23 Vectors 4 APQ Soln (RVHS)

30 Additional Practice Questions 1. N92/II/15 (partial) The equation of a plane is 6 2 5,x y z   and the point A has coordinates (3,–12,1). Write down (i) a vector perpendicular to , (ii) a vector equation for the straight line  which passes through A and is perpendicular to . Find the coordinates of the point of intersection of  and , and hence show that the perpendicular distance from A to  is 2 41.

31 2. N90/II/15 The point A has coordinates (3, 1 ,5) and the line  has equation 8 6 0 1 1 4 t                  r . Find the coordinates of the point B on  such that AB is perpendicular to . The plane  has equation 1 1 15 3        r . Find the coordinates of the point C where  intersects . Find a vector perpendicular to the plane ABC. Hence show that the acute angle between  and the plane ABC is 68, correct to the nearest degree.

32 3. VJC/2018 MYE/Q6 The diagram above shows the vertical cross-section of a slab of glass in the form of a trapezoidal prism, where the top surface is a plane 1p and the left side of the glass is a plane 2p . C and D are points in 2p . The light from a particle placed at C travels in a straight line to B in the glass. The light is refracted at B and travels in a straight line to A (3,1,2) in the air. To an observer at A, the particle at C appears to be at D (0,7,4 ). The plane 1p has equation 0z and AB is parallel to 2 2 i j k . (i) Find the coordinates of B. [3] BC  is in the direction of 4 a  i j k , where a is a positive constant. The line BC makes an angle of 1 1cos 9      with the normal to 1p at B. (ii) Find the value of a. [3] (iii) Given that the distance BC is 18, show that the position vector of C is 6 19 2  i j k . [2] The plane 2p has equation 34 r n . (iv) Given that 2p is perpendicular to a plane with equation 2 7x y  , find n. [4] (i) Equation of p1 is 0 0 0 1 r        and equation on line AB is 3 1 1 2 , 2 2 r s s                     Solving them, 3 1 0 1 2 0 0 2 2 1 2 2 0 1 s s s                                       3 1 2 1 2 3 2 2 0 OB                                The coordinates of B is (2, 3, 0). A D C B 1p 2p

33 (ii)   1 2 2 2 2 4 0 0 1 11cos cos 9 4 1 1 1Realise that the scalar product is positive. Else we should use .9 1 1 9 17 17 81 8 or 8 NA 0 a a a a a a                                          (iii) Method 1: 4 8118 8 1616 64 1 1 2 2 8 6 3 16 19 0 2 2 BC OC OB BC                                                                 Method 2: 4 4 8 , 0 because is in the direction of 8 1 1 BC k k BC                       4 8 9 18 2 1 BC k k k           2 4 6 3 2 8 19 0 1 2 OC OB BC                                      (iv) Method 1: 0 6 6 1 7 19 12 6 2 4 2 6 1 CD                                              A normal vector to the plane = 1 2 1 2 1 2 1 0 5                             

34 2 1 0 1 : 2 7 2 34 5 4 5 p r                                , 1 i.e. 2 34 5 r        1 2 5 n        Method 2: Let x n y z       . 2 2 2 1 0 2 0 0 0 7 34 7 4 34 (since is in ) 4 6 19 34 6 19 2 34 (since is in ) 2 x y x y z x y y z D p z x y x y z C p z                                                              By GC, x = -1, y = 2, z = -5 1 2 5 n       

35 *4 TPJC 10/II/4 The figure below shows a cuboid positioned on level ground so that it rests on one of its vertices, O. The vectors i and j are on the ground. Given that 3 12 3OA  i j k  , 2 2OB i + j k  and 2 2OC i k  (i) Show that the position vector of X is 12 5 i j k . (ii) State the height of X above the ground. Hence find the angle between OX and the level ground, giving your answer to nearest 0.1o. (iii) Find the equation of plane BDX in the form of d r n (iv) Find the acute angle between planes BDX and OBDC. 4 (i) 3 1 2 2 1 12 3 4 ; 1 ; 0 2 0 3 1 2 2 1 OA OB OC                                                            3 2 1 12 0 12 3 2 5 OX OA AX OA OC                                          (ii) Height of X above the ground = 5 units Let  be the required angle. Hence, 5 5sin 1 144 25 22.5o OX         

36 (iii) 1 2 0 1 1 2 1 12 1 13 5 2 3 BD OC BX                                            1 1 13 0 13 2 1 3 13                              Therefore, equation of plane 13 2 13 : 2 1 2 54 13 2 13 BDX                              r (iv) Normal vector to plane OBDC is 1 2 4 1 OA          Let  be the required angle. 13 1 2 4 13 1cos 342 18                  76.7o 

37 5. JJC/2012/Prelim II/4 The planes 1 and 2 have equations 2 4  r i j k and 3 ( ) (4 )         r i j k i j k i j respectively, and meet in a line l1. (i) Find the acute angle between 1 and 2 . [3] (ii) Find a vector equation of l1. [2] (iii) The points A and B have coordinates 6, 3, 5 and 2, 3, 1 respectively. Find the length of projection of AB  onto the line l1. [2] The line 2l passes through the point C with position vector (2 1) 3p p  i j k and is parallel to 3 3q q i j k , where p and q are positive constants. Given that the perpendicular distance from C to 1 is 15 6 and that the acute angle between 2l and 1 is 1 2sin 6      , find the values of p and q. [6] (i) 2 1 4 1 1 1 4 1 0 5 n                              Acute angle between 1 and 2 1 1 1 2 4 1 5cos 1 1 2 4 1 5                                   1 14cos 6 42      28.1  (to 1 dp) (ii) 3 1 1 4 12 1 5                   2 : 1 4 12 4 5 12 5 x y z             r

38 1 : 2 4x y z   2 : 4 5 12x y z    1 4 3 Using GC, eqn of is 4 2 , where 0 1 l                      r . (iii) Given 6 3 5 OA         and 2 3 1 OB         , 2 6 4 3 3 0 1 5 6 AB                               Length of projection of AB  onto the line 1l 2 2 2 4 3 0 2 6 1 3 2 1                   6 14  3 14 7 1 : 1 2 4 1        r l2 : 3 2 1 3 3 p q p t q                    r Let D be a point on the plane 1 . Since 4 1 0 2 4 0 1                   , D is (4, 0, 0).  4 4 2 1 0 2 1 3 0 3 p p DC p p                                  Perpendicular distance from C to 1 = Length of projection of DC  onto the normal of 1

39 1 4 1 2 2 1 2 1 3 1 15 1 6 6 2 1 p DC p                                      . . ( 4) 2(2 1) 3 15 3 9 15 3 9 15 or