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

2  Higher Math 1st Paper Chapter-2 4| †f±i c×wZ‡Z 3 i ^  j ^  k ^ Ges 2i ^  2j ^ – k ^ Gi †Q`we›`yMvgx mij‡iLvi mgxKiY wbY©q Ki| [BUET 17-18] mgvavb: 3i ^ + j ^ + k ^ Ges 2i ^ + 2j ^ – k ^ Gi †Q`we›`yMvgx mij‡iLvi †f±i mgxKiY: r  = 3i ^ + j ^ + k ^ + (2i ) ^ + 2j ^ – k ^ – 3i ^ – j ^ – k ^  xi ^ + yj ^ + zk ^ = (3 – )i ^ + (1 + )j ^ + (1 – 2)k ^  x = 3 –   x – 3 = –   x – 3 – 1 =  Ges y = 1 +    = y – 1 1 Ges z = 1 – 2   = z – 1 – 2  wb‡Y©q mij‡iLvi mgxKiY: x – 3 – 1 = y – 1 1 = z – 1 – 2 =  (Ans.) 5| hw` a   b   c   0 Ges |a  |  3, |b  |  5, |c  |  7 nq, Zvn‡j a  Ges b  Gi ga ̈eZ©x †KvY wbY©q Ki| [BUET 17-18] mgvavb: awi, a  Ges b  Gi ga ̈eZ©x †KvY | †`Iqv Av‡Q, a  + b  + c  = 0  a  + b  = – c   c  = – (a )  + b   c  .c  = { – (a )}  + b  { – (a )}  + b   |c|  2 = |a|  2 + |b|  2 + 2a  .b   2|a|  . |b|  cos = |c|  2 – |a|  2 – |b|  2  cos = 7 2 – 3 2 – 5 2 2  3  5 [⸪ | a| ]  = 3 |b|  = 5 |c|  = 7  cos = 1 2 = cos60   = 60 (Ans.) 6| ABC wÎfz‡Ri kxl©Îq A(1, 2, 3), B (2, 3, 1) Ges C (3, 1, 2) n‡j, †f±i c×wZ‡Z ABC wÎfz‡Ri †ÿÎdj wbY©q Ki| [BUET 16-17] mgvavb: OA  = i ^ + 2j ^ + 3k ^ , OB  = 2i ^ + 3j ^ + k ^ , OC  = 3i ^ + j ^ + 2k ^ AB  = OB  – OA  = (2 – 1)i ^ + (3 – 2)j ^ + (1 – 3)k ^ = i ^ + j ^ – 2k ^ AC  = OC  – OA  = (3 – 1)i ^ + (1 – 2)j ^ + (2 – 3)k ^ = 2i ^ – j ^ – k ^  ABC Gi †ÿÎdj = 1 2 |AB |   AC  AB   AC  =      i  ^ 1 2 j ^ 1 – 1 k ^ – 2 – 1 = i ^ (– 1 – 2) – j ^ (– 1 + 4) + k ^ (– 1 – 2) = – 3i ^ – 3j ^ – 3k ^  ABC Gi †ÿÎdj = 1 2  (– 3) 2 + (– 3) 2 + (– 3) 2 eM© GKK = 3 3 2 eM© GKK (Ans.) 7| A   i   j  – k  , B   3 i   j   k  , C   5 i   j  – k  Ges D   – i   2j   k  , |AB  | Gi Dci BC  Gi Awf‡ÿc wbY©q Ki? [BUET 15-16] mgvavb: AB  = (3 – 1)i ^ + (1 – 1)j ^ + (1 + 1)k ^ = 2i ^ + 2k ^ Ges BC  = (5 – 3)i ^ + (1 – 1)j ^ + (– 1 – 1)k ^ = 2i ^ – 2k ^  AB  Gi Dci BC  Gi Awf‡ÿc = AB  .BC  |AB|  = 2.2 + 2.(– 2) (2) 2 + (2) 2 = 0 (Ans.) 8| B  = 2i ^ – 4j ^ + 3k ^ †f±i Gi j¤^ w`K eivei A  = – i ^ – j ^ + 2k ^ †f±i Gi AskK W  wbY©q Ki| AZ:ci W  †f±‡ii Dci A  †f±‡ii Awf‡ÿc wbY©q Ki| [BUET 14-15] mgvavb: B  Gi j¤^ w`K eivei A  Gi AskK, W  = A  – A  .B  |B|   B  ...... (i) B  = B  |B|  = 2i ^ – 4j ^ + 3k ^ 29 (i) †_‡K cvB, W  = (– i ) ^ – j ^ + 2k ^ – (– 2 + 4 + 6) 29  (2i ) ^ – 4j ^ + 3k ^ 29 = (– i ) ^ – j ^ + 2k ^ – 8 29 (2i ) ^ – 4j ^ + 3k ^ = – 45 29i ^ + 3 29j ^ + 34 29k ^ (Ans.) W  †f±‡ii Dci A  Gi Awf‡ÿc = W  .A  |W|  = 110 29 110 29 = 110 29 (Ans.) 9| AB  = 3i ^ + 3j ^ – k ^ Ges AC  = 5i ^ – j ^ + 2k ^ n‡j, AB I AC †K mwbœwnZ evû a‡i Aw1⁄4Z mvgvšÍwi‡Ki †ÿÎdj wbY©q Ki| [BUET 04-05] mgvavb: AB   AC  =      i  ^ 3 5 j ^ 3 – 1 k ^ – 1 2 = i ^ (6 – 1) – j ^ (6 + 5) + k ^ (– 3 – 15) = 5i ^ – 11j ^ – 18k ^
†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.)