Vectors (GCSE Maths)
Skill Check
ABCD is a rhombus. Opposite sides are parallel.
BE is an extension of AB, such that AB : BE = 4 : 3.
The point F is halfway along CD.
Find FE in terms of a and b.
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Step 1 · Find vector $\overrightarrow{FC}$
- Because $ABCD$ is a rhombus with parallel opposite sides, the vector for the bottom edge is the same as the top edge, meaning $\overrightarrow{DC} = \boldsymbol{a}$.
- Since point $F$ is halfway along $\overrightarrow{CD}$, the vector $\overrightarrow{FC}$ is exactly half of $\boldsymbol{a}$.
Step 2 · Find vector $\overrightarrow{BE}$
The given ratio for the extension is $AB : BE = 4 : 3$. This indicates that the vector $\overrightarrow{BE}$ is equal to $\frac{3}{4}$ of the vector $\boldsymbol{a}$.
Step 3 · Construct a path from $F$ to $E$
- To determine $\overrightarrow{FE}$, construct a path from $F$ to $E$: $\overrightarrow{FE} = \overrightarrow{FC} + \overrightarrow{CB} + \overrightarrow{BE}$.
- The vector $\overrightarrow{CB}$ is the reverse of $\overrightarrow{BC}$, so $\overrightarrow{CB} = -\boldsymbol{b}$.
Step 4 · Simplify the expression
Combine the terms with $\boldsymbol{a}$ $\left(\frac{2}{4}\boldsymbol{a} + \frac{3}{4}\boldsymbol{a}\right)$ to reach the final simplified expression.
Final Answer
$\overrightarrow{FE} = \frac{5}{4}\boldsymbol{a} - \boldsymbol{b}$
EFG is a scalene triangle.
EG = a - 5b, EF = 3a + 4b
GH, not shown, is 3 times the length of and parallel to GF.
Find GH in terms of a and b.
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Step 1 · Find the vector $\overrightarrow{GF}$
Determine the path from $G$ to $F$ via $E$:
Since $\overrightarrow{GE} = -\overrightarrow{EG}$, we reverse the signs of $\overrightarrow{EG}$:
Simplify the expression:
Step 2 · Calculate the vector $\overrightarrow{GH}$
Since $\overrightarrow{GH}$ is 3 times the length of and parallel to $\overrightarrow{GF}$, multiply the vector $\overrightarrow{GF}$ by 3:
Final Answer
$6\mathbf{a} + 27\mathbf{b}$
KLM is a scalene triangle.
Point P is half way along ML.
MK = 2a, KL = b.
Write the vector ML in terms of a and b.
Find MP in terms of a and b.
A new point N is to the right of L. MN is 4 times the length of MP.
Write MN in terms of a and b.
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Step 1 · Find vector $\vec{ML}$
To find the vector $\vec{ML}$, trace the path from $M$ to $K$, and then from $K$ to $L$.
Substitute the given vectors into the expression:
Step 2 · Find vector $\vec{MP}$
Since point $P$ is halfway along the line segment $ML$, the vector $\vec{MP}$ is exactly half of the vector $\vec{ML}$.
Substitute the answer from part (a):
Expand the brackets to simplify:
Step 3 · Find vector $\vec{MN}$
Set up the relationship given in the question.
Substitute the expression for $\vec{MP}$ found in part (b).
Expand the brackets by multiplying both terms by $4$.
Final Answer
$\text{(a) } 2\mathbf{a} + \mathbf{b} \quad \text{(b) } \mathbf{a} + \frac{1}{2}\mathbf{b} \quad \text{(c) } 4\mathbf{a} + 2\mathbf{b}$
Problem Solving
ABC is an isosceles triangle. Point M is halfway along the triangle base, AC.
Point D is positioned such that AC and AD are collinear. Point E is positioned such that AB and AE are collinear.
AB : BE = 3 : 2.
MA = b, AB = 3a, AD = 2AC.
Find DE in terms of a and b.
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Step 1 · Determine the vector $\overrightarrow{DA}$
- Because $M$ is the midpoint of $AC$ and $\overrightarrow{MA} = \mathbf{b}$, the vector $\overrightarrow{CA}$ is twice $\overrightarrow{MA}$, so $\overrightarrow{CA} = 2\mathbf{b}$.
- Using the given relationship $\overrightarrow{AD} = 2\overrightarrow{AC}$, we can determine that $\overrightarrow{DA} = 2\overrightarrow{CA}$.
- Therefore, $\overrightarrow{DA} = 4\mathbf{b}$.
Step 2 · Determine the vector $\overrightarrow{BE}$
Using the ratio $AB : BE = 3 : 2$ and the given vector $\overrightarrow{AB} = 3\mathbf{a}$, it follows that:
Step 3 · Calculate the total vector $\overrightarrow{AE}$
Combining $\overrightarrow{AB}$ and $\overrightarrow{BE}$:
Step 4 · Find the vector $\overrightarrow{DE}$
Taking the path from $D$ to $A$ and then $A$ to $E$:
Final Answer
$4\mathbf{b} + 5\mathbf{a}$
The diagram below shows a regular hexagon ABCDEF.
X is the point at the centre of the hexagon.
XB = a, ED = b.
Write down the vector BC in terms of a and b.
Write down the vector BE in terms of a and b.
Write down the vector AE in terms of a and b.
Write down the vector FB in terms of a and b.
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Step 1 · Determine vector $\overrightarrow{XC}$
In a regular hexagon, the vector from the center to a vertex is parallel to and the same length as the opposite edge. Therefore, $\overrightarrow{XC} = \overrightarrow{ED} = \boldsymbol{b}$.
Step 2 · Set up the path for $\overrightarrow{BC}$
The path from $B$ to $C$ is:
Step 3 · Substitute values to find $\overrightarrow{BC}$
- Since $\overrightarrow{BX} = -\overrightarrow{XB}$, we have $\overrightarrow{BX} = -\boldsymbol{a}$.
- Substitute the known vectors into the path equation:
Final Answer
$-\boldsymbol{a} + \boldsymbol{b}$
Step 1 · Set up the path for $\overrightarrow{BE}$
- The vector from $B$ to $E$ passes straight through the center $X$.
- The path is:
Step 2 · Determine vector $\overrightarrow{XE}$
Because it's a regular hexagon, $\overrightarrow{XE}$ is equal in magnitude and direction to $\overrightarrow{BX}$. So, $\overrightarrow{XE} = -\boldsymbol{a}$.
Step 3 · Substitute and calculate $\overrightarrow{BE}$
Add the two vectors:
Final Answer
$-2\boldsymbol{a}$
Step 1 · Set up the path for $\overrightarrow{AE}$
Find a path from $A$ to $E$. We can use:
Step 2 · Determine vector $\overrightarrow{AB}$
In a regular hexagon, $\overrightarrow{AB}$ is parallel and equal to $\overrightarrow{ED}$, so $\overrightarrow{AB} = \boldsymbol{b}$.
Step 3 · Substitute and calculate $\overrightarrow{AE}$
- We already found $\overrightarrow{BE} = -2\boldsymbol{a}$ in part (b).
- Substitute the known vectors:
Final Answer
$-2\boldsymbol{a} + \boldsymbol{b}$
Step 1 · Set up the path for $\overrightarrow{FB}$
Find a path from $F$ to $B$. We can use:
Step 2 · Determine vector $\overrightarrow{FX}$
Due to the properties of a regular hexagon, $\overrightarrow{FX}$ is parallel and equal to $\overrightarrow{ED}$, so $\overrightarrow{FX} = \boldsymbol{b}$.
Step 3 · Substitute and calculate $\overrightarrow{FB}$
- The vector $\overrightarrow{XB}$ is given as $\boldsymbol{a}$.
- Add the two vectors:
Final Answer
$\boldsymbol{a} + \boldsymbol{b}$
KLMN is an irregular quadrilateral.
ML = 2a, MN = 2a + 2b, NK = 3a + b.
Point P is positioned such that LP : PK = 1 : 2.
Find the vector LK in terms of a and b.
Show that MP = NK.
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Step 1 · Define the vector path
To find the vector $\overrightarrow{LK}$, determine the path from $L$ to $K$ via $M$ and $N$:
Step 2 · Substitute the known vectors
Because $\overrightarrow{LM}$ is the reverse of $\overrightarrow{ML}$, its sign changes to $-2\boldsymbol{a}$.
Step 3 · Simplify the expression
Combine the terms with $\boldsymbol{a}$ and the terms with $\boldsymbol{b}$ to simplify.
Final Answer
$\overrightarrow{LK} = 3\boldsymbol{a} + 3\boldsymbol{b}$
Step 1 · Determine the vector $\overrightarrow{LP}$
The ratio $LP : PK = 1 : 2$ means that $\overrightarrow{LP}$ is $\frac{1}{3}$ of the total vector $\overrightarrow{LK}$.
Step 2 · Define and substitute the vector $\overrightarrow{MP}$
Next, define the vector $\overrightarrow{MP}$ by tracing the path from $M$ to $P$ via $L$: $\overrightarrow{MP} = \overrightarrow{ML} + \overrightarrow{LP}$. Substitute the known vectors into the equation.
Step 3 · Simplify the expression for $\overrightarrow{MP}$
Simplify the expression to find $\overrightarrow{MP}$.
Step 4 · Compare with vector $\overrightarrow{NK}$
Since it is given that $\overrightarrow{NK} = 3\boldsymbol{a} + \boldsymbol{b}$, we have shown that $\overrightarrow{MP} = \overrightarrow{NK}$.
Final Answer
$\overrightarrow{MP} = \overrightarrow{NK}$
Exam-Style Questions
On the diagram below:
AD = a
DB = 3b - a
AC and AB are straight lines.
AD : DY : YC = 1 : 1 : 1
AX : XB = 2 : 1
Show that XY is parallel to BC.
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Step 1 · Find the vector $\overrightarrow{BC}$
Establish the path to find $\overrightarrow{BC}$:
- Because $AD : DY : YC = 1 : 1 : 1$ and $\overrightarrow{AD} = \boldsymbol{a}$, both $\overrightarrow{DY}$ and $\overrightarrow{YC}$ are also equal to $\boldsymbol{a}$.
- The vector $\overrightarrow{BD}$ is the reverse of $\overrightarrow{DB}$, so $\overrightarrow{BD} = -(3\boldsymbol{b} - \boldsymbol{a}) = \boldsymbol{a} - 3\boldsymbol{b}$.
- Substitute these into the path for $\overrightarrow{BC}$:
Step 2 · Find the vector $\overrightarrow{AB}$
Before finding $\overrightarrow{XY}$, we need to find $\overrightarrow{AB}$:
Step 3 · Find the vector $\overrightarrow{XY}$
- The ratio $AX : XB = 2 : 1$ means $\overrightarrow{AX}$ is $\frac{2}{3}$ of $\overrightarrow{AB}$, so $\overrightarrow{AX} = \frac{2}{3}(3\boldsymbol{b}) = 2\boldsymbol{b}$.
- Therefore, $\overrightarrow{XA} = -2\boldsymbol{b}$.
- Trace the path for $\overrightarrow{XY} = \overrightarrow{XA} + \overrightarrow{AD} + \overrightarrow{DY}$:
Step 4 · Prove the vectors are parallel
Compare the two vectors we found:
Since $\overrightarrow{BC}$ is a scalar multiple of $\overrightarrow{XY}$ $\left(\overrightarrow{BC} = \frac{3}{2}\overrightarrow{XY}\right)$, the lines $BC$ and $XY$ are parallel.
Final Answer
$\text{Shown}$
On the diagram below:
AB = BC = a
AD = 3b
AE is a straight line. CBEF is a parallelogram.
AD : BE : CF = 3 : 2 : 2
Find expressions for the vectors below in terms of a and b.
DC
FD
DE is extended upwards until it hits the line CF. The point of intersection is X.
What is the ratio CX : XF?
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Step 1 · Establish the vector path
The path from $D$ to $C$ is:
Step 2 · Substitute given vectors and simplify
Substitute the known vectors into the equation:
- $\overrightarrow{DA} = -3\boldsymbol{b}$
- $\overrightarrow{AB} = \boldsymbol{a}$
- $\overrightarrow{BC} = \boldsymbol{a}$
Final Answer
$\overrightarrow{DC} = 2\boldsymbol{a} - 3\boldsymbol{b}$
Step 1 · Establish vector $\overrightarrow{CF}$
First, find the vector $\overrightarrow{CF}$ using the given ratios:
- We are given the ratio $AD : BE : CF = 3 : 2 : 2$.
- Knowing that $\overrightarrow{AD} = 3\boldsymbol{b}$, it follows that $\overrightarrow{CF} = 2\boldsymbol{b}$.
Step 2 · Determine the path for $\overrightarrow{FD}$
The path from $F$ to $D$ is:
Step 3 · Substitute vectors and calculate
Substitute the known vectors into the path equation:
- $\overrightarrow{FC} = -2\boldsymbol{b}$ (since $\overrightarrow{CF} = 2\boldsymbol{b}$)
- $\overrightarrow{CB} = -\boldsymbol{a}$
- $\overrightarrow{BA} = -\boldsymbol{a}$
- $\overrightarrow{AD} = 3\boldsymbol{b}$
Combine the terms:
Final Answer
$\overrightarrow{FD} = -2\boldsymbol{a} + \boldsymbol{b}$
Step 1 · Find vector $\overrightarrow{DE}$
The path for $\overrightarrow{DE}$ is $\overrightarrow{DA} + \overrightarrow{AB} + \overrightarrow{BE}$:
- Given the ratio $AD : BE = 3 : 2$, the vector $\overrightarrow{BE} = 2\boldsymbol{b}$.
- Substitute the vectors into the path.
Step 2 · Express $\overrightarrow{CX}$ in terms of a multiplier
Let the extension to $X$ be represented as $\overrightarrow{EX} = x(\overrightarrow{DE})$. Trace the path from $C$ to $X$:
- Because $X$ lies on line $CF$, the vector $\overrightarrow{CX}$ must be a multiple of $\boldsymbol{b}$.
- The path is $\overrightarrow{CX} = \overrightarrow{CB} + \overrightarrow{BE} + \overrightarrow{EX}$.
Step 3 · Solve for $x$ and find $\overrightarrow{CX}$
For $\overrightarrow{CX}$ to lie on line $CF$ (which is parallel to $\boldsymbol{b}$):
- The $\boldsymbol{a}$ components must cancel out, which requires $x = 1$.
- Substitute $x = 1$ back into the path equation for $\overrightarrow{CX}$.
Step 4 · Determine the ratio $CX : XF$
- The total length of $\overrightarrow{CF}$ is $2\boldsymbol{b}$.
- Since $\overrightarrow{CX} = \boldsymbol{b}$, the remaining length $\overrightarrow{XF}$ must also be $\boldsymbol{b}$.
- Therefore, $\overrightarrow{CX} = \overrightarrow{XF} = \boldsymbol{b}$, making the ratio $1 : 1$.
Final Answer
$1 : 1$
ABC is a scalene triangle.
AC = 3b, AB = 2a.
Point D is positioned such that BD : DC = 1 : 3.
Point E is positioned such that AE : ED = 1 : 1.
Find AE in terms of a and b.
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Step 1 · Find the vector $\overrightarrow{BC}$
- Follow the path from $B$ to $A$ and then $A$ to $C$: $\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC}$.
- Since $\overrightarrow{BA}$ is the reverse of $\overrightarrow{AB}$, it becomes $-2\mathbf{a}$.
Step 2 · Determine the vector $\overrightarrow{BD}$
The given ratio $BD : DC = 1 : 3$ means that $\overrightarrow{BD}$ is $\dfrac{1}{4}$ of the full vector $\overrightarrow{BC}$.
Step 3 · Find the vector $\overrightarrow{AD}$
Follow the path from $A$ to $B$ and then $B$ to $D$: $\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BD}$.
Step 4 · Find the vector $\overrightarrow{AE}$
The point $E$ splits $AD$ such that $AE : ED = 1 : 1$, which means $E$ is the midpoint of $AD$, so $\overrightarrow{AE}$ is half of $\overrightarrow{AD}$.
This can also be written in factored form as $\dfrac{3}{4}\left(\mathbf{a} + \dfrac{\mathbf{b}}{2}\right)$.
Final Answer
$\overrightarrow{AE} = \dfrac{3}{4}\mathbf{a} + \dfrac{3}{8}\mathbf{b}$
On the diagram below: AE is a straight line.
AB = a, AC = b, AC = CE, DE = 1/4CE.
Find expressions for the vectors below in terms of a and b.
BC
DB
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Step 1 · Find vector $\overrightarrow{BC}$
- To find the vector $\overrightarrow{BC}$, use the path from $B$ to $A$ and then $A$ to $C$: $\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC}$.
- Since $\overrightarrow{BA}$ is the reverse of $\overrightarrow{AB}$, we have $\overrightarrow{BA} = -\mathbf{a}$.
Step 2 · Find vector $\overrightarrow{CE}$
First, identify the vector for $\overrightarrow{CE}$. Since $AC = CE$ and they lie on the same straight line, they are equal vectors.
Step 3 · Establish the vector $\overrightarrow{DE}$
Use the given relationship $DE = \frac{1}{4}CE$ to establish the vector $\overrightarrow{DE}$.
Step 4 · Find the vector $\overrightarrow{CD}$
Find the vector $\overrightarrow{CD}$ by subtracting $\overrightarrow{DE}$ from $\overrightarrow{CE}$.
Step 5 · Determine the vector $\overrightarrow{DB}$
Finally, determine the vector $\overrightarrow{DB}$ by following the path from $D$ to $C$ and then $C$ to $B$. This is calculated as $\overrightarrow{DC} + \overrightarrow{CB}$. Since $\overrightarrow{DC} = -\frac{3}{4}\mathbf{b}$ and $\overrightarrow{CB} = -\mathbf{b} + \mathbf{a}$ (which is $-\overrightarrow{BC}$), we combine these terms.
Final Answer
(a) $-\mathbf{a} + \mathbf{b}$
(b) $-\frac{7}{4}\mathbf{b} + \mathbf{a}$Feel confident with Vectors (GCSE Maths)?
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