Tiêu đề Vector Varsity Daily-2 (Set-A) Solution.pdf ✅

1 Daily-02 [Set-A (Solve Sheet)] wm‡jevm: †f±i c~Y©gvb: 30 †b‡MwUf gvK©: 0.25 mgq: 20 wgwbU 1. A   B  = 0 n‡j, A  I B  Gi ga ̈eZ©x †KvY,  = ? [If A   B  = 0 what is the angle between A  and B  , α?] sin–1 (–1) tan–1 (1) sin–1 (1) cos–1 (–1) DËi: cos–1 (–1) e ̈vL ̈v: A   B  = 0 hLb  = 0 Zvi gv‡b A  I B  †f±i ci ̄ú‡ii mv‡_ 0 †KvY •Zwi K‡i, A  B  hw`, A  I B  ci ̄úi wecixZ w`‡K wμqvkxj nq, A  B  A_©vr A  I B  †f±i ci ̄ú‡ii mv‡_ 180 †KvY Drcbœ K‡i, |A |   B  = |A|  |B|  sin 180 = 0 2. `ywU †f±‡ii † ̄‹jvi ̧Yd‡ji gvb 6 3 Ges †f±i ̧Yd‡ji gvb 18, †f±i؇qi ga ̈eZ©x †KvY KZ? [If the dot product of two vectors is √ and the cross product of the same two vectors is 18, what is the angle between the two vectors?] 30 60 90 45 DËi: 60 e ̈vL ̈v: A  .B  = 6 3  ABcos = 6 3 .... (i) A   B  = 18 ABsin = 18 ........ (ii) (ii)  (i) tan = 3  = 60 Shortcut : ga ̈eZ©x †Kv‡Yi m~Î :  = tan–1 †f±i ̧Yd‡ji gvb † ̄‹jvi ̧Yd‡ji gvb = tan–1 18 6 3 = tan–1 3 = 60 †KvY g‡b ivLvi Dcvq : 2 4 0 4 1 4 3 4 4 4 4 0 1 2 3 Step 1. cÖ_‡g 5wU AvOzj‡K 0 n‡Z 4 ch©šÍ ch©vq μ‡g count Ki‡Z n‡e| 2. Gici ch©vqμ‡g 4 w`‡q fvM K‡i, 0 4 , 1 4 , 2 4 , 3 4 , 4 4 Gici eM©g~j Ki‡j, 0 4 , 1 4 , 2 4 , 3 4 , 4 4 h_vμ‡g, sin0, sin30, sin45, sin60 I sin90 Gi gvb cv‡ev| ch©vqμ‡g Dëv‡j cos Gi gvb cv‡ev| Gici sin cos Ki‡j tan cv‡ev| 3. (P )  . Q  2 – |P |   Q  2 = ? P 2Q 2 P 2Q 2 cos2 0 P 2Q 2 sin2 DËi: P 2Q 2 cos2 e ̈vL ̈v: {|P| }  |Q|  cos 2 – {|P| }  |Q|  sin 2 = P2Q 2 cos2  – P 2Q 2 sin2  = P2Q 2 (cos2  – sin2 ) = P2Q 2 cos2 [cos2  – sin2  = cos2] 4. P  = 2i  + 3j  + k  †f±iwUi Q  = 2i  + 4k  †f±‡ii Awfgy‡L AskK wbY©q Ki| [If vector P = 2i + 3j + k and vector Q = 2i + 4k, find the projection of P onto Q.] 4 5 (i )  + j  + k  1 2 (2i )  + 3j  + k  2 5 (2i )  + 4j  2 5 (2i )  + 4k  DËi: 2 5 (2i )  + 4k 