Question 1
KLM is a scalene triangle.
Point P is half way along \overrightarrow{ML}.
\overrightarrow{MK} = \textbf{2a},
\overrightarrow{KL} = \textbf{b}.
1(a):
Write the vector \overrightarrow{ML} in terms of \textbf{a} and \textbf{b}.
ANSWER: Multiple Choice (Type 1)
A: -\textbf{b} + \textbf{a}
B: 2\textbf{a} + 2\textbf{b}
C: \textbf{b} - 2\textbf{a}
D: 2\textbf{a} + \textbf{b}
Answer: D
Workings:
The vector of \overrightarrow{ML} requires a movement of 2\textbf{a} followed by a movement of \textbf{b}.
Marks = 1
1(b):
Find \overrightarrow{MP} in terms of \textbf{a} and \textbf{b}.
ANSWER: Multiple Choice (Type 1)
A: \textbf{a} + \dfrac{1}{2}\textbf{b}
B: 2\textbf{a} + 2\textbf{b}
C: \textbf{a} + \textbf{b}
D: \dfrac{1}{2}\textbf{a} + \textbf{b}
Answer: A
Workings:
\dfrac{1}{2}(2\textbf{a} + \textbf{b}) = \textbf{a} + \dfrac{\textbf{b}}{2}
Marks = 1
1(c):
A new point N is to the right of L.
\overrightarrow{MN} is 4 times the length of \overrightarrow{MP}.
Write \overrightarrow{MN} as a column vector.
ANSWER: Multiple Choice (Type 2)
A: 2\begin{pmatrix}2\textbf{a}\\\textbf{b}\end{pmatrix}
B: \begin{pmatrix}2\textbf{a}\\\textbf{b}\end{pmatrix}
C: 2\begin{pmatrix}\textbf{a}\\2\textbf{b}\end{pmatrix}
D: 2\begin{pmatrix}4\textbf{a}\\2\textbf{b}\end{pmatrix}
Answer: A
Workings:
4\begin{pmatrix}\textbf{a}\\\dfrac{\textbf{b}}{2}\end{pmatrix} = \begin{pmatrix}4\textbf{a}\\2\textbf{b}\end{pmatrix}\\
= 2\begin{pmatrix}2\textbf{a}\\\textbf{b}\end{pmatrix} = 2 \overrightarrow{ML}
Marks = 1
Question 2
EFG is a scalene triangle.
\overrightarrow{EG} = \textbf{a} - 5\textbf{b},
\overrightarrow{EF} = 3\textbf{a} + 4\textbf{b}.
\overrightarrow{GH}, not shown, is 3 times the length of and parallel to \overrightarrow{GF}.
Find \overrightarrow{GH} in terms of \textbf{a} and \textbf{b}.
ANSWER: Multiple Choice (Type 2)
A: 6\textbf{a} + 27\textbf{b}
B: 5\textbf{a} + 24\textbf{b}
C: 4\textbf{a} + 21\textbf{b}
D: 3\textbf{a} + 18\textbf{b}
Answer: A
Workings:
\overrightarrow{GF} = 3\textbf{a} - \textbf{a} + 5\textbf{b} + 4\textbf{b}
\overrightarrow{GF} = 2\textbf{a} + 9\textbf{b}
\overrightarrow{GH} = 3(2\textbf{a} + 9\textbf{b}) = 6\textbf{a} + 27\textbf{b}
Marks = 3
Question 3
ABCD is a rhombus. Opposite sides are parallel.
\overrightarrow{BE} is an extension of \overrightarrow{AB}, such that AB : BE = 4 : 3
The point F is halfway along \overrightarrow{CD}
Find \overrightarrow{FE} in terms of \textbf{a} and \textbf{b}.
ANSWER: Multiple Choice (Type 1)
A: \dfrac{5}{4}\textbf{a} - \textbf{b}
B: \textbf{a} - \dfrac{\textbf{b}}{2}
C: \dfrac{3\textbf{a}}{2} + \textbf{b}
D: \dfrac{3\textbf{a}}{4} + \textbf{2\textbf{b}}
Answer: A
Workings:
\overrightarrow{FC} = \dfrac{\textbf{a}}{2}
\overrightarrow{BE} = \dfrac{3}{4}\textbf{a}
\overrightarrow{FE} = \dfrac{5}{4}\textbf{a} - \textbf{b}
Marks = 3
Question 4
KLMN is an irregular quadrilateral.
\overrightarrow{ML} = 2\textbf{a},
\overrightarrow{MN} = 2\textbf{a} + 2\textbf{b},
\overrightarrow{NK} = 3\textbf{a} + \textbf{b}Point P is positioned such that LP : PK = 1 : 2
4(a):
Find the vector \overrightarrow{LK} in terms of \textbf{a} and \textbf{b}.
ANSWER: Multiple Choice (Type 1)
A: \textbf{a} + \textbf{b}
B: 2\textbf{a} + 2\textbf{b}
C: 3\textbf{a} + 3\textbf{b}
D: 4\textbf{a} + 4\textbf{b}
Answer: C
Workings:
\overrightarrow{LM} = -2\textbf{a},
\overrightarrow{MN} = 2\textbf{a} + 2\textbf{b}
\overrightarrow{NK} = 3\textbf{a} + \textbf{b}
Marks = 2
4(b):
Find the vector \overrightarrow{MP} in terms of \textbf{a} and \textbf{b}
ANSWER: Multiple Choice (Type 1)
A: 3\textbf{a} + \textbf{b}
B: 4\textbf{a} + 2\textbf{b}
C: 5\textbf{a} + 3\textbf{b}
D: 6\textbf{a} + 4\textbf{b}
Answer: A
Workings:
\overrightarrow{LP} = \dfrac{1}{3}(3\textbf{a} + 3\textbf{b})
\overrightarrow{LP} = \textbf{a} + \textbf{b}
\overrightarrow{MP} = \overrightarrow{ML} + \overrightarrow{LP} = 2\textbf{a} + \textbf{a} + \textbf{b}
\overrightarrow{MP} = 3\textbf{a} + \textbf{b} = \overrightarrow{NK}
Marks = 4
Question 5
The diagram below shows a regular hexagon ABCDEF.
X is the point at the centre of the hexagon.
\overrightarrow{XB} = \textbf{a}\\ \overrightarrow{ED} = \textbf{b}
Write down the following vectors in terms of \textbf{a} and \textbf{b}.
5(a):
\overrightarrow{BC}ANSWER: Multiple Choice (Type 1)
A: \textbf{b} - \textbf{a}
B: 2\textbf{b} - 2\textbf{a}
C: \textbf{a} + \textbf{b}
D: 2\textbf{a} - 2\textbf{b}
Answer: A
Workings:
\overrightarrow{XC} = \overrightarrow{ED}
\overrightarrow{BC} = \overrightarrow{BX} + \overrightarrow{XC} = -\textbf{a} + \textbf{b}
Marks = 1
5(b):
\overrightarrow{BE}ANSWER: Multiple Choice (Type 1)
A: -2\textbf{a}
B: 2\textbf{b}
C: 2\textbf{a}
D: -2\textbf{b}
Answer: A
Workings:
\overrightarrow{XE} = \overrightarrow{BX}
\overrightarrow{BE} = \overrightarrow{BX} + \overrightarrow{XE} = -\textbf{a} -\textbf{a} = -2\textbf{a}
Marks = 1
5(c):
\overrightarrow{AE}ANSWER: Multiple Choice (Type 1)
A: \textbf{b} - 2\textbf{a}
B: \textbf{b} + 2\textbf{a}
C: \textbf{b} + 2\textbf{a}
D: \textbf{b} + 2\textbf{a}
Answer: A
Workings:
\overrightarrow{AF} = \overrightarrow{BX}, \overrightarrow{FX} = \overrightarrow{ED}, \overrightarrow{XE} = \overrightarrow{BX}
\overrightarrow{AE} = \overrightarrow{AF} + \overrightarrow{FX} + \overrightarrow{XE} = -\textbf{a} + \textbf{b} -\textbf{a} = -2\textbf{a} + \textbf{b}
Marks = 1
5(d):
\overrightarrow{FB}ANSWER: Multiple Choice (Type 1)
A: \textbf{a} + \textbf{b}
B: \textbf{a} - \textbf{b}
C: -\textbf{a} + \textbf{b}
D: -\textbf{a} - \textbf{b}
Answer: A
Workings:
\overrightarrow{FA} = \overrightarrow{XB}, \overrightarrow{AB} = \overrightarrow{ED}
\overrightarrow{FB} = \overrightarrow{FA} + \overrightarrow{AB} = -\textbf{a} + \textbf{b}
Marks = 1
Question 6
ABC is an isosceles triangle.
Point M is halfway along the triangle base, AC.
Point D is positioned such that \overrightarrow{AC} and \overrightarrow{AD} are collinear.
Point E is positioned such that \overrightarrow{AB} and \overrightarrow{AE} are collinear.
AB : BE = 3 : 2
\overrightarrow{MA} = \textbf{b},
\overrightarrow{AB} = 3\textbf{a},
\overrightarrow{AD} = 2\overrightarrow{AC}.
Find \overrightarrow{DE} in terms of \textbf{a} and \textbf{b}.
ANSWER: Multiple Choice (Type 1)
A: 5\textbf{a} + 4\textbf{b}
B: 3\textbf{a} + 4\textbf{b}
C: 4\textbf{a} + 5\textbf{b}
D: 4\textbf{a} + 3\textbf{b}
Answer: A
Workings:
\overrightarrow{DA} = 4\textbf{b}
\overrightarrow{BE} = 2\textbf{a}
\overrightarrow{AE} = 5\textbf{a}
\overrightarrow{DE} = 4\textbf{b} + 5\textbf{a}
Marks = 4
Question 7
On the diagram below:
AE is a straight line[/latex]
\overrightarrow{AB} = \textbf{a}\\ \overrightarrow{AC} = \textbf{b}\\ AC = CE\\ DE = \dfrac{1}{4}CE
Find expressions for the vectors below in terms of \textbf{a} and \textbf{b}.
7(a):
\overrightarrow{BC}ANSWER: Multiple Choice (Type 1)
A: -\textbf{a} + \textbf{b}
B: -\textbf{a} - \textbf{b}
C: \textbf{a} + \textbf{b}
D: \textbf{a} - \textbf{b}
Answer: A
Workings:
\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC} = -\textbf{a} + \textbf{b}
Marks = 1
7(b):
\overrightarrow{DB}ANSWER: Multiple Choice (Type 1)
A: \textbf{a} - \dfrac{7}{4}\textbf{b}
B: -\textbf{a} + \dfrac{7}{4}\textbf{b}
C: -\textbf{a} - \dfrac{7}{4}\textbf{b}
D: \textbf{a} + \dfrac{7}{4}\textbf{b}
Answer: A
Workings:
\overrightarrow{AC} = \overrightarrow{CE} = \textbf{b}
\overrightarrow{DE} = \dfrac{1}{4}\overrightarrow{CE} = \dfrac{1}{4}\textbf{b}
\overrightarrow{CD} = \overrightarrow{CE} - \overrightarrow{DE} = \textbf{b} - \dfrac{1}{4}\overrightarrow{b} = \dfrac{3}{4}\textbf{b}
\overrightarrow{DB} = \overrightarrow{DC} + \overrightarrow{CB} = -\dfrac{3}{4}\textbf{b} -\textbf{b} + \textbf{a} = -\dfrac{7}{4}\textbf{b} + \textbf{a}
Marks = 4
Question 8
ABC is a scalene triangle.
\overrightarrow{AC} = 3\textbf{b} , \overrightarrow{AB} = 2\textbf{a}
Point D is positioned such that BD:DC = 1:3
Point E is positioned such that AE:ED = 1:1
Find \overrightarrow{AE} in terms of \textbf{a} and \textbf{b}.
ANSWER: Multiple Choice (Type 1)
A: \dfrac{3}{4}(\textbf{a} + \dfrac{1}{2}\textbf{b})
B : \dfrac{3}{8}(\textbf{a} + \textbf{b})
C: \dfrac{3}{4}(\dfrac{1}{2}\textbf{a} + \dfrac{1}{2}\textbf{b})
D: \dfrac{3}{4}(\dfrac{1}{2}\textbf{a} + \textbf{b})
Answer: A
Workings:
\overrightarrow{BC} = -2\textbf{a} + 3\textbf{b}
\overrightarrow{BD} = - \dfrac{\textbf{a}}{2} + \dfrac{3}{4}\textbf{b}
\overrightarrow{AD} = 2\textbf{a}-\dfrac{\textbf{a}}{2} + \dfrac{3}{4}\textbf{b} = \dfrac{3}{2}\textbf{a} \dfrac{3}{4}\textbf{b}
\overrightarrow{AE} = \dfrac{1}{2}(\dfrac{3}{2}\textbf{a} + \dfrac{3}{4}\textbf{b})
= \dfrac{3}{4}\textbf{a} + \dfrac{3}{8}\textbf{b} = \dfrac{3}{4}(\textbf{a}+\dfrac{\textbf{b}}{2})
Marks = 5
Question 9
On the diagram below:
\overrightarrow{AB} = \overrightarrow{BC} = \textbf{a}\\ \overrightarrow{AD} = 3\textbf{b}AE is a straight line
CBEF is a parallelogram
AD : BE : CF = 3 : 2 : 2
Find expressions for the vectors below in terms of \textbf{a} and \textbf{b}.
9(a):
\overrightarrow{DC}ANSWER: Multiple Choice (Type 1)
A: 2\textbf{a} - 3\textbf{b}
B: \textbf{a} + 3\textbf{b}
C: 3\textbf{a} + 2\textbf{b}
D: -2\textbf{a} - 3\textbf{b}
Answer: A
Workings:
\overrightarrow{DC} = \overrightarrow{DA} + \overrightarrow{AB} + \overrightarrow{BC}
= -3\textbf{b} + \textbf{a} + \textbf{a} = 2\textbf{a} - 3\textbf{b}
Marks = 1
9(b):
\overrightarrow{FD}ANSWER: Multiple Choice (Type 1)
A: -2\textbf{a} + \textbf{b}
B: -\textbf{a} + 2\textbf{b}
C: 3\textbf{b}
D: \textbf{a} + 4\textbf{b}
Answer: A
Workings:
AD : BE : CF = 3 : 2 : 2
AD : BE : CF = 3\textbf{b} : 2\textbf{b} : 2\textbf{b}
CF = 2\textbf{b}\\ \overrightarrow{FD} = \overrightarrow{FC} + \overrightarrow{CB} + \overrightarrow{BA} + \overrightarrow{AD}\\ = -2\textbf{b} -\textbf{a}-\textbf{a}+3\textbf{b}\\ =-2\textbf{a}+\textbf{b}Marks =
9(c):
DE is extended upwards until it hits the line CF.
The point of intersection is X.
What is the ratio CX : XF?
ANSWER: Simple Text Answer
Answer: 1 : 1
Workings:
\overrightarrow{DE} = -3\textbf{b}+\textbf{a}+2\textbf{b}
=\textbf{a}-\textbf{b}
\overrightarrow{EX} = x\overrightarrow{DE}
Need 1 lot of \overrightarrow{EX} to reach \overrightarrow{CF}, and gives:
\overrightarrow{CX} = \overrightarrow{XF} = \textbf{b}
CF : XF = 1 : 1
Marks = 3
Question 10
On the diagram below:
\overrightarrow{AD} = \textbf{a}\\ \overrightarrow{DB} = 3\textbf{b} - \textbf{a}AC and AB are straight lines
AD : DY : YC = 1 : 1 : 1
AX : XB = 2 : 1
10(a):
Find an expression for the vector XY in terms of \textbf{a} and \textbf{b}.
ANSWER: Multiple Choice (Type 1)
A: 2(\textbf{a} - \textbf{b})
B: 2\textbf{a} - \textbf{b}
C: \textbf{a} - 3\textbf{b}
D: 2(2\textbf{a} - \textbf{b})
Answer: A
Workings:
\overrightarrow{AD} : \overrightarrow{DY} : \overrightarrow{YC} = 1 : 1 : 1
\overrightarrow{AD} = \overrightarrow{DY} = \overrightarrow{YC} = \textbf{a}
\overrightarrow{BC} = \overrightarrow{BD} + \overrightarrow{DY} + \overrightarrow{YC}
= -(3\textbf{b} - \textbf{a}) + \textbf{a} + \textbf{a}
=\textbf{a} - 3\textbf{b} + \textbf{a} + \textbf{a}
= 3\textbf{a} - 3\textbf{b}
= 3(\textbf{a}-\textbf{b})
\\
\overrightarrow{AB} = \overrightarrow{AD} + \overrightarrow{DB}
= \textbf{a} + 3\textbf{b} - \textbf{a}
= 3\textbf{b}
\overrightarrow{AX} : \overrightarrow{XB} = 2 : 1
\overrightarrow{AX} = 2\textbf{b}
\\
\overrightarrow{XY} = \overrightarrow{XA} + \overrightarrow{AD} + \overrightarrow{DY} = -2\textbf{b} + \textbf{a} + \textbf{a}
=2\textbf{a}-2\textbf{b}
=2(\textbf{a}-\textbf{b})
Marks = 3
10(b):
Is XY parallel to BC?
ANSWER: Multiple Choice (Type 2)
A: Yes
B: No
Answer: A
Workings:
\overrightarrow{BC} is a multiple of \overrightarrow{XY}, so they are going in the same direction.
Therefore they are parallel.
Marks = 1