TMJC H2 Chapter 6A 3D Vector Geometry Lines Assignment Solutions 2024

Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 1 of 5 H2 Mathematics (9758) Chapter 6A 3D Vector Geometry (Lines) Assignment Solutions 1 N2008/I/11(modified) The cartesian equation for two lines are 2 5 1 3 4 and .1 2 1 1 3 1 x y z x y z           (i) Show that the lines intersect and state the point of intersection. [4] (ii) Find the acute angle between the lines. [2] Q1 Solutions (i) 0 1 2 5 2 21 2 1 1 , 5 x y z                              r  1 1 1 3 4 3 ,31 3 1 4 1 x y z                               r  At the point of intersection, 1 2 2 3 3 for some , 5 4                                 1 (1) 2 2 3 3 (2) 5 4 (3)                Solving using GC, 4, 3   Since there is a consistent value of and   satisfying the equations, the two lines intersect at the point. Point of intersection:  4,6,1 (ii) Let  be the acute angle between the two lines. 1 1 1 1 2 3 1 1 8cos cos 10.0 6 11 6 11                         (1 d.p.)

Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 2 of 5 2 2011/SRJC/II/2(modified) With respect to origin O, the position vector of the point A is 2i – 2j – 16k and the line l1 has equation 1 2 0 2 , 4 3                        r  . (i) Find the exact length of projection of OA   onto line l1 . [2] (ii) Find the position vector of the foot of the perpendicular from the point A to the line l1. [4] (iii) Find the perpendicular distance from the point A to the line l1. [1] (iv) Find the position vector of the point A’, which is the reflection of point A in the line l1. [2] Q2 Solutions (i) Required length of projection   22 2 2 2 2 2 16 3 2 2 3                       48 17  (ii) Let F be the foot of the perpendicular from A to l1. Since F lies on l1 1 2 2 1 2 2 2 2 2 4 3 16 12 3 AF OF OA                                                   1 2 2 2 2 2 0 12 3 3                         (iii) The perpendicular distance from A to l1  1 2 2 ,for some 4 3 OF                 0·AF d AF d            2 4 36 4 4 9 0 2           1 2(2) 5 2(2) 4 4 3(2) 10 OF                         5 2 3 4 2 6 10 16 6 AF OF OA                                             2 2 23 6 6 81 = 9 unitsAF       F A (2, -2, -16) l2

Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 3 of 5 (iv) F is the mid-point of AA’ So, ' 2 OA OAOF       5 2 8 ' 2 2 4 2 10 10 16 4 OA OF OA                                        

Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 4 of 5 3 N2019/II/5 With reference to the origin O, the points A, B, C and D are such that OA a   , OB b   , 2 4OC a b   and 5OD b a   . The lines BD and AC cross at X (see diagram). (i) Express OX   in terms of a and b. [4] The point Y lies on CD and is such that the points O, X and Y are collinear. (ii) Express OY   in terms of a and b and find the ratio  :OX OY . [6] Q3 Solutions (i)  5 5 2 4 4 BD AC          b a b a a b a a b       , 4 : : , BD AC l l           r b r a a a b   Since the lines BD and AC cross at X, we equate the two lines:  4 .    b a a a b Since a and b are non-zero and non-parallel vectors, 11 4 4 51 4             Substitute 5 4   into BDl to be the position vector of X which is the point of intersection of the two lines. 5 4 OX  b a   O B A X C D a b Note the presentation of equation of lines: - Label it clearly so that you are able to identify it easily for future uses - BDl  b a is wrong presentation. Concept: Point of intersection between 2 intersecting lines Learning Point: If a and b are non-zero and non- parallel vectors and ba   for some ,     , then 0  and 0  .

Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 5 of 5 (ii)  5 2 4 3 3 CD      b a a b a b    ,: 2 4CDl      r a b a b  Since Y lies on CDl ,   for some 2 4OY      a b a b    . Given that points O, X and Y are collinear, let 5 4 OY OX        b a      5 2 44          b a a b a b Since a and b are non-zero and non-parallel vectors,     4 1 5 2 24           Using GC, 8 4,3 3 8 5 3 4 8 3 : 3: 8 OY OY OX OX OY            b a       Alternative: Observe that point Y is the point of intersection between line CD and line OX.  ,: 2 4CDl      r a b a b  5 ,4:OXl        b ar  Since the lines CD and OX cross at Y, we equate the two lines:  5 2 44          b a a b a b Since a and b are non-zero and non-parallel vectors,     4 1 5 2 24           Using GC, 8 4,3 3 8 10 8 5 3 3 3 4 8 3 : 3:8 OY OX OX OY               b a b a     O B A X C D a b Y Learning Point: Points A, B, and C are collinear  AB AC     , for some , 0   . Take note of presentation: There is no division of vectors, hence writing 5 4 8 10 3 3 3 8 OX OY OX OY     b a b a         is WRONG