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

†f±i  CQ & MCQ Practice Sheet Solution (HSC 26) 1 02 †f±i Vector WRITTEN 1| a  = 2 i ^ + 3 j ^ – k ^ , b  = i ^ – 2 j ^ , c  = i ^ + p j ^ + 2 k ^ d  = 3i ^ – j ^ + 2k ^ [Xv. †ev. 19] (K) A(2, – 3, 1) Ges B(– 1, 0, 4) n‡j AB  wbY©q Ki| (L) b  I c  Gi ga ̈eZ©x †KvY 45 n‡j p Gi gvb wbY©q Ki| (M) a  Ges d  †h mgZ‡j Aew ̄’Z Zvi Dci j¤^ GKK †f±i wbY©q Ki| mgvavb: K OB  = – i ^ + 4k ^ OA  = 2i ^ – 3j ^ + k ^  AB  = OB  – OA  = (– i ^ + 4k ^ ) – (2i ^ – 3j ^ + k ^ ) = (– 1 – 2)i ^ + (0 + 3)j ^ + (4 – 1)k ^ = – 3i ^ + 3j ^ + 3k ^ (Ans.) L †`Iqv Av‡Q, b  I c  Gi ga ̈eZ©x †KvY,  = 45 Ges b  = i ^ – 2j ^ , c  = i ^ + pj ^ + 2k ^ Avgiv Rvwb, b  .c  = |b  ||c  |cos  (i ^ – 2j ^ ).(i ^ + pj ^ + 2k ^ ) = ( 1 + 4)  ( 1 + p ) 2 + 4 cos45  1 + (– 2)  p + 0  2 = 5  5 + p2  1 2  (1 – 2p)2 =      5  5 + p  2 2 2 [eM© K‡i]  1 – 4p + 4p2 = 5(5 + p2 ) 2  2 – 8p + 8p2 = 25 + 5p2  3p2 – 8p – 23 = 0  p = – (– 8)  (– 8) 2 – 4  3  (– 23) 2  3 = 8  2 85 6  p = 4  85 3 (Ans.) M †`Iqv Av‡Q, a  = 2i ^ + 3j ^ – k ^ Ges d  = 3i ^ – j ^ + 2k ^  a  Ges d  †h mgZ‡j Aew ̄’Z Zvi Dci j¤^ GKK †f±i =  a   d  |a   d  | GLb, a   d  =          i 2 3  j 3 – 1  k – 1 2 = i ^ (6 – 1) – j ^ (4 + 3) + k ^ (– 2 – 9) = 5i ^ – 7j ^ – 11k ^ Avevi, |a   d  | = (5) 2 + (– 7) 2 + (– 11) 2 = 195  wb‡Y©q GKK †f±i =  (5i ^ – 7j ^ – 11k ^ ) 195 =  1 195 (5i ^ – 7j ^ – 11k ^ ) (Ans.) 2| P  = 3i ^ – 3j ^ + 4k ^ , Q  = 3i ^ – 2j ^ + 4k ^ Ges R  = i ^ – j ^ + 2k ^ . [iv. †ev. 17] (K) P  we›`yMvgx Ges Q  †f±‡ii mgvšÍivj mij‡iLvi †f±i mgxKiY wbY©q Ki| (L) †`LvI †h, P  – Q  †f±iwU P  Ges Q  †f±i Øviv MwVZ mgZ‡ji Dci j¤^ †f±‡ii mv‡_ j¤^| (M) DÏxc‡K DwjøwLZ †f±i ̧wji i ^ , j ^ , k ^ Gi mnM Øviv MwVZ g ̈vwUa· A n‡j A –1 wbY©q Ki| mgvavb: K †`Iqv Av‡Q, P  = 3i ^ – 3j ^ + 4k ^ Ges Q  = 3i ^ – 2j ^ + 4k ^  P  we›`yMvgx Ges Q  †f±‡ii mgvšÍivj mij‡iLvi †f±i mgxKiY, r  = P  + tQ  = (3i ^ – 3j ^ + 4k ^ ) + t(3i ^ – 2j ^ + 4k ^ ) = (3 + 3t)i ^ – (3 + 2t)j ^ + (4 + 4t)k ^ (Ans.) L †`Iqv Av‡Q, P  = 3i ^ – 3j ^ + 4k ^ ; Q  = 3i ^ – 2j ^ + 4k ^  P  – Q  = (3 – 3)i ^ + (– 3 + 2)j ^ + (4 – 4)k ^ = – j ^  P  Ges Q  †f±i Øviv MwVZ mgZ‡ji Dci j¤^ †f±i,

†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.)