 2j ^ + 5k ^ ) = 8  4 + 1  (– 2) + (– 6)  5 = 0  A  Ges B  †f±iØq ci ̄úi j¤^| (Showed) 20| A I B we›`yi Ae ̄’vb †f±i h_vμ‡g, 2i ^ + 3j ^ – 4k ^ I 4i ^ – 3j ^ + 2k ^ | AB  Gi gvb Ges AB  eivei GKK †f±i wbY©q Ki| [BUTex 06-07] mgvavb: AB  = OB  – OA  = (4i ^ – 3j ^ + 2k ^ ) – (2i ^ + 3j ^ – 4k ^ ) = 2i ^ – 6j ^ + 6k ^  |AB  | = 4 + 36 + 36 = 76 (Ans.) awi, AB  eivei GKK †f±i = a ^  a ^ = AB  |AB  | = 2i ^ – 6j ^ + 6k ^ 76 (Ans.) 21| †`LvI †h, r  = i ^ + j ^ + k ^ †f±iwU Aÿ·qi mv‡_ mgvb †Kv‡Y AvbZ| [BUTex 05-06] mgvavb: awi, x Aÿ eivei GKK †f±i, a  = i ^ GLb, |  a| = 1  r  = i ^ + j ^ + k ^ Ges |r  | = 1 2 + 12 + 12 = 3 Avevi, cos = r  .a  |r  |  |a  |  cos = 1  1 3  1 = 1 3   = cos–1     1 3 Abyiƒcfv‡e, y I z A‡ÿi †ÿ‡Î cÖgvY Kiv hvq,  = cos–1 1 3 (Showed) 22| A  = 3i ^ – 2j ^ + k ^ , B  = i ^ – 3j ^ + 5k ^ I C  = 2i ^ + j ^ – 4k ^ †f±img~n Øviv GKwU mg‡KvYx wÎfzR ˆZwi Kiv wK m¤¢e? [BUTex 03-04] mgvavb: GLv‡b, B  + C  = i ^ – 3j ^ + 5k ^ + 2i ^ + j ^ – 4k ^ = 3i ^ – 2j ^ + k ^ = A   A  , B  Ges C  wÎfzR MVb K‡i| GLb, |A  | 2 = 9 + 4 + 1 = 14 Avevi, |B  | 2 = 1 + 9 + 25 = 35 Avevi, |C  | 2 = 4 + 1 + 16 = 21  |B  | 2 = |A  | 2 + |C  | 2 = 35 Avevi, A  . C  = 6 – 2 – 4 = 0  †h‡nZz †f±i ؇qi DU ̧Ydj k~b ̈ Ges B 2 = A2 + C2 ; mg‡KvYx wÎfzR MVb Kiv m¤¢e| (Ans.)