23 Vectors 4 BMQ Soln (RVHS)

1 Tutorial 8D: Vectors IV (Planes & 3D Geometry Problems) Basic Mastery Questions 1. Find the vector equations in parametric form, vector equations in scalar product form and the Cartesian equations of the planes containing (i) the point (0, 1, 1) and the two vectors 2 i j k and  i k; (ii) the points (1, 0, 1), (1, 2, 1) and (1, 1, 0); (iii) the point (1 , 2, 3) and the line with equation 2 7 2     r i k i j k ; (iv) the lines r = k + s(i – 3j) and r = k + t(2j + 5k).

2 2. A line  has equation 2 2    r j k i j k . Find the vector equation of the plane containing the point A(1, 3, 1), perpendicular to the plane OXZ and parallel to . 3. N89/II/15(partial) The plane  has equation 3 2 1 0x y z    and the line  has Cartesian equation 4 3 7 .1 2 1 x y z    Show that  lies in . Method 1 Method 2 If  lies in , then 4 1 3 2 7 1                    r must satisfy 3 2 1 1        r for all . Substituting 4 1 3 2 7 1                    r into 3 2 1 1        r : 4 1 3 4 3 3 2 2 3 2 2 7 1 1 7 1                                                                  12 3 6 4 7 1         which is true for all .  lies in  (shown).

3 4. IJC/2018/CT/10 In a particular experiment, Scott shoots a laser beam from point A with coordinates 9, 1, 5 towards a plane Π with equation 5 8 4x y z   . The laser beam travels in a line L with equation 4 3 65 2 x y z    . Find (i) the acute angle between L and Π. [3] (ii) the coordinates of the point where the laser beam meets the plane and deduce the shortest distance from A to Π. [5] Immediately after the laser beam meets the plane, it is being reflected as line M such that the angle between L and Π equals to the angle between M and Π. Find the equation of the line M. [5] Q4 Solution (i) Equation of line L: 4 3 65 2 4 5 3 2 , 6 1 x y z                           r  2 2 2 2 2 2 1 Let be the acute angle between and . 5 5 2 1 1 8 15sin 27005 2 1 5 1 8 15sin 16.8 (to 1 d.p.)2700 o L                                  

4 (ii) Let be the point of intersection between and 4 5 3 2 for some (1) 6 1 4 5 5 3 2 1 4 6 1 8 71 15 4 5 4 5 3 5 2 6 1 B L OB OB                                                                                  21 13 11 i.e. coordinates of is 21,13, 11 .B                2 2 2 21 9 30 13 1 12 11 5 6 Shortest distance sin 1530 12 6 2700 215 5 3 10 9.49 (3.s.f ) AB AB                                              Alternative:

5 Shortest distance = 5 5 5 1 6 2 1 8 1 8 90 90 6( 25 2 8) 90 90 90 3 10 AB                                     (iii) Let be the foot of perpendicular from to 9 5 1 1 for some (1) 5 8 9 5 5 1 1 1 4 5 8 8 86 90 4 1 9 5 1 1 5 8 F A OF OF                                                                                     4 0 3        

6 Let be the reflection of in 2 2 4 9 1 2 0 1 1 3 5 11 21 1 20 10 13 1 14 2 7 11 11 22 11 Equation A A OA OAOF OA OF OA OA A B                                                                                       d of line : 1 10 21 10 1 7 , 13 7 , 11 11 11 11 M OR                                                   r r   OR 2 BA BABF    2 4 21 30 2 0 13 12 3 11 6 10 2 7 11 BA BF BA                                              