Title Vector Varsity Daily-2 (Home Practice)-Solution.pdf ✅

1 Daily-02 [Home Practice (Solve Sheet)] wm‡jevm: †f±i c~Y©gvb: 30 †b‡MwUf gvK©: 0.25 mgq: 20 wgwbU 1. k  . (i )  + j  = ? 2 0 4 1 DËi: 0 e ̈vL ̈v: k  . i  = |k|  |i|  cos90 = 0 Ges k  . j  = |k|  |j|  cos90 = 0 k  .i  + k  .j  = 0 + 0 = 0 †h‡nZz i  , j  I k  h_vμ‡g x, y I z Aÿ eivei GKK †f±i Ges x, y I z Aÿ ci ̄ú‡ii mv‡_ j¤^fv‡e Aew ̄’Z| ZvB k  I i  Gi †ÿ‡Î  = 90 Ges k  I j  Gi †ÿ‡Î  = 90| 2. wZbwU †f±i, [p, q, r] x-A‡ÿi abvZ¥K w`‡Ki mv‡_ 60, 120, 90 †KvY Drcbœ K‡i Ges p  + q  + r  = n  n‡j, n  Gi w`K KZ? (p )   q   r  Gi gvb 50 [Three vectors, [p, q, r] x-make angles of 60°, 120°, and 90° with the positive x-axis, respectively. If p  + q  + r  = n  what is the direction of n  ? (Given that, (p )   q   r  = 50 ] 130 120 – 50 None of these DËi: None of these e ̈vL ̈v: 120 r  60 q  p  rcos = Pcos60 + qcos120 + r cos90 = 50  1 2 – 50  1 2 + 0 = 25 – 25 + 0 = 0 rsin = Psin60 + qsin120 + rsin90 = 50  3 2 + 50  3 2 + 50 = 50( 3 + 1) tan = ( 3 + 1) (50) 0  = tan–1    50 ( 3 + 1) 0 hv Awb‡Y©q| 3. A  = 2i  – 3j  Ges B  = – i  + 4j  n‡j, |A |  – B  = ? A  = 2i  – 3j  and B  = – i  + 4j  then |A |  – B  = ? 50 22 58 14 DËi: 58 e ̈vL ̈v: A  – B  = (2i )  – 3j  – (–i )  + 4j  = i  (2 + 1) – j  (3 + 4) = 3i  – 7j  |A |  – B  = (3) 2 + (–7) 2 = 58 (Ans.) 4. `ywU †f±‡ii †hvMdj I we‡qvMd‡ji gvb ci ̄úi mgvb n‡j, †f±iØq ci ̄ú‡ii Kx n‡e? [If the magnitude of the sum and difference of two vectors are equal, what is the relationship between the vectors? ] mgvšÍivj (Parallel) j¤^ (Perpendicular) wecixZ (Antiparallel) †Kv‡bvwU bq (None of the above) DËi: j¤^ (Perpendicular) e ̈vL ̈v: awi, `ywU †f±i A  , B  Gi ga ̈eZ©x †KvY  |A |  + B  = |A |  – B  A  B    A 2 + B2 + 2AB cos = A2 + B2 – 2ABcos [eM© K‡i|]  2AB cos + 2ABcos = 0  4ABcos = 0  cos = 0 4AB  cos = 0  cos = cos90   = 90   =  2 (j¤^) 5. ci ̄úi  2 †Kv‡Y wμqviZ P I Q e‡ji gvb mgvb n‡j, Zv‡`i jwä Q e‡ji mv‡_ KZ †KvY Drcbœ Ki‡e? [If the angle between two vectors P and Q is  2 , and their magnitudes are equal, what is the angle between P and the resultant of P and Q?]  2 4  2  3  4 DËi:  4
2 e ̈vL ̈v: tan = Psin 2 Q + Pcos 2 = sin 2 1 + cos 2 = 2sin  4 cos  4 2cos2  4  tan = tan  4   =  4 6. GK e ̈w3 m~h©v‡ ̄Íi w`‡K 5 m hvIqvi ci `wÿY w`‡K 12 m †Mj| Zvi `~iZ¡ I mi‡Yi AbycvZ KZ? [A person walks 5 m west and then 12 m south. What is the ratio of his distance and displacement?] 13 17 1 17 13 2 DËi: 17 13 e ̈vL ̈v: 5 m 12 m c: `: D: c~: miY, R = A 2 + B2 [when,  = 90] = 5 2 + 122 [A = 5, B = 12-a‡i] = 25 + 144 = 169 = 13 m `~iZ¡ = 12 + 5 = 17 m AbycvZ = 17 13 7. wb‡Pi †Kvb †f±‡ii cv`we›`y I kxl©we›`y GKB? [Which vector has its tail and head at the same point?] mg‡iL †f±i (Scalar vector) GKK †f±i (Zero vector) bvj †f±i (Null vector) mgZjxq †f±i (Unit vector) DËi: bvj †f±i (Null vector) 8. (A )  . B  2  |A |   B  2 = ? tan2  A 2B 2 cot cot2  DËi: cot2  e ̈vL ̈v: (A )  . B  2  |A |   B  2  A 2B 2 cos2  A 2B 2 sin2  = cot2  9. wP‡Î V1  Gi mv‡c‡ÿ V2  Gi †e‡Mi gvb KZ? [Given the vectors V1  and V2  in the diagram, what is the magnitude of V2 relative to V1?] V2  = 4ms–1 30 V1  = 3ms–1 5 – 2 3 10 + 12 3 73 37 DËi: 37 e ̈vL ̈v: V1  = 3j  V2  = 4cos30i  + 4sin30j  V21  = V2  – V1  = 4  3 2 i  + 4  1 2 j  – 3j  V21  = 2 3i  + 5j  |V | 21  = (2 3) 2 + (5) 2 = 12 + 25 = 37 10. b  I c  `ywU †f±i GKB mgZ‡j wμqvkxj Ges b   c  Gi mv‡_ a  Aci GKwU †f±i ‘’ †Kv‡Y wμqvkxj| GLb, a   (b )   c  Gi †f±i †Kvb Z‡j Ae ̄’vb Ki‡e? [Given three vectors b  , c  that lie in the same plane, and vector a makes an angle  with the plane containing b  and c  what is the direction of the vector a   (b )   c  ] a  †h Z‡j Av‡Q †mB Z‡j (a  lies in the same plane as b  and c  ) a   c  †h Z‡j Av‡Q †mB Z‡j (a   c  lies in the same plane as b  and c  ) b  I c  †h Z‡j Av‡Q †mB Z‡j (b  and c  lies in the same plane as b  and c  ) †KvbwUB bq (None of these) DËi: b  I c  †h Z‡j Av‡Q †mB Z‡j (b  and c  lies in the same plane as b  and c  ) e ̈vL ̈v: b   c  Gi gvb b  I c  Gi mv‡_ j¤^fv‡e Ae ̄’vb K‡i| GLb, b   c  Gi mv‡_  †Kv‡Y a  Ae ̄’vb Ki‡j, G‡`i μm †cÖvWv± Gi gvb Avevi b  I c  Gi Z‡j †diZ Avm‡e!
3 11. a  = 3 2 b  ; c  = 1 2 a  n‡j, a  . c  Gi gvb KZ? [If a  = 3 2 b  ; c  = 1 2 a  , then what is the value of a  . c  ?] 3 4 3 4 b 2 9 4 b 2 9 8 b 2 DËi: 9 8 b 2 e ̈vL ̈v: a  . c  = 3 2 b  . 1 2 a  = 3 2 b  .     1 2  3 2 b  = 3 2 b  . 3 4 b  = 3 2  3 4  b  .b  = 9 8 b 2 12. hw` GKwU †bŠKv d cÖ‡ ̄’i †Kv‡bv b`x †mvRvmywR cvi n‡Z Pvq Z‡e, wb‡Pi †KvbwU mwVK? (u = † ̄av‡Zi †eM, v = †bŠKvi †eM) [If a boat needs to cross a river of width d directly, what should be the angle at which it should be steered with respect to the current? (u = speed of the current, v = speed of the boat)] t = d v cos     – u v t = d sin    cos–1 u v t = d v sin     sin–1 v 2 – u 2 v t = d V DËi: t = d v sin     sin–1 v 2 – u 2 v e ̈vL ̈v:  = cos–1     – u v u d w v  t = d v sin     cos–1     – u v v –u v 2 – (–u) 2 = v 2 – u 2  cos–1     –u v = sin–1 v 2 – u 2 v = d v sin     sin–1    v  2 – u 2 v 13. `ywU e‡ji jwäi gvb 3 N| ej `ywUi g‡a ̈ †QvU ejwUi gvb 2N n‡j, hvi mv‡_ jwä 90 †KvY Drcbœ K‡i| Zvn‡j, eo ejwUi gvb KZ? [Two forces of magnitude √3 N and 2N act at a point. If the resultant force makes an angle of 90° with the smaller force, what is the magnitude of the larger force?] 2– 3 2 2 + 3 2 2 2  3 2 None of these DËi: None of these e ̈vL ̈v: 3 N 2N P  tan90 = Psin 2 + Pcos  1 0 = Psin 2 + Pcos  Pcos = – 2 R 2 = P2 + Q2 + 2PQ cos  ( 3) 2 = P2 + 22 + 2  (– 2)  2  3 = P2 – 4  P 2 = 7  P =  7 14. †KvbwU mwVK? [Which of the following is correct?] a  . b  = a  (a )  Gi w`‡K b  Gi j¤^ Awf‡ÿc (a  . b  = a ) (projection of b  onto a  ) a  . b  = b   (a )  Gi w`‡K b  Gi j¤^ Awf‡ÿc (a  . b  = b   (projection of b  onto a  )) a  . b  = ab cos sin a  . b  = a   b  hLb  = 180 (a  . b  = a   b  then  = 180 DËi: a  . b  = a  (a )  Gi w`‡K b  Gi j¤^ Awf‡ÿc (a  . b  = a ) (projection of b  onto a  ) e ̈vL ̈v: a  . b  = a(bcos)  a  (a )  Gi w`‡K b  Gi j¤^ Awf‡ÿc 15. V  = ax2 i  + bxy2 j  + cyz3 k  , V  Gi Kvj© k~b ̈ n‡j, wb‡Pi †KvbwU mwVK? [Given the vector field, V  = ax2 i  + bxy2 j  + cyz3 k  , if the curl of V is zero, which of the following is correct?] AN~Y©bkxj, AmsiÿYkxj (Irrotational, non-conservative) N~Y©bkxj, AmsiÿYkxj (Rotational, non-conservative) N~Y©bkxj, msiÿYkxj (Rotational, conservative) AN~Y©bkxj, msiÿYkxj (Irrotational, conservative) DËi: AN~Y©bkxj, msiÿYkxj (Irrotational, conservative)
4 e ̈vL ̈v: Curl V  =    V  = 0 n‡j, V   AN~Y©bkxj, msiÿYkxj Curl V  =    V   0 n‡j, V   N~Y©bkxj, AmsiÿYkxj Curl V  =    V  > 0, V   N~Y©bkxj (Counter clock wise) Curl V  =    V  < 0, V   N~Y©bkxj (clock wise) 16. `ywU UavK ci ̄úi †_‡K 60 †Kv‡Y h_vμ‡g 50 kmh–1 Ges 40 kmh–1 †e‡M MwZkxj Zv‡`i jwä †eM KZ? [Two trucks are moving at 50 kmh–1 and 40 kmh–1 at an angle of 60° to each other. What is the magnitude of their resultant velocity?] 10 61 kmh–1 15 50 kmh–1 20 3 kmh–1 30 3 kmh–1 DËi: 10 61 kmh–1 e ̈vL ̈v: jwä = (A) 2 + (B) 2 + 2AB cos60 = 502 + 402 + 2  50  40 cos 60 = 6100 = 61  100 = 10 61 kmh–1 GLv‡b, A  = 50 kmh–1 B  = 40 kmh–1  = 60 R  = ? 17. V  = axi  – 4yj  + 8x3 y 2 k  , †f±i †ÿÎwU mwjbqWvj n‡j, a Gi gvb KZ? [Given the vector field V  = axi  – 4yj  + 8x3 y 2 k  , if the vector field is solenoidal, what is the value of a?] 3 2 5 4 DËi: 4 e ̈vL ̈v: Avgiv Rvwb, GKwU †f±i †ÿÎ ZLbB mwjbqWvj n‡e hw`, div V  = 0 nq|    b ̈vejv   = i   x + j   y + k   z V  = axi  – 4yj  + 8x3 y 2 k  div V  =   . V  = 0      i   x + j   y + k   z . (axi )  – 4yj  + 8x3 y 2 k  = 0  i  .i   x ax – j  .j   y 4y + k  .k   z 8x3 y 2 = 0  a – 4 + 0 = 0 [i  .i  = j  .j  = k  .k  = 1]  a – 4 = 0      x x n = nxn–1  a = 4 18. P  = (x + 3y)i  + (3y – 5z)j  + (9x + 5z)k  n‡j, †f±iwUi divergence Kx n‡e? [Given the vector field P  = (x + 3y)i  + (3y – 5z)j  + (9x + 5z)k  , what is the divergence of P?] –9 8 –8 9 DËi: 9 e ̈vL ̈v: P  = (x + 3y)i  + (3y – 5z)j  + (9x + 5z)k  div P  =   .P  =     i   x + j   y + k   z {(x + 3y)i }  + (3y – 5z)j  + (9x + 5z) k  = 1 + 3 + 5 = 9 19. a  – b  (b )  .a  b 2 †f±iwU, b  †f±‡ii mv‡_ KZ †KvY Drcbœ K‡i? [a  – b  (b )  .a  b 2 what is the angle between a and the projection of b  onto a  ?] 0 90 Error 180 DËi: 90 e ̈vL ̈v: †f±i `yBwUi WU ̧Ydj = b  .         a  – b  (b )  .a  b 2 = b  .a  – (b )  . b  (b )  .a  b 2 = b  .a  – b 2 (b )  .a  b 2 = b  .a  – b  .a  = 0 A_©vr †f±iØq ci ̄úi j¤^| 20. A   B  = 0 Ges A  = 2i  + 2j  + 2k  n‡j, B  = 3i  + mj  + nk  n‡j, m I n Gi gvb KZ? [Given, A   B  = 0 and, A  = 2i  + 2j  + 2k  , B  = 3i  + mj  + nk  . what are the values of m and n?] 4, 2 3, 3 3, 2 6, 6 DËi: 3, 3 e ̈vL ̈v:        i   2 3 j  2 m k  2 n = 0  (2n – 2m)i  + (6 – 2n)j  + (2m – 6)k  = 0 2n – 2m = 0 6 = 2n 2m – 6 = 0 m = n n = 3 m = 3 Trick : A   B  = 0 Abyiƒc mnM ̧‡jvi AbycvZ mgvb| 2 3 = 2 m = 2 n Zvn‡j, m I n Gi gvb 3, 3 n‡Z n‡e|