10.Vectors Solutions 2023 (RVHS)

River Valley High School, Mathematics Department 2023 Vectors T1 Solutions (Vectors) 1 (i) 02 1 , 0 0 02 11 x z y r                         Vector parallel to 1 1 0 1 1 0 1 0 1 1                                 1 1 2 1 1 0 3 1 1 2 1 0 1 1 : 3 0 3 3 2 2 1 2 2 n rr                                                                         (ii) Let foot of perpendicular from Q to 1 be N. 31 13 3 62 ON                   31 13 3 3 2 6 2 2 4                          1 1 2 ON   (iii) 1 3 5 ' ' 2 1 13 112 2 6 10 OQ OQON OQ                                  22 5 1 6 3 ' 11 1 10 2 5 10 0 10 5 1 3 5 3 : 1 5 OR : 11 5 0 5 10 5 PQ l r l r                                                                                         

River Valley High School, Mathematics Department 2023 Vectors T2 (iv) 12 1 : 3 2 : 6 24 a r r b                     1 3 2    = 62 4 a ka    Method 1: 1 2 2 1 1 1 : 3 2 : 2 3 : 3 22 2 2 br r b r                                    Distance between the 2 planes =  22 224 1 9 4 b   108 or -116b Method 2: Distance QN =         22 2 3 1 13 1 6 2 224        2 3 2 2 5 2 6 13 6 108 OR ' 6 11 6 116 4 6 4 4 10 4 b OQ b OQ                                                                     Method 3 1 11 12: 3 2 3 14 1422 rr                      2 1 12: 3 224 14 142 1 3 2 56 54 or 58 108 or 116 2 r rb                (v) 2,ab  or a = 2, b = –4

River Valley High School, Mathematics Department 2023 Vectors T3 2 '4 tan ' 3 4 ' 433 FF FF F F k       ' ' 3 4 10OF OB BF F F j k i      (Shown) 10 0 6 3 04 DB DE                     10 0 24 12 6 3 40 2 20 0 4 30 15 n                                            12 10 20 3 15 4sin 769 125 50.7228.. 50.7                     3 (i)           4 4 4 4               a b a c a b a c 0 a b a c 0 a b c 0 These are the three possible conclusion that can be drawn from the above equation. 1) a = 0 2)  4bc = 0 3) a is parallel to b – 4c , hence b – 4c =  a We will present no. 3) as we are expected to show the given expression in (i). O B C D