23 Vectors 1 BMQ Soln (RVHS)

1 Tutorial 8A: Vectors I (Vectors in Two and Three Dimensions) Basic Mastery Questions 1. If P is the mid-point of AB  and Q is the mid-point of XY , show that 2PQ AX BY      . PQ  = PA AX XQ        = 1 1 2 2BA AX XY        2PQ  = 2BA AX XY       = ( )AX BA AX XY          = AX BY    (shown) Or PQ  = 1 2 PX PY    2PQ  = PA AX PB BY          = AX BY    (shown) 2. ABCDEF is a regular hexagon inscribed in a circle centred at O. Given that AB  = p and BC = q, express the following in terms of p and q : (i)AD , (ii)CD , (iii)AE , (iv)FA  . Each interior angle of the regular hexagon. = 6 2 180 1206     (i) ~ 2AD q  (ii) CD  = BA AO     = ~ ~ p q  (iii) AE  = AD DE    = ~ ~ 2q p (iv) FA   = DC  = ~ ~ p q A B X Y P Q Or PQ  = OQ OP    = 1 2 OX OY    – 1 2 OA OB    2PQ  = OX OY    – OA OB    = OX OA    + OY OB    = AX BY    (shown) 60o 60o p q O A B F E D C

2 3. A, B, C, D are points with position vectors 2j k ,  i j , 4i k and 3 2 i j k respectively. Prove that the triangle ABC is right-angled and that triangle ABD is isosceles. [Note: The question can be solved using Dot Product (in Vectors II).] ~ 0 1 2 a       , ~ 1 1 0 b        , ~ 4 0 1 c       , ~ 3 1 2 d       AB  = ~ ~b a = 1 2 2        AB  = 1 4 4  = 3 BC  = ~ ~c b = 5 1 1        BC  = 25 1 1  = 27 AC  = ~ ~c a = 4 1 1        AC  = 16 1 1  = 18 AB 2 + AC 2 = 32 + 182 = 9 + 18 = 27 = BC 2  triangle ABC is right-angled (shown) AD  = ~ ~d a = 3 0 0        AD  = 3 Since AB  = AD   triangle ABD is isosceles (shown)

3 Tutorial Questions 1. The points A, B, C, D have coordinates (0, 1, 3), (4, 5, –5), ( –3, 0, –1) and (7, 5, 4) respectively. Show that the point P which divides AB in the ratio 1 : 3 also divides CD in the ratio 2 : 3. P divides AB in the ratio 1:3: 1 34OP OA OB        = 0 41 3 1 54 3 5                       = 1 2 1       CP   = 1 3 2 0 1 1                 = 4 2 2       PD  = 7 1 5 2 4 1                 = 6 3 3       3 2PD CP     //PD CP    Hence, is on (since is a common point) and :2:3P CD P CP PD  (shown) 2. ABCD is a parallelogram with E as the midpoint of AB and F lies on DE such that 3DE FE   . Prove that A, F and C are collinear and 3AC AF   . 1 23AF AD AE       1 3 AD AB     1 3 AD DC     ( AB DC    ) 1 3AC   //AF AC     A, F and C are collinear since A is a common point, & 3AC AF    (shown) A B P 1 : 3 A B D C E F 1 2 C D P 2 : 3