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               