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 E F i j k

River Valley High School, Mathematics Department 2023 Vectors T4 (ii)       1 1262 1 4 1262 126 2 4 126 23 3 1264 2 378 2           ab ac ac bc c b c c c bc (iii) Area of parallelogram with adjacent sides OB and OC. (iv) (b – 4c) . (b – 4c) = 3|a|2 |b|2 – 8 b . c + 16|c|2 = 3|a|2 b . c = 10 8 0 10 . c 8cos | || | 1(2) 128.7 and not 51.3      b bc 4 Question (i)     a b c 0  ,,a b c lie on the same plane         a a b c a 0              a a a b a c 0        0 a b a c       a b c a (shown) … (A) Similarly         b a b c b 0              b a b b b c 0       b a b c       b a c b … (B) Question (ii) Consider,

River Valley High School, Mathematics Department 2023 Vectors T5 sin ... (1) sin ... (2) sin ... (3) BOC COA AOB          b c b c c a c a a b a b Take sin (2) :(3) sin COA AOB   c a c a a b a b sin from (A) : sin sin sin COA AOB COA AOB           ab ab Similarly, taking sin(1) :(3) sin BOC AOB   b c b c a b a b sinfrom (B) : sin sin sin BOC AOB BOC AOB           ba ab Thus, sin sin sinCOA BOC AOB       (proven) Q5 solutions    2 3 2 3PQ a b a b a b          2 2 3PR a b a b a b      Area of triangle PQR             1 2 1 332 1 3 9 32 4 (Since 0, 0, and ) PQ PR a b a b a a a b b a b b a b a a b b b a a b                         Since area of triangle PQR = b , 4 a b b . 4 sin 1sin (since 0, is obtuse)4 165.5 a b b ab           2 3 2 2 2 3 43 3 OP OROM a b a b ab     

River Valley High School, Mathematics Department 2023 Vectors T6       22 2 2 43 303 4 3 3 0 4 12 3 9 0 4 9 9 0 9 9 cos 4 0 ab ab a b a b a a b a b b a b b bb                 Since 1 15sin ,cos44 θθ 2 9 159 4 0 4bb   By GC, 1.31b  or 0.340b  Since 0,b  0.340b  6 1 12       17 25 1       7 0 (1) 5 2 (2) 2 2 (3)       Solving (1) and (2): 1, 7 (3): 2(7)+1=15 2 No unique solution and the lines are not parallel. Therefore, they are skew lines. Let 1 12 OM       for some . 11 1 0 2 1 0 2 2 2 2 AM                                     2 4 4 0 1          1 | | 1 2 0 AM      units

River Valley High School, Mathematics Department 2023 Vectors T7 OR Let B = (1, 0, 1). 0 2 2 BA    Shortest distance = 0 1 2 11 2 1 2 2 1 1 4 6 2 2 2                                  7(i)     22 22 2 a 2 b 16 36 64 2 4 9 16 80 36 4 13 16 28 28 1 or 1 Reject 0 1 pp pp p p p p p               (ii) Length of projection of a on b (iii)   222 24 36 481 42 292 3 4                       bab 8(i) Since ABba and A, B and C are collinear, 2k 2 2 3 2OC OB BC     b+ b a= b a

River Valley High School, Mathematics Department 2023 Vectors T8 (ii) 2 2 3 4 3 44 16 3 4 4 3 33 .    aab a a a a Alternatively,   ^^ 2 2 2 2 3 4 43 4 3 44 16 3 4 4 3 33        b a aa b a a a aa a a a a (iii)   222 0 b 2 2 2 2 3 a.b 4 1cos a b 2 43 233 or 603              (iv) A B C O N D E

River Valley High School, Mathematics Department 2023 Vectors T9     22 2 32 233 2 DC DADE        a a b b = Alternatively,  22 3 2 2 233 2 3 2 2 2 2 2 AE AE DE DA AE CO AE                  b a a b a a b b a b 9(i) 2 0 2 1 1 2 4 1 3 AB                                2 0 2 1 1 2 0 1 1 AC                                 2 2 8 2 2 2 4 4 1 3 1 8 2 AB AC                                                Choose normal vector pn for plane p = 2 1 2   . 2 0 2 : 1 1 1 3 2 1 2 p                                 r A cartesian equation of the plane is 2 2 3p x y z    (ii) Let the acute angle between l and p be  . The angle between the normal vector pn (for plane p) and the direction vector lm (for line l),

River Valley High School, Mathematics Department 2023 Vectors T10 11 21 10 22 6cos cos 26.56521 35 10 22                                         90 26.565 63.4 to 1d.p.     or 1.11 rad Alte