Tiêu đề Vector CQ & MCQ Practice Sheet Solution [HSC 26].pdf ✅

2  Higher Math 1st Paper Chapter-2 P   Q  =      i  ^ 3 3 j ^ – 3 – 2 k ^ 4 4 = i ^ (– 12 + 8) – j ^ (12 – 12) + k ^ (– 6 + 9) = – 4i ^ + 3k ^ Ges (P  – Q  ).(P   Q  ) = (– j ^ ).(– 4i ^ + 3k ^ ) = 0  (P  – Q  ) †f±iwU P  Ges Q  †f±i Øviv MwVZ mgZ‡ji Dci j¤^ †f±‡ii mv‡_ j¤^ (Showed) M †`Iqv Av‡Q, P  = 3i ^ – 3j ^ + 4k ^ , Q  = 3i ^ – 2j ^ + 4k ^ , R  = i ^ – j ^ + 2k ^ †f±i ̧wj i ^ , j ^ , k ^ Gi mnM Øviv MwVZ g ̈vwUa· A n‡jÑ A =       3 3 1 – 3 – 2 – 1 4 4 2  |A| =       3 3 1 – 3 – 2 – 1 4 4 2 = 3(– 4 + 4) + 3(6 – 4) + 4(– 3 + 2) = 3.0 + 3.2 + 4.(– 1) = 2  0  †h‡nZz|A|  0; A g ̈vwUa· wecixZ‡hvM ̈| A Gi mn ̧YK mg~n: A11 = (– 1)1+1    – 2 – 1 4 2 = (– 4 + 4) = 0 A12 = (– 1)1+2    3 1 4 2 = – (6 – 4) = – 2 A13 = (– 1)1+3    3 1 – 2 – 1 = (– 3 + 2) = – 1 A21 = (– 1)2+1    – 3 – 1 4 2 = – (– 6 + 4) = 2 A22 = (– 1)2+2    3 1 4 2 = (6 – 4) = 2 A23 = (– 1)2+3    3 1 – 3 – 1 = – (– 3 + 3) = 0 A31 = (– 1)3+1    – 3 – 2 4 4 = (– 12 + 8) = – 4 A32 = (– 1)3+2    3 3 4 4 = – (12 – 12) = 0 A33 = (– 1)3+3    3 3 – 3 – 2 = (– 6 + 9) = 3 GLb, adjA =       A11 A21 A31 A12 A22 A32 A13 A23 A33 T =       0 2 – 4 – 2 2 0 – 1 0 3 T =       0 – 2 – 1 2 2 0 – 4 0 3  A –1 = adjA |A| =       0 – 2 – 1 2 2 0 – 4 0 3 2 =         0 1 – 2 – 1 1 0 – 1 2 0 3 2 (Ans.) 3| A  = 2i ^ + 3j ^ – k ^ , B  = i ^ + 2j ^ – k ^ , C  = i ^ + bj ^ + 3k ^ .[Xv. †ev. 22] (K) Ae ̄’vb †f±i ej‡Z wK eyS? (L) A  †f±i eivei B  †f±‡ii Dcvsk C  †f±‡ii mv‡_ j¤^ n‡j b Gi gvb wbY©q Ki| (M) A  + B  Ges A   B  †f±i؇qi ga ̈eZ©x †KvY wbY©q Ki| mgvavb: K Ae ̄’vb †f±i: hw` †Kv‡bv we›`y P Gi Ae ̄’vb‡K g~jwe›`y O(0, 0) Gi mv‡c‡ÿ (x, y, z) Øviv wb‡`©k Kiv nq Z‡e OP  †K O we›`yi mv‡c‡ÿ P we›`yi Ae ̄’vb †f±i ejv nq| †h‡Kv‡bv we›`y P(x, y, z) Gi Ae ̄’vb †f±i‡K cÖKvk Kiv nq: P(x, y, z) X Y Z O OP  = r  = xi ^ + yj ^ + zk ^ Øviv| L †`Iqv Av‡Q, A  = 2i ^ + 3j ^ – k ^ , B  = i ^ + 2j ^ – k ^ , C  = i ^ + bj ^ + 3k ^ GLb, A  .B  = 2  1 + 3  2 + (– 1)  (– 1) = 9  A  †f±i eivei B  †f±‡ii Dcvsk = A  .B  |A  | .A ^ = A  .B  |A  | 2 .A  = 9 14(2i ^ + 3j ^ – k ^ ) = 9 7 i ^ + 27 14j ^ – 9 14k ^ Avevi, DcvskwU C  †f±‡ii mv‡_ j¤^ n‡j,     9 7 i ^ + 27 14 j ^ – 9 14 k ^ .(i ^ + bj ^ + 3k ^ ) = 0  9 7 + 27 14b – 27 14 = 0  27b 14 = 9 14  b = 1 3 (Ans.)
†f±i  CQ & MCQ Practice Sheet Solution (HSC 26) 3 M †`Iqv Av‡Q, A  = 2i ^ + 3j ^ – k ^ , B  = i ^ + 2j ^ – k ^ A  + B  = 3i ^ + 5j ^ – 2k ^ A   B  =      i  ^ 2 1 j ^ 3 2 k ^ – 1 – 1 = i ^ (– 3 + 2) – j ^ (– 2 + 1) + k ^ (4 – 3) = – i ^ + j ^ + k ^ GLb, (A  + B  ).(A   B  ) = (3i ^ + 5j ^ – 2k ^ ).(– i ^ + j ^ + k ^ ) = – 3 + 5 – 2 = 0 †h‡nZz(A  + B  ) Ges (A   B  ) Gi WU ̧Ydj 0, myZivs †f±iØq ci ̄úi j¤^ Ges †f±i؇qi AšÍf©~3 †KvY,  = 90 (Ans.) 4| A  = 2i ^ – 3j ^ – k ^ , B  = – i ^ – 4j ^ + 7k ^ Ges wZbwU we›`yi ̄’vbvsK P(– 3, – 2, – 1), Q(4, 0, – 3) Ges S(6, – 7, 8)| [wm. †ev. 22] (K) D`vniYmn GKK †f±i Gi msÁv `vI| (L) DÏxc‡Ki Av‡jv‡K A eivei B Gi Dcvsk wbY©q Ki| (M) DÏxc‡Ki Av‡jv‡K PQS Gi †ÿÎdj wbY©q Ki| mgvavb: K GKK †f±i: †h †f±‡ii •`N© ̈ ev gvb 1, †mB †f±iwU‡K GKK †f±i e‡j|  b  †f±iwU GKK †f±i n‡j, |b  | = 1 n‡e| D`vniY: awi, b  GKwU †f±i ivwk Ges |b  | Gi w`K eivei GKK †f±i b ^ |  b ^ = b  |b  | †hLv‡b, |b  |  0 L A  = 2i ^ – 3j ^ – k ^ , B  = – i ^ – 4j ^ + 7k ^ A  .B  = 2  (– 1) + (– 3)  (– 4) + (– 1)  7 = – 2 + 12 – 7 = 3 |A  | = (2) 2 + (– 3) 2 + (– 1) 2 = 4 + 9 + 1 = 14  A  eivei B  Gi Dcvsk = A  .B  |A  | .A ^ = A  .B  |A  | 2 .A  = 3 14(2i ^ – 3j ^ – k ^ ) (Ans.) M †`Iqv Av‡Q, P  = – 3i ^ – 2j ^ – k ^ , Q  = 4i ^ – 3k ^ , S  = 6i ^ – 7j ^ + 8k ^ PQ  = (4 + 3)i ^ + (0 + 2)j ^ + (– 3 + 1)k ^ = 7i ^ + 2j ^ – 2k ^ PS  = (6 + 3)i ^ + (– 7 + 2)j ^ + (8 + 1)k ^ = 9i ^ – 5j ^ + 9k ^  PQ   PS  =      i  ^ 7 9 j ^ 2 – 5 k ^ – 2 9 = i ^ (18 – 10) – j ^ (63 + 18) + k ^ (– 35 – 18) = 8i ^ – 81j ^ – 53k ^  QPS Gi †ÿÎdj = 1 2 |PQ   PS  | = 1 2  (8) 2 + (– 81) 2 + (– 53) 2 = 48.56 eM© GKK (Ans.) 5| A  = 4i ^ + 7j ^ – 3k ^ Ges B  = 3i ^ + 4j ^ + 7k ^ . [g. †ev. 19] (K) U  = 2i ^ + 5k ^ Ges V  = 3j ^ + 2k ^ Gi jwä †f±‡ii gvb wbY©q Ki| (L) DÏxc‡Ki Av‡jv‡K A  I B  †f±i؇qi ga ̈eZ©x †KvY wbY©q Ki| (M) DÏxc‡Ki Av‡jv‡K A  I B  †f±i؇qi j¤^ GKK †f±i wbY©q Ki| mgvavb: K †`Iqv Av‡Q, U  = 2i ^ + 5k ^ , V  = 3j ^ + 2k ^ jwä †f±i, R  = U  + V  = (2 + 0)i ^ + (0 + 3)j ^ + (5 + 2)k ^ = 2i ^ + 3j ^ + 7k ^  jwä †f±‡ii gvb, |R  | = (2) 2 + (3) 2 + (7) 2 = 62 (Ans.) L †`Iqv Av‡Q, A  = 4i ^ + 7j ^ – 3k ^ Ges B  = 3i ^ + 4j ^ + 7k ^ Avgiv Rvwb, A  .B  = |A  | |B  | cos   = cos–1      A   .B  |A  | |B  | = cos–1      4  3 + 7  4 + (– 3)  7  4 2 + 72 + (– 3) 2  (3) 2 + (4) 2 + (7) 2 = cos–1     19 74 (Ans.) M †`Iqv Av‡Q, A  = 4i ^ + 7j ^ – 3k ^ , B  = 3i ^ + 4j ^ + 7k ^ A   B  =      i  ^ 4 3 j ^ 7 4 k ^ – 3 7 = i ^ (49 + 12) – j ^ (28 + 9) + k ^ (16 – 21) = 61i ^ – 37j ^ – 5k ^  A  I B  †f±i؇qi j¤^ GKK †f±i =  A   B  |A |   B  =  1 5115  (61i ^ – 37j ^ – 5k ^ ) (Ans.)
4  Higher Math 1st Paper Chapter-2 6| A  = 3i ^ + j ^ – 5k ^ , B  = 2i ^ – 6j ^ + 3k ^ , C  = 5i ^ – 3j ^ + 6k ^ . (K) A  I B  †f±i `yBwUi ga ̈eZ©x †KvY m~2‡KvY n‡j,  Gi gvb wbY©q Ki| (L) B  I C  †Kv‡bv wÎfz‡Ri evû n‡j wÎfzRwUi †ÿÎdj wbY©q Ki| (M) A  Gi w`‡K GKK †f±i A ^ wbY©q Ki hv B  I C  Gi mv‡_ mgZjxq| mgvavb: K †`Iqv Av‡Q, A  = 3i ^ + j ^ – 5k ^ , B  = 2i ^ – 6j ^ + 3k ^ A  I B  †f±iØq ga ̈eZ©x †KvY m~2‡KvY n‡j, A  .B  > 0  6 – 6 – 15 > 0  – 6 – 9 > 0  – 6 > 9   < – 3 2 (Ans.) L †`Iqv Av‡Q, B  = 2i ^ – 6j ^ + 3k ^ , C  = 5i ^ – 3j ^ + 6k ^ GLb, B   C  =      i  ^ 2 5 j ^ – 6 – 3 k ^ 3 6 = i ^ (– 36 + 9) – j ^ (12 – 15) + k ^ (– 6 + 30) = – 27i ^ + 3j ^ + 24k ^  wÎfz‡Ri †ÿÎdj = 1 2 |B   C  | = 1 2  (– 27) 2 + (3) 2 + (24) 2 = 1 2 1314 eM© GKK (Ans.) M †`Iqv Av‡Q, A  = 3i ^ + j ^ – 5k ^ , B  = 2i ^ – 6j ^ + 3k ^ , C  = 5i ^ – 3j ^ + 6k ^  A  Gi w`‡K †f±i hv B  I C  Gi mv‡_ mgZjxq †f±i, r  = (2i ^ – 6j ^ + 3k ^ ) + (5i ^ – 3j ^ + 6k ^ )  r  = (2 + 5)i ^ + (– 6 – 3)j ^ + (3 + 6)k ^ ...... (i) r  Ges A  Gi w`K GKB n‡j, 2 + 5 3 = – 6 – 3  = 3 + 6 – 5 GLb, 2 + 5 3 = 3 + 6 – 5  – 10 – 25 = 9 + 18  – 19 = 43   = – 43 19 A  Gi w`‡K †f±i =       2     –  43 19  + 5 i ^ +       – 6     –  43 19  – 3 j ^ +       3     –  43 19  + 6 k ^ = 9 19i ^ + 201 19 j ^ – 15 19k ^  A  Gi w`‡K GKK †f±i, A  = 9 19i ^ + 201 19 j ^ – 15 19k ^     9 19 2 +     201 19 2 +    –  15 19 2 = 9i ^ + 201j ^ – 15k ^  (9) 2 + (201) 2 + (15) 2 = 9i ^ + 201j ^ – 15k ^ 40707 (Ans.) 7| OA I OB Gi ga ̈we›`y h_vμ‡g P  I Q  . B(3, 2, – 1) A(2, 2, 2) O(0, 0, 0) (K) AB Gi †f±i mgxKiY wbY©q Ki| (L) †f±i c×wZ‡Z cÖgvY Ki †h, PQ || AB. (M) OA  I OB  †K mvgvšÍwi‡Ki mwbœwnZ evû a‡i mvgvšÍwiKwUi †ÿÎdj wbY©q Ki| mgvavb: K awi, OA  = 2i ^ + 2j ^ + 2k ^ = a  OA  = 3i ^ + 2j ^ – k ^ = b   AB  Gi †f±i mgxKiY, r  = a  + (b  – a  ) = 2i ^ + 2j ^ + 2k ^ + t(3i ^ + 2j ^ – k ^ – 2i ^ – 2j ^ – 2k ^ ) = 2i ^ + 2j ^ + 2k ^ + t(i ^ – 3k ^ ) (Ans.) L †`Iqv Av‡Q, OA I OB Gi ga ̈we›`y h_vμ‡g P  I Q  B(3, 2, – 1) O(0, 0, 0) P(1, 1, 1) A(2, 2, 2) Q     3 2  1 – 1 2 OP  = 1 2 OA  = 1 2 (2i ^ + 2j ^ + 2k ^ ) = i ^ + j ^ + k ^