Nội dung text 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.)
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