T-Leng Tuition · mugcornersg.com
Total: 30 marksSuggested time: 45 minutes
Name: ______________________________ Class: ____________ Date: ____________
Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.
Section A 1 (a) P Q → = 2 a + 5 b \overrightarrow{PQ} = 2\mathbf{a} + 5\mathbf{b} P Q = 2 a + 5 b and R S → = 4 a + 10 b \overrightarrow{RS} = 4\mathbf{a} + 10\mathbf{b} R S = 4 a + 10 b . Find the value of k k k such that R S → = k P Q → \overrightarrow{RS} = k\overrightarrow{PQ} R S = k P Q .
Answer (a) [2]
(b) P Q → = 4 a + b \overrightarrow{PQ} = 4\mathbf{a} + \mathbf{b} P Q = 4 a + b and R S → = a + 4 b \overrightarrow{RS} = \mathbf{a} + 4\mathbf{b} R S = a + 4 b . State whether P Q PQ P Q and R S RS R S are parallel.
Answer (b) [1]
2 (a) P Q → = a − 3 b \overrightarrow{PQ} = \mathbf{a} - 3\mathbf{b} P Q = a − 3 b and R S → = 2 a − 6 b \overrightarrow{RS} = 2\mathbf{a} - 6\mathbf{b} R S = 2 a − 6 b . Find the value of k k k such that R S → = k P Q → \overrightarrow{RS} = k\overrightarrow{PQ} R S = k P Q .
Answer (a) [2]
(b) P Q → = 3 a + 2 b \overrightarrow{PQ} = 3\mathbf{a} + 2\mathbf{b} P Q = 3 a + 2 b and R S → = − 6 a − 4 b \overrightarrow{RS} = -6\mathbf{a} - 4\mathbf{b} R S = − 6 a − 4 b . State whether P Q PQ P Q and R S RS R S are parallel.
Answer (b) [1]
3 (a) a = ( − 10 8 ) \mathbf{a} = \begin{pmatrix} -10 \\ 8 \end{pmatrix} a = ( − 10 8 ) . Find − 1 2 a -\dfrac{1}{2}\mathbf{a} − 2 1 a .
Answer (a) [2]
(b) a = ( − 6 2 ) \mathbf{a} = \begin{pmatrix} -6 \\ 2 \end{pmatrix} a = ( − 6 2 ) and b = ( − 2 5 ) \mathbf{b} = \begin{pmatrix} -2 \\ 5 \end{pmatrix} b = ( − 2 5 ) . Find ∣ 3 a − 2 b ∣ |3\mathbf{a} - 2\mathbf{b}| ∣3 a − 2 b ∣ , correct to 3 significant figures.
Answer (b) [2]
4 (a) O A → = ( − 4 − 3 ) \overrightarrow{OA} = \begin{pmatrix} -4 \\ -3 \end{pmatrix} O A = ( − 4 − 3 ) and A B → = ( − 1 4 ) \overrightarrow{AB} = \begin{pmatrix} -1 \\ 4 \end{pmatrix} A B = ( − 1 4 ) . Find the coordinates of B B B .
Answer (a) [2]
(b) O A → = ( 2 6 ) \overrightarrow{OA} = \begin{pmatrix} 2 \\ 6 \end{pmatrix} O A = ( 2 6 ) and A B → = ( 4 − 7 ) \overrightarrow{AB} = \begin{pmatrix} 4 \\ -7 \end{pmatrix} A B = ( 4 − 7 ) . Find the coordinates of B B B .
Answer (b) [2]
Section B 5 a = ( 5 3 ) \mathbf{a} = \begin{pmatrix} 5 \\ 3 \end{pmatrix} a = ( 5 3 ) and b = ( − 3 − 4 ) \mathbf{b} = \begin{pmatrix} -3 \\ -4 \end{pmatrix} b = ( − 3 − 4 ) . (a) Find 3 a − 2 b 3\mathbf{a} - 2\mathbf{b} 3 a − 2 b .
Answer (a) [1]
(b) Find ∣ 3 a − 2 b ∣ |3\mathbf{a} - 2\mathbf{b}| ∣3 a − 2 b ∣ , correct to 3 significant figures.
Answer (b) [2]
(c) Given that c = ( h 9 ) \mathbf{c} = \begin{pmatrix} h \\ 9 \end{pmatrix} c = ( h 9 ) is parallel to a \mathbf{a} a , find the value of h h h .
Answer (c) [1]
6 In the diagram, O A → = a \overrightarrow{OA} = \mathbf{a} O A = a and O B → = b \overrightarrow{OB} = \mathbf{b} O B = b . P P P lies on A B AB A B such that A P : P B = 1 : 2 AP : PB = 1 : 2 A P : P B = 1 : 2 , and Q Q Q is the midpoint of O A OA O A . (a) Express A B → \overrightarrow{AB} A B in terms of a \mathbf{a} a and b \mathbf{b} b .
Answer (a) [1]
(b) Express O P → \overrightarrow{OP} O P in terms of a \mathbf{a} a and b \mathbf{b} b , as simply as possible.
Answer (b) [2]
(c) Express Q P → \overrightarrow{QP} QP in terms of a \mathbf{a} a and b \mathbf{b} b , as simply as possible.
Answer (c) [2]
7 Relative to the origin O O O , the position vectors of the points A A A and B B B are O A → = ( 2 2 ) \overrightarrow{OA} = \begin{pmatrix} 2 \\ 2 \end{pmatrix} O A = ( 2 2 ) and O B → = ( − 3 − 8 ) \overrightarrow{OB} = \begin{pmatrix} -3 \\ -8 \end{pmatrix} O B = ( − 3 − 8 ) . (a) Find A B → \overrightarrow{AB} A B .
Answer (a) [1]
(b) Find ∣ A B → ∣ |\overrightarrow{AB}| ∣ A B ∣ , correct to 3 significant figures.
Answer (b) [2]
(c) The point P P P lies on A B AB A B such that A P : P B = 3 : 2 AP : PB = 3 : 2 A P : P B = 3 : 2 . Find the position vector of P P P .
Answer (c) [2]
(d) The point E E E has coordinates ( 12 , h ) (12, h) ( 12 , h ) , and A A A , B B B and E E E lie on a straight line. Find the value of h h h .
Answer (d) [2]
END OF PAPER
Question 1
(a) Answer: 2 × 5 = 10 2 \times 5 = 10 2 × 5 = 10 [2] k × 2 = 4 k \times 2 = 4 k × 2 = 4 k = 2 k = 2 k = 2 2 × 5 = 10 2 \times 5 = 10 2 × 5 = 10 (b) Answer: NO [1]
1 4 \dfrac{1}{4} 4 1 4 1 = 4 \dfrac{4}{1} = 4 1 4 = 4 Parallel? No. Question 2
(a) Answer: 2 × ( − 3 ) = − 6 2 \times (-3) = -6 2 × ( − 3 ) = − 6 [2] k × 1 = 2 k \times 1 = 2 k × 1 = 2 k = 2 k = 2 k = 2 2 × ( − 3 ) = − 6 2 \times (-3) = -6 2 × ( − 3 ) = − 6 (b) Answer: YES [1]
− 6 3 = − 2 \dfrac{-6}{3} = -2 3 − 6 = − 2 − 4 2 = − 2 \dfrac{-4}{2} = -2 2 − 4 = − 2 Parallel? Yes. Question 3
(a) Answer: − 1 2 a = ( 5 − 4 ) -\dfrac{1}{2}\mathbf{a} = \begin{pmatrix} 5 \\ -4 \end{pmatrix} − 2 1 a = ( 5 − 4 ) [2] − 1 2 × ( − 10 ) = 5 -\dfrac{1}{2} \times (-10) = 5 − 2 1 × ( − 10 ) = 5 − 1 2 × 8 = − 4 -\dfrac{1}{2} \times 8 = -4 − 2 1 × 8 = − 4 − 1 2 a = ( 5 − 4 ) -\dfrac{1}{2}\mathbf{a} = \begin{pmatrix} 5 \\ -4 \end{pmatrix} − 2 1 a = ( 5 − 4 ) (b) Answer: ∣ 3 a − 2 b ∣ = 14.6 |3\mathbf{a} - 2\mathbf{b}| = 14.6 ∣3 a − 2 b ∣ = 14.6 (3 s.f.) [2] 3 × ( − 6 ) + ( − 2 ) × ( − 2 ) = − 14 3 \times (-6) + (-2) \times (-2) = -14 3 × ( − 6 ) + ( − 2 ) × ( − 2 ) = − 14 3 × 2 + ( − 2 ) × 5 = − 4 3 \times 2 + (-2) \times 5 = -4 3 × 2 + ( − 2 ) × 5 = − 4 ∣ 3 a − 2 b ∣ = ( − 14 ) 2 + ( − 4 ) 2 |3\mathbf{a} - 2\mathbf{b}| = \sqrt{(-14)^2 + (-4)^2} ∣3 a − 2 b ∣ = ( − 14 ) 2 + ( − 4 ) 2 ∣ 3 a − 2 b ∣ = 212 |3\mathbf{a} - 2\mathbf{b}| = \sqrt{212} ∣3 a − 2 b ∣ = 212 ∣ 3 a − 2 b ∣ = 14.6 |3\mathbf{a} - 2\mathbf{b}| = 14.6 ∣3 a − 2 b ∣ = 14.6 (3 s.f.) Question 4
(a) Answer: B = ( − 5 , 1 ) B = (-5, 1) B = ( − 5 , 1 ) [2] O B → = O A → + A B → \overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} O B = O A + A B x : − 4 + ( − 1 ) = − 5 x: -4 + (-1) = -5 x : − 4 + ( − 1 ) = − 5 y : − 3 + 4 = 1 y: -3 + 4 = 1 y : − 3 + 4 = 1 B = ( − 5 , 1 ) B = (-5, 1) B = ( − 5 , 1 ) (b) Answer: B = ( 6 , − 1 ) B = (6, -1) B = ( 6 , − 1 ) [2] O B → = O A → + A B → \overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} O B = O A + A B x : 2 + 4 = 6 x: 2 + 4 = 6 x : 2 + 4 = 6 y : 6 + ( − 7 ) = − 1 y: 6 + (-7) = -1 y : 6 + ( − 7 ) = − 1 B = ( 6 , − 1 ) B = (6, -1) B = ( 6 , − 1 ) Question 5
(a) Answer: ( 21 17 ) \begin{pmatrix} 21 \\ 17 \end{pmatrix} ( 21 17 ) [1] 3 a − 2 b = ( 15 9 ) − ( − 6 − 8 ) 3\mathbf{a} - 2\mathbf{b} = \begin{pmatrix} 15 \\ 9 \end{pmatrix} - \begin{pmatrix} -6 \\ -8 \end{pmatrix} 3 a − 2 b = ( 15 9 ) − ( − 6 − 8 ) = ( 21 17 ) = \begin{pmatrix} 21 \\ 17 \end{pmatrix} = ( 21 17 ) (b) Answer: 27.0 27.0 27.0 [2] ∣ 3 a − 2 b ∣ = 21 2 + 17 2 |3\mathbf{a} - 2\mathbf{b}| = \sqrt{21^2 + 17^2} ∣3 a − 2 b ∣ = 2 1 2 + 1 7 2 = 730 = \sqrt{730} = 730 = 27.0 = 27.0 = 27.0 c = 3 a \mathbf{c} = 3\mathbf{a} c = 3 a (comparing the bottom components) h = 3 × 5 h = 3 \times 5 h = 3 × 5 = 15 = 15 = 15 Question 6
(a) Answer: − a + b -\mathbf{a} + \mathbf{b} − a + b [1] A B → = A O → + O B → \overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} A B = A O + O B = − a + b = -\mathbf{a} + \mathbf{b} = − a + b (b) Answer: 2 3 a + 1 3 b \dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b} 3 2 a + 3 1 b [2] O P → = O A → + A P → \overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} O P = O A + A P = a + 1 3 ( − a + b ) = \mathbf{a} + \dfrac{1}{3}(-\mathbf{a} + \mathbf{b}) = a + 3 1 ( − a + b ) = 2 3 a + 1 3 b = \dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b} = 3 2 a + 3 1 b (c) Answer: 1 6 a + 1 3 b \dfrac{1}{6}\mathbf{a} + \dfrac{1}{3}\mathbf{b} 6 1 a + 3 1 b [2] Q P → = Q O → + O P → \overrightarrow{QP} = \overrightarrow{QO} + \overrightarrow{OP} QP = QO + O P = − 1 2 a + 2 3 a + 1 3 b = -\dfrac{1}{2}\mathbf{a} + \dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b} = − 2 1 a + 3 2 a + 3 1 b = 1 6 a + 1 3 b = \dfrac{1}{6}\mathbf{a} + \dfrac{1}{3}\mathbf{b} = 6 1 a + 3 1 b Question 7
(a) Answer: ( − 5 − 10 ) \begin{pmatrix} -5 \\ -10 \end{pmatrix} ( − 5 − 10 ) [1] A B → = O B → − O A → \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} A B = O B − O A = ( − 3 − 8 ) − ( 2 2 ) = \begin{pmatrix} -3 \\ -8 \end{pmatrix} - \begin{pmatrix} 2 \\ 2 \end{pmatrix} = ( − 3 − 8 ) − ( 2 2 ) = ( − 5 − 10 ) = \begin{pmatrix} -5 \\ -10 \end{pmatrix} = ( − 5 − 10 ) (b) Answer: 11.2 11.2 11.2 units [2] ∣ A B → ∣ = ( − 5 ) 2 + ( − 10 ) 2 |\overrightarrow{AB}| = \sqrt{(-5)^2 + (-10)^2} ∣ A B ∣ = ( − 5 ) 2 + ( − 10 ) 2 = 125 = \sqrt{125} = 125 = 11.2 = 11.2 = 11.2 (c) Answer: O P → = ( − 1 − 4 ) \overrightarrow{OP} = \begin{pmatrix} -1 \\ -4 \end{pmatrix} O P = ( − 1 − 4 ) [2] A P → = 3 5 A B → \overrightarrow{AP} = \dfrac{3}{5}\overrightarrow{AB} A P = 5 3 A B = ( − 3 − 6 ) = \begin{pmatrix} -3 \\ -6 \end{pmatrix} = ( − 3 − 6 ) O P → = O A → + A P → \overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} O P = O A + A P = ( 2 2 ) + ( − 3 − 6 ) = \begin{pmatrix} 2 \\ 2 \end{pmatrix} + \begin{pmatrix} -3 \\ -6 \end{pmatrix} = ( 2 2 ) + ( − 3 − 6 ) = ( − 1 − 4 ) = \begin{pmatrix} -1 \\ -4 \end{pmatrix} = ( − 1 − 4 ) (d) Answer: h = 22 h = 22 h = 22 [2] A E → = k A B → \overrightarrow{AE} = k\overrightarrow{AB} A E = k A B ( 10 h − 2 ) = k ( − 5 − 10 ) \begin{pmatrix} 10 \\ h - 2 \end{pmatrix} = k\begin{pmatrix} -5 \\ -10 \end{pmatrix} ( 10 h − 2 ) = k ( − 5 − 10 ) k = − 10 5 k = -\dfrac{10}{5} k = − 5 10 = − 2 = -2 = − 2 h − 2 = − 2 × ( − 10 ) h - 2 = -2 \times (-10) h − 2 = − 2 × ( − 10 ) h = 22 h = 22 h = 22