23 Vectors 3 APQ Soln (RVHS)

13 Additional Practice Questions 1. (J92/I/6) The lines 1l and 2l have equations 3 1 1 2 , 0 4 t t                   r and 1 2 1 1 , 1 1 s s                    r  Show that 1l passes through the point 2, 1, 4  but that 2l does not pass through this point. Find the acute angle between 2l and the line joining the points 1, 1, 1 and 2, 1, 4  , giving your answers correct to the nearest degree. [Solution] Since 2 3 1 1 1 2 1 4 0 4 t t                               , 1l passes through the point 2, 1, 4  . Since we cannot find a value s such that 2 1 2 1 1 1 4 1 1 s                               , thus 2l does not pass through the point 2, 1, 4  . Acute angle between 2l and the line joining the points 1, 1, 1 and 2, 1, 4  1 1 2 11 1cos 1 06 261 5 7cos 55.9 566 26                          2. (N89/I/16) The points A and B have position vectors 4 2 4 i j k and 3 6 6 i j k respectively, relative to the origin O. The point P is such that 1 2OP OA    and the point Q is such that 1 3OQ OB   . The point C is such that OACB is a parallelogram. (i) Find the length AB and the cosine of the angle between OA  and OB . (ii) Find a vector equation, in parametric form, of the line PQ. (iii) Find the position vector of the point of intersection of PQ and OC. [Solution]

14 4 2 4 OA         , 3 6 6 OB         (i) 1 4 2 AB         . Thus 21AB  . cos 48 8 936 81 OA OBAOB OA OB             (ii) 4 21 2 12 4 2 OP                    , 3 11 6 23 6 2 OQ                    1 1 0 PQ         Vector equation of line PQ is 2 1 1 1 , 2 0                     r  (iii) 7 8 10 OC OA OB                Vector equation of line OC is 7 8 , 10          r  2 1 7 1 1 8 2 0 10                              2 7 3 11 8 , 5 52 10                 position vector of the point of intersection is 71 85 10       .

15 . Relative to an origin O, points A and B have position vectors 2i j and 3 bj k respectively, where b > 0. Given that the angle between a and b is 120, find the exact value of b. For the following questions, take b = 2. Write down a vector equation of the line passing through A and B. (i) Find the position vector of the point P on the line AB such that OP  is perpendicular to .AB  (ii) A point Q is on AB produced. Find the position vector of Q such that AB BQ . Solution: 1 2 0 OA         , 0 3OB b         2 2 2 2 2 1 0 2 3 1 4 0 0 9 cos120 0 6 5 9 ( 0.5) 12 5(9 ) 99 5 99 since 05 b b b b b b b                                   0 3 2 OB         , 1 5 2 AB         1 1 : 2 5 , 0 2 ABl t t                    r  (i) Since P is on lAB, 1 2 5 2 t OP t t           for some t. 0 1 1 2 5 5 0 2 2 1 10 25 4 0 30 11 11 30 OP AB t t t t t t t t                                  

16 19 30 191 1 or 56 30 2211 15 OP                          (ii) Since Q is on lAB, 1 2 5 2 t OQ t t           for some t. AB = BQ:   2 2 2 2 2 2 2 1 1 25 4 2 5 2 30 1 ( 5 5 ) (2 2) 30 1 2 25 50 25 4 8 4 30 60 0 ( 2) 0 t t t t t t t t t t t t t t t t                               So, t = 2 or t = 0 (rej. as it gives point A) 1 8 4 OQ           3. Relative to the origin O, the points A, B and C have position vectors 6i + 5j + 11k, –3i + 5j – k, –4i + 10j + 6k, respectively. (i) Find the Cartesian equation of the line AB. (ii) Find the length of projection of AC  onto the line AB. (iii) Hence, or otherwise, find the perpendicular distance from C to the line AB, and the position vector of the foot N of the perpendicular from C to the line AB. (iv) The point D lies on the line CN produced and is such that N is the mid-point of CD. Find the position vector of D. Solution: (i) 9 3 0 3 0 12 4 AB OB OA                          Hence, a vector equation of lAB: r = 6 3 5 0 11 4                  ,   Cartesian equation of lAB: 6 11 , 53 4 x z y   .

17 (ii) length of projection of AC  onto the line AB = ACm  10 3 1 15 0 ( 30 0 20) 1059 0 165 4                         units (iii) AC = 100 25 25 150   ACN is a right angled triangle. 2 2 2 2 210 150 CN AN AC CN      perpendicular dist.(CN) = 2150 10 50 5 2   units ( 10) 6 3 0 15 10 0 5511 4 3 AN AC ON OA ON                                          m m m           and N(0, 5, 3) . Otherwise method Since N is a point on the line ABl , So 6 3 5 0 11 4 ON                     for some value of to be determined. Then 4 6 3 10 3 10 5 0 5 6 11 4 5 4 NC                                                      and since NC  and 3 0 4       are perpendicular, so 3 0 0 4 NC          10 3 3 5 0 0 5 4 4                        50 25 0    2 Thus 6 3 0 5 2 0 5 11 4 3 ON                               and N(0, 5, 3) . A N C lAB AB  10 D

18  perpendicular dist.(CN) = 4 5 50 3 NC           5 2 units. (iv) Take note that point D is the reflection of point C about lAB. 2 4 0 4 4 10 2 5 10 0 6 3 6 0 OD OC CN                                                      . ACJC/2017/Prelim Paper2/Q4(a) (i) The unit vector d makes angles of 60 with both the x- and y-axes, and  with the z-axis, where 0 90   . Show that d is parallel to 2 i j k . (ii) The line m is parallel to d and passes through the point with coordinates (2, 1, 0) . Find the coordinates of the point on m that is closest to the point with coordinates (3, 2, 0). Solution: (i)   2 2 2 2 1 1 1 4 4 2 1 2 cos 60 cos 60 cos cos 60 cos 60 cos 1 cos 1 cos is acute                     d i j k  1 1 1 2 2 2 // 2    d i j k i j k (a)(ii) 2 1 1 1 0 2 :m                  r 22 2 2 1 3 1 1 1 1 1 1 2 1 3 1 1 0 0 0 2 2 2 2 0 0 ( 1 3) (1 1 2 ) 0 1                                                                                                    Therefore position vector of point is 2 1 3 1 1 0 0 2 2                           Coordinates = 3,0, 2 OR

19   3 2 1 1 2 1 1 1 0 0 2 2 1 1 2 1 1 2 1 1 2 2 1 3 1 1 0 0 2 2 AN AP ON OA AN                                                                                           d d          *4. (modified from IJC/2011/II/4) Relative to origin O, the position vectors of two points A and B are a and b respectively. The vector a is a unit vector which is perpendicular to a + 3b. The angle between a and