Title Vector Varsity Practice Sheet_With Solve-25.pdf ✅

†f±i  Varsity Practice Sheet ......................................................................................................................................... 1 weMZ mv‡j DU-G Avmv cÖkœvejx 1. x = e –2t, y = 2 cos3t Ges z = 2 sin2t Øviv ewY©Z eμc‡_ GKwU KYv ågY K‡i| t = 0 s mg‡q KYvwUi †eM KZ n‡e? [DU-A 24-25] – 2  i + 4  k – 2  i + 6  j + 4  k –  i + 2  i 2  i + 6  j + 4  k DËi: – 2  i + 4  k e ̈vL ̈v: miY, r  = x  i + y  j + z  k  †eM, v  = d r  dt = dx dt  i + dy dt  j + dz dt  k = – 2e–2t  i – 6 sin3t  j + 4 cos2t  k  v  | t=0 = – 2  i + 4  k 2. wZbwU †f±i A  = 6  i – 8  j , B  = – 8  i + 3  j Ges C  = 26  i + 19  j †`Iqv Av‡Q| hw` aA  + bB  + C  = 0  nq, Z‡e a I b Gi m¤¢ve ̈ gvb ̧‡jv n‡eÑ [DU-A 24-25] a = 7, b = 5 a = 5, b = 7 a = 5, b = 8 a = 3, b = 7 DËi: a = 5, b = 7 e ̈vL ̈v: aA  + bB  + C  = 0   6a  i – 8a  j – 8b  i + 3b  j + 26  i + 19  j = 0   (6a – 8b + 26)  i + (– 8a + 3b + 19)  j = 0   6a – 8b + 26 = 0, – 8a + 3b + 19 = 0  a = 5, b = 7 3. `ywU †f±i A  Ges B  -Gi †hvMdj Zv‡`i cv_©‡K ̈i Ici j¤^| wb‡Pi †Kvb wee„wZwU Aek ̈B mZ ̈? [DU 23-24] |A|  = |B|  A  . B  = 0 A   B  = 0 A  = – 2B  DËi: |A|  = |B|  e ̈vL ̈v: (A  + B  ) . (A  – B  ) = 0  A 2 + AB – AB – B 2 = 0  A = B A_©vr, |A|  = |B|  4. P  = 2i  + 2j  – k  Ges Q  = 6i  + 3j  – 3k  †f±i؇qi Df‡qi Ici j¤^ w`‡K GKwU GKK †f±i †KvbwU n‡e? [DU 22-23] – i  – 2k  – 3i  – 6k  –3(i  + 2k  ) 45 –3(i  – 2k  ) 45 DËi: –3(i  + 2k  ) 45 e ̈vL ̈v: B  A  A  I B  †f±‡ii j¤^ w`‡K GKK †f±i   n‡j,   =  A   B  |A |   B  A   B  =        i   2 6 j  2 3 k  –1 –3 = – 3i  – 6k   j¤^ GKK †f±i =  A   B  |A |   B  =  –3i  – 6k  (–3) 2 + (–6) 2 =  –3i  – 6k  45 5. GKwU 3 gv‡bi †f±i‡K GKwU 4 gv‡bi †f±‡ii mv‡_ †hvM Ki‡j jwä †f±‡ii gvb wb‡Pi †KvbwU n‡e bv? [DU 21-22] 0 1 3 5 DËi: 0 e ̈vL ̈v: †f±i؇qi jwäi m‡e©v”P gvb, Rmax = 3 + 4 = 7 Ges me©wb¤œ gvb, Rmin = 4 – 3 = 1 ZvB jwäi gvb KL‡bvB 1 Gi †P‡q †QvU A_ev 7 Gi †P‡q eo n‡e bv| 6. a Gi gvb KZ n‡j A  = 2i  + 2j  – k  Ges B  = ai  + j  †f±iØq ci ̄úi j¤^ n‡e? [DU 20-21; JU 20-21; JnU 17-18, 16-17] 0 7 4 –1 2 DËi: –1
2 .........................................................................................................................................  Physics 1st Paper Chapter-2 e ̈vL ̈v: A  B  j¤^ n‡Z n‡j, †f±i؇qi WU MyYb k~b ̈ n‡Z n‡e| (2i )  + 2j  – k  .(ai )  + j  = 0  2a + 2 = 0  a = –1 7. `yBwU †f±i A  = 3i  – 3j  Ges B  = 5i  + 5k  Gi ga ̈eZ©x †KvY KZ? [DU 19-20; JU 16-17; DU 14-15] 60 30 45 90 DËi: 60 e ̈vL ̈v:  = cos–1 A  .B  AB = cos–1 (3i )  – 3j  . (5i )  + 5k  3 2 + (–3) 2  5 2 + 52 = cos–1 15 3 2  5 2 = cos–1 1 2 = 60 8. wZbwU †f±i a  , b  I c  , hv‡`i gvb h_vμ‡g 4, 3 Ges 5; †hvM Ki‡j k~b ̈ nq A_©vr a  + b  + c  = 0| Zvn‡j | c |   (a   b  ) Gi gvb n‡jvÑ [DU 18-19] 12 60 25 15 DËi: 60 e ̈vL ̈v: a  + b  + c  = 0 | a |  + b  = |–c|  | a|  = 4 |b|  = 3 | c|  = 5 b  a  c  c 2 = a2 + b2 + 2ab cos  5 2 = 32 + 42 + 2  3  4 cos   = 90  |c |   (a )   b  = 5  4  3 sin  sin90 = 5  4  3 sin2 90 = 60 9. †f±i A  , B  I C  Gi gvb h_vμ‡g 12, 5 I 13 Ges A  + B  = C  | A  I B  †f±i؇qi ga ̈eZ©x †Kv‡Yi gvb KZ? [DU 17-18]  2  3  4  6 DËi:  2 e ̈vL ̈v: C  = 13 B  = 5  A  A  = 12 A  + B  = C   |A |  + B  = |C|   A 2 + B2 + 2AB cos = C  122 + 52 + 2  12  5 cos = 132  2  12  5 cos = 0  cos = 0   =  2 10. GKwU KYvi Dci F  = (10i  + 10j  + 10k  ) N ej cÖ‡qvM Ki‡j KYvwUi miY nq r  = (2i  + 2j  – 2k  ) m| ej KZ...©K m¤úvw`Z KvR KZ n‡e? [DU 17-18] 20 J 30 J 10 J 40 J DËi: 20 J e ̈vL ̈v: W = F  .r  = (10i )  + 10j  + 10k  .(2i )  + 2j  – 2k  = 20 + 20 – 20 = 20 J 11. hw` A  = i  – j  + k  Ges B  = i  + j  – k  GKwU mvgvšÍwi‡Ki `yBwU mwbœwnZ evû wb‡`©k K‡i, Zvn‡j Dchy3 GK‡K mvgvšÍwi‡Ki †ÿÎdj wbY©q Ki| [DU 15-16] 2 2 2 1 2 DËi: 2 2 e ̈vL ̈v: mvgvšÍwi‡Ki †ÿÎdj = |A |   B  A   B  =        i   1 1 j  –1 1 k  1 –1 B  A  = i  (1 – 1) – j  (–1 – 1) + k  (1 + 1) = 2j  + 2k   †ÿÎdj = |A |   B  = 2 2 + 22 = 2 2 eM© GKK 12. j  .(2i )  – 3j  +[] k  -Gi gvb KZ? [DU 14-15] 2 –3 1 –2 DËi: –3 e ̈vL ̈v: j  .(2i )  – 3j  +[ ] k  = –3
†f±i  Varsity Practice Sheet ........................................................................................................................................ 3 weMZ mv‡j GST-G Avmv cÖkœvejx 1.  A = 3  i +  j + 2  k Ges  B = 2  i – 2  j + 4  k | Dfq †f±‡ii Dci Awfj¤^ †f±iwU n‡jvÑ [GST 23-24] –8  i – 8  j + 8  k 8  i – 8  j – 8  k 8  i – 8  j + 8  k 8  i + 8  j + 8  k DËi: 8  i – 8  j – 8  k e ̈vL ̈v:  A ×  B =          i  j  k 3 1 2 2 –2 4 =  i(4 + 4) –  j(12 – 4) +  k(–6 – 2) = 8  i – 8  j – 8  k 2. GK e ̈w3 m~‡h©v`‡qi w`‡K 4 m hvIqvi c‡i `wÿY w`‡K 3 m hvq| Zvi AwZμvšÍ `~iZ¡ I mi‡Yi cv_©K ̈ KZ m? [GST 21-22] 2 4 1 7 DËi: 2 e ̈vL ̈v: cwðg c~e© `wÿY DËi 3 m 4 m †gvU `~iZ¡ d n‡j, d = 4 + 3 = 7 m miY = 4 2 + 32 = 5 m cv_©K ̈ = `~iZ¡ – miY = (7 – 5) m = 2 m 3. XY mgZ‡j 6i  + 8j  – 5k  †f±iwUi ˆ`N© ̈ KZ GKK? [GST 21-22] 6 10 0 5 5 DËi: 10 e ̈vL ̈v: X Z 6i  + 8j  – 5k  Y XY Z‡j 6i  + 8j  – 5k  †f±iwU n‡”Q 6i  + 8j  XY Z‡j ˆ`N© ̈ = 6 2 + 82 = 10 GKK weMZ mv‡j Agri-G Avmv cÖkœvejx 1. 7 kg f‡ii †Kvb e ̄‘i Ici F  = 2i  – 3j  + 6k  N ej cÖhy3 n‡j e ̄‘wU KZ Z¡iY cÖvß n‡e? [Cluster Agri 24-25] 1.4 ms–1 1.5 ms–2 1.0 ms–2 7.0 ms–2 DËi: 1.0 ms–2 e ̈vL ̈v: a  = F  m = 2i  – 3j  + 6k  7 = 2 7 i  – 3 7 j  + 6 7 k   |a|  =     2 7 2 +     – 3 7 2 +     6 7 2 = 4 + 9 +36 49 = 1 ms–2 2. `ywU mggv‡bi †f±i GKwU we›`y‡Z wμqvkxj| G‡`i jwäi gvb †h‡Kv‡bv GKwU †f±‡ii gv‡bi mgvb| †f±i `ywUi ga ̈eZ©x †KvY KZ? [Agri. Guccho 20-21; DU 18-19] 120 180 90 0 DËi: 120 e ̈vL ̈v: P 2 = P2 + P2 + 2  P  Pcos   = 120 P   P  P  3. `ywU e‡ji jwäi gvb 40 N, ej `yÕwUi g‡a ̈ †QvU ejwUi gvb 30 N Ges GwU jwä e‡ji j¤^ eivei wμqv K‡i| eo ejwUi gvb KZ? [Agri. Guccho 20-21; BRUR 19-20; CU 15-16] 40 N 45 N 50 N 60 N DËi: 50 N e ̈vL ̈v: P 2 + R2 = Q2 |R  | = 40 N |P  | = 30 N Q  Q   Q = 302 + 402 = 50N 4. 10 N gv‡bi GKwU ej Ab ̈ GKwU ARvbv e‡ji mv‡_ 120 †Kv‡Y AvbZ| ej `ywUi jwä ARvbv e‡ji mv‡_ 90 †Kv‡Y Aew ̄’Z| ARvbv ejwUi gvbÑ [Agri. Guccho 21-22] 8 N 7 N 6 N 5 N DËi: 5 N e ̈vL ̈v: R  10 N 60 P  30 120 30 10 N cos60 = P 10  P = 10  0.5 = 5 N