2. 2022 Vectors (Lines) Discussion Solution

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: VECTORS Page | 1 H2 MATHEMATICS (9758) TOPIC VECTORS (LINES) 2022/JC1 DISCUSSION 1 Find a vector equation and cartesian equation of the following lines: (a) passing through the point with position vector 7i + 2j – 4k and parallel to to i – 3j + k (b) passing through the points (2, –2, 1) and (0, 4, 9 ) (c) passing through the point with position vector 7 i and parallel to the line r = (2 – p) i + pj + 5k. Solution: (a) Vector equation of line: 71 r 2 3 , 41          = + −      −    Cartesian equation of line: 7 2 ( 4) 1 3 1 274 3 − − − −==− −− = = +− x y z yxz (b) 2 0 2 1 Direction of line = 2 4 6 2 3 1 9 8 4 −               − − = − =−              −        Equation of line: 21 r 2 3 , 14   −       = − +          Cartesian equation of line: 2 ( 2) 1 1 3 4 212 34 − − − −==− +−− = = x y z yzx (c) Equation of parallel line: 2 2 2 1 r 0 0 1 5 5 0 5 0 − − −                   = = + = +                            pp p p p 1 Direction of line = 1 0 −   Equation of line: 71 r 0 1 , 00   −       =+          Cartesian equation of line: 70 , 011 7 , 0 −− ==− − = = xy z x y z

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: VECTORS Page | 2 2 Convert the equations of following lines to their vector equation form: (a) 13 2 6 2 −= − = zyx (b) 5 2 4 , 3+ = − =x y z Solution: (a) 13 2 6 2 −= − = zyx 2 6 3 1 1 1 2 3 0 1 11 2 23 −− == − − −== x y z x y z Vector equation of line: 1 2 '' 1 3 3 r 0 , 12   =+     33 r 0 2 , 1 12          =+          (b) 5 2 4 , 3+ = − =x y z ( 5) 4 2 , 311 1 ( 5) 2 , 311 4 − − −== − −−− == − xy z yx z Vector equation of line: ''11 24 51 r , 30   −       = + −          1 2 54 r 1 , 30   −       = + −         

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: VECTORS Page | 3 3 Given three points ( )0,2,7A , ( )5, 3, 2B − and ( )1, 1, 1C , find the position vector of the point R on AB such that CR is perpendicular to AB . Hence find the perpendicular distance from C to AB and the position vector of the reflection of C in AB . Solution: 5 0 5 1 3 2 5 5 1 2 7 5 1 AB −               = − − = − =−              −        Equation of line AB : 01 r 2 1 , 71  −       =+           Since R lies on line AB , 01 2 1 2 7 1 7 OR    −−           = + = +           +      for a  value. 11 2 1 1 7 1 6 CR OR OC    − − −           = − = + − = +          ++      Since line CR AB⊥ , 11 1 1 0 61    − − −       +  =      +    6 3 1 1 0 8     + + + + = =− +  ( ) ( ) 8 3 8 3 8 3 8 122 37 13 OR    = + − = −  +−  Perpendicular distance from C to AB ( ) ( ) ( ) 8 3 8 3 8 3 11 5511 33 6u 62 nitsCR − − −    = = + − = − =  +−  Let 'C be the reflection of C in line AB . Using mid-point theorem, ( ) 1 '2OR OC OC=+ '2 8 1 13 112 2 1 733 13 1 23 OC OR OC=−              = − − = −                  R C 1 1 1 B 5 −3 2 A 0 2 7 C’

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: VECTORS Page | 4 4 Given a line with vector equation ( ) ( )2 3 2 5 , = + + + ++r i j k i j k , (i) show that point ( )3, 6,13A lies on the line, (ii) find the perpendicular distance from point ( )1, 7, 4B to the line. Solution: (i) 3 1 1 6 2 2 , 13 3 5             =+                 3 21 + == 6 2 2 2=+ = 13 3 5 2=+ =  point A lies on the line. (ii) Let point C be 1 2 3    and 1 m2 5  = . 1 1 0 7 2 5 4 3 1 CB            =−=                0 5 301 0 1 5 1 23 1 1 5 1 555 3 m 1 1 2 5 1 2 30 5 30 30 units7 2 h CB=  =   =   = −        = = ______________________________________________________________________________

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: VECTORS Page | 5 5 Find whether the following pairs of lines are parallel, intersecting o r skew. If they intersect, find the point of intersection and the acute angle between the lines. (a) 13 1 4 , 11         = − +          r and 19 2 12 , 33         =+          r . (b) 12 0 1 , 31 ss        =+          r and 21 1 2 , 10 tt        = − + −          r . (c) ( )8 ,4 23 = + + + + + ri jkj ik and ( )67 6 5 4 5 ,= + + + + + r i j ki j k . (d) 42 2 1 , 12         =+       −   r and z-axis. [Hint: The z-axis has equation 0 0, 1   =     r .] Solution: (a) Since 93 12 3 4 31        =          , the two lines are parallel lines. (b) Since 21 12 10 k         −      for any k, the 2 lines are not parallel. Assuming that the two lines intersect, 1 2 2 1 0 1 1 2 3 1 1 0 st                + = − + −                      11 2 2 2st st+ = +  − = ⎯ (1) 1 2 2 1s t s t=− −  + =− ⎯ (2) 3 1 2ss+ =  =− ⎯ (3) Solving (2) and (3): 2s=− and 1 2t = . Substitute 2s=− and 1 2t = into (1): ( ) 19L.H.S. 2 2 22 R.H.S. L H.S.=1 . = − − =− . Since the two lines are not parallel and they do not intersect, they are skew lines.

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: VECTORS Page | 6 (c) Since 16 24 15 k                    for any k, the 2 lines are not parallel. Assuming that the two lines intersect, 4 1 7 6 8 2 6 4 3 1 5 5                 + = +                      346 67   ++ −= = ⎯ (1) 8 2 6 4 2 4 2   + = +  − =− ⎯ (2) 3 5 5 5 2   + = +  − = ⎯ (3) Solving (1) and (2): 3 =− and 1 =− Substitute 3 =− and 1 =− into (3): ( )L.H.S. 3 5 1 2 R.H.S. 2 L.H.S. =− − − = == Hence the two lines intersect. Position vector of intersection is 4 1 1 8 3 2 2 3 1 0            −=                . Point of intersection is ( )1,2,0 . Let the angle between two lines be  . ( ) ( ) 1 1 16 11cos 2 4 6 7715 19co 6 77 0.487 rad to 3 s.f. or 27.9 to 1 d.p. s  − −                        =  = = The angle between the two lines is 27.9 .

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: VECTORS Page | 7 (d) Since 20 10 21 k               −    for any k, the 2 lines are not parallel. Let the vector equation of the line on the z-axis be 0 r 0 , 1   =     Assuming that the pair of lines intersects, -axis 4 2 0 2 1 0 12 1 z             +=          −      4 02+ = ⎯ (1) 20 += ⎯ (2) 12 −= ⎯ (3) Solving (1): 2=− Solving (