Hard A-Level Vector geometry Questions

Challenging, exam-style A-Level Vector geometry questions with worked solutions. Stretch yourself on the hardest vector-line, intersection, section-formula, ratio problems.

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A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
OABCOABC is a parallelogram with OA→=a\overrightarrow{OA}=\mathbf{a} and OC→=c\overrightarrow{OC}=\mathbf{c} (so OB→=a+c\overrightarrow{OB}=\mathbf{a}+\mathbf{c}). MM is the midpoint of BCBC. The line OMOM meets the diagonal ACAC at XX. Find the ratio OX:XMOX:XM.
Show worked solution

Worked solution

  1. Understand what is being asked

    a1+td1=a2+sd2\mathbf{a}_1+t\mathbf{d}_1=\mathbf{a}_2+s\mathbf{d}_2

    Before calculating, we identify the goal and choose the method: equating the two line equations and solving. Having a plan stops us getting lost part-way through.

  2. Recall the key result we will use

    a1+td1=a2+sd2\mathbf{a}_1+t\mathbf{d}_1=\mathbf{a}_2+s\mathbf{d}_2

    Set the two position vectors equal to find where the lines meet.

  3. Set up base vectors

    OA→=a,OC→=c\overrightarrow{OA}=\mathbf{a},\quad \overrightarrow{OC}=\mathbf{c}

    Use the two sides from OO as non-parallel base vectors.

  4. Position vector of BB

    OB→=a+c\overrightarrow{OB}=\mathbf{a}+\mathbf{c}

    The opposite corner of the parallelogram is the sum of the two sides.

  5. Find MM, the midpoint of BCBC

    OM→=12(OB→+OC→)=12a+c\overrightarrow{OM}=\tfrac{1}{2}(\overrightarrow{OB}+\overrightarrow{OC})=\tfrac{1}{2}\mathbf{a}+\mathbf{c}

    Average the position vectors of BB and CC: 12((a+c)+c)\tfrac{1}{2}((\mathbf{a}+\mathbf{c})+\mathbf{c}).

  6. Describe line OMOM

    OX→=λ(12a+c)\overrightarrow{OX}=\lambda\left(\tfrac{1}{2}\mathbf{a}+\mathbf{c}\right)

    Any point on OMOM is a scalar multiple λ\lambda of OM→\overrightarrow{OM}.

  7. Describe diagonal ACAC

    OX→=a+μ(c−a)=(1−μ)a+μc\overrightarrow{OX}=\mathbf{a}+\mu(\mathbf{c}-\mathbf{a})=(1-\mu)\mathbf{a}+\mu\mathbf{c}

    A point on ACAC is a fraction μ\mu of the way from AA to CC.

  8. Equate coefficients of a\mathbf{a}

    λ2=1−μ\tfrac{\lambda}{2}=1-\mu

    Match the a\mathbf{a}-parts of the two expressions.

  9. Equate coefficients of c\mathbf{c}

    λ=μ\lambda=\mu

    Match the c\mathbf{c}-parts; this gives λ=μ\lambda=\mu directly.

  10. Solve

    λ2=1−λ ⇒ 3λ2=1 ⇒ λ=23\tfrac{\lambda}{2}=1-\lambda\ \Rightarrow\ \tfrac{3\lambda}{2}=1\ \Rightarrow\ \lambda=\tfrac{2}{3}

    Substitute μ=λ\mu=\lambda and solve. Then OX:XM=λ:(1−λ)=23:13=2:1OX:XM=\lambda:(1-\lambda)=\tfrac{2}{3}:\tfrac{1}{3}=2:1.

  11. Why this method works

    a1+td1=a2+sd2\mathbf{a}_1+t\mathbf{d}_1=\mathbf{a}_2+s\mathbf{d}_2

    Equating gives two equations in two unknowns, which we solve simultaneously.

  12. Link to earlier work

    a1+td1=a2+sd2\mathbf{a}_1+t\mathbf{d}_1=\mathbf{a}_2+s\mathbf{d}_2

    Relies on solving simultaneous equations. Connecting new work to things you already know makes it far easier to remember.

  13. Write down what we are working with

    list the given vectors / points before substituting\text{list the given vectors / points before substituting}

    Carefully noting the given information first is a habit that prevents careless substitution errors.

  14. Check the answer

    OX:XM=2:1OX:XM=2:1

    Check the found point satisfies both original equations. Checking your work is how you catch small mistakes before they cost marks.

  15. Interpret the result

    OX:XM=2:1OX:XM=2:1

    The answer is the crossing point of the two lines.

  16. Watch out for a common slip

    a1+td1=a2+sd2\mathbf{a}_1+t\mathbf{d}_1=\mathbf{a}_2+s\mathbf{d}_2

    A frequent mistake here is rushing the signs or the order of subtraction; going slowly on those parts keeps the work accurate.

Answer
OX:XM=2:1OX:XM=2:1
Question 2
8 markschallenging
In triangle OABOAB, DD is the midpoint of OBOB, and EE lies on OAOA with OE:EA=1:2OE:EA=1:2. The lines ADAD and BEBE intersect at XX. Find the ratio AX:XDAX:XD.
Show worked solution

Worked solution

  1. Understand what is being asked

    OP→=a+mm+nAB→\overrightarrow{OP}=\mathbf{a}+\tfrac{m}{m+n}\overrightarrow{AB}

    Before calculating, we identify the goal and choose the method: the section (ratio) formula along the line. Having a plan stops us getting lost part-way through.

  2. Recall the key result we will use

    OP→=a+mm+nAB→\overrightarrow{OP}=\mathbf{a}+\tfrac{m}{m+n}\overrightarrow{AB}

    A point dividing ABAB in the ratio m:nm:n is that fraction of the way from AA to BB.

  3. Set up position vectors

    OA→=a,OB→=b\overrightarrow{OA}=\mathbf{a},\quad \overrightarrow{OB}=\mathbf{b}

    Choose a\mathbf{a} and b\mathbf{b} as base vectors; every point can be written using them.

  4. Note that a\mathbf{a} and b\mathbf{b} are non-parallel

    a∦b\mathbf{a}\nparallel\mathbf{b}

    Because they are not parallel, if two combinations are equal their coefficients must match — this is the key idea.

  5. Find OD→\overrightarrow{OD} and OE→\overrightarrow{OE}

    OD→=12b,OE→=13a\overrightarrow{OD}=\tfrac{1}{2}\mathbf{b},\quad \overrightarrow{OE}=\tfrac{1}{3}\mathbf{a}

    DD is the midpoint of OBOB; EE is 13\tfrac{1}{3} of the way along OAOA.

  6. Write the first line through the intersection

    OX→=a+t(12b−a)=(1−t)a+t2b\overrightarrow{OX}=\mathbf{a}+t\left(\tfrac{1}{2}\mathbf{b}-\mathbf{a}\right)=(1-t)\mathbf{a}+\tfrac{t}{2}\mathbf{b}

    Line ADAD starts at AA and moves a fraction tt towards DD.

  7. Describe line BEBE

    OX→=b+s(13a−b)\overrightarrow{OX}=\mathbf{b}+s\left(\tfrac{1}{3}\mathbf{a}-\mathbf{b}\right)

    Line BEBE starts at BB and moves a fraction ss towards EE.

  8. Write the second line

    OX→=s3a+(1−s)b\overrightarrow{OX}=\tfrac{s}{3}\mathbf{a}+(1-s)\mathbf{b}

    Expand line BEBE into a\mathbf{a} and b\mathbf{b} parts.

  9. Set the two expressions for OX→\overrightarrow{OX} equal

    λ(…)=μ(…) form\lambda(\ldots)=\mu(\ldots)\ \text{form}

    At the intersection XX both expressions describe the same point, so we equate them.

  10. Compare the coefficients of a\mathbf{a}

    1−t=s31-t=\tfrac{s}{3}

    Equate a\mathbf{a}-coefficients.

  11. Compare the coefficients of b\mathbf{b}

    t2=1−s\tfrac{t}{2}=1-s

    Equate b\mathbf{b}-coefficients.

  12. Eliminate one unknown

    s=1−t2 ⇒ 1−t=13(1−t2)s=1-\tfrac{t}{2}\ \Rightarrow\ 1-t=\tfrac{1}{3}\left(1-\tfrac{t}{2}\right)

    Rearrange the b\mathbf{b}-equation for ss and substitute into the a\mathbf{a}-equation.

  13. Solve for λ\lambda

    1−t=13−t6 ⇒ 5t6=23 ⇒ t=451-t=\tfrac{1}{3}-\tfrac{t}{6}\ \Rightarrow\ \tfrac{5t}{6}=\tfrac{2}{3}\ \Rightarrow\ t=\tfrac{4}{5}

    Simplify and solve for tt.

  14. Back-substitute for the other scalar

    s=1−12⋅45=35s=1-\tfrac{1}{2}\cdot\tfrac{4}{5}=\tfrac{3}{5}

    Back-substitute to find ss.

  15. Interpret λ\lambda as a ratio

    AX:XD=t:(1−t)=45:15AX:XD=t:(1-t)=\tfrac{4}{5}:\tfrac{1}{5}

    XX is a fraction tt of the way from AA to DD.

  16. State the ratio

    AX:XD=4:1AX:XD=4:1

    Clearing fifths gives the ratio.

Answer
AX:XD=4:1AX:XD=4:1
Question 3
7 markschallenging
OACBOACB is a parallelogram with OA→=a=(60)\overrightarrow{OA}=\mathbf{a}=\begin{pmatrix} 6 \\ 0 \end{pmatrix} and OB→=b=(24)\overrightarrow{OB}=\mathbf{b}=\begin{pmatrix} 2 \\ 4 \end{pmatrix}. MM is the midpoint of the side ACAC. Find OM→\overrightarrow{OM}.
Show worked solution

Worked solution

  1. Understand what is being asked

    AB→=DC→\overrightarrow{AB}=\overrightarrow{DC}

    Before calculating, we identify the goal and choose the method: the property that opposite sides are equal vectors. Having a plan stops us getting lost part-way through.

  2. Recall the key result we will use

    AB→=DC→\overrightarrow{AB}=\overrightarrow{DC}

    In a parallelogram opposite sides are equal and parallel as vectors.

  3. Position vector of CC

    OC→=a+b=(60)+(24)=(84)\overrightarrow{OC}=\mathbf{a}+\mathbf{b}=\begin{pmatrix} 6 \\ 0 \end{pmatrix}+\begin{pmatrix} 2 \\ 4 \end{pmatrix}=\begin{pmatrix} 8 \\ 4 \end{pmatrix}

    In parallelogram OACBOACB, CC is reached by adding both side vectors.

  4. Position vector of AA

    OA→=(60)\overrightarrow{OA}=\begin{pmatrix} 6 \\ 0 \end{pmatrix}

    This is one endpoint of the side ACAC.

  5. Find AC→\overrightarrow{AC}

    AC→=OC→−OA→=(84)−(60)=(24)\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=\begin{pmatrix} 8 \\ 4 \end{pmatrix}-\begin{pmatrix} 6 \\ 0 \end{pmatrix}=\begin{pmatrix} 2 \\ 4 \end{pmatrix}

    The side ACAC is parallel to OBOB, as expected in a parallelogram.

  6. Midpoint of ACAC from AA

    AM→=12AC→=(12)\overrightarrow{AM}=\tfrac{1}{2}\overrightarrow{AC}=\begin{pmatrix} 1 \\ 2 \end{pmatrix}

    Halve AC→\overrightarrow{AC} to reach its midpoint from AA.

  7. Add to OA→\overrightarrow{OA}

    OM→=OA→+AM→=(60)+(12)\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM}=\begin{pmatrix} 6 \\ 0 \end{pmatrix}+\begin{pmatrix} 1 \\ 2 \end{pmatrix}

    Travel from OO to AA, then halfway along ACAC.

  8. Simplify

    OM→=(72)\overrightarrow{OM}=\begin{pmatrix} 7 \\ 2 \end{pmatrix}

    Add the components.

  9. State the answer

    OM→=(72)\overrightarrow{OM}=\begin{pmatrix} 7 \\ 2 \end{pmatrix}

    So MM has position vector (72)\begin{pmatrix} 7 \\ 2 \end{pmatrix}.

  10. Why this method works

    AB→=DC→\overrightarrow{AB}=\overrightarrow{DC}

    Equal opposite sides let us reach a missing vertex by adding a known displacement.

  11. Link to earlier work

    AB→=DC→\overrightarrow{AB}=\overrightarrow{DC}

    This uses the vector addition and equality ideas from earlier. Connecting new work to things you already know makes it far easier to remember.

  12. Write down what we are working with

    list the given vectors / points before substituting\text{list the given vectors / points before substituting}

    Carefully noting the given information first is a habit that prevents careless substitution errors.

  13. Check the answer

    (72)\begin{pmatrix} 7 \\ 2 \end{pmatrix}

    Check both pairs of opposite sides come out equal. Checking your work is how you catch small mistakes before they cost marks.

  14. Interpret the result

    (72)\begin{pmatrix} 7 \\ 2 \end{pmatrix}

    The four points form a genuine parallelogram.

  15. Watch out for a common slip

    AB→=DC→\overrightarrow{AB}=\overrightarrow{DC}

    A frequent mistake here is rushing the signs or the order of subtraction; going slowly on those parts keeps the work accurate.

  16. State the final answer clearly

    (72)\begin{pmatrix} 7 \\ 2 \end{pmatrix}

    Always finish by writing the answer plainly so it is easy to read and mark.

Answer
(72)\begin{pmatrix} 7 \\ 2 \end{pmatrix}
Question 4
7 markschallenging
In triangle OABOAB, OA→=a=(42)\overrightarrow{OA}=\mathbf{a}=\begin{pmatrix} 4 \\ 2 \end{pmatrix} and OB→=b=(26)\overrightarrow{OB}=\mathbf{b}=\begin{pmatrix} 2 \\ 6 \end{pmatrix}. MM and NN are the midpoints of OAOA and OBOB. Find MN→\overrightarrow{MN} and verify it equals 12AB→\tfrac{1}{2}\overrightarrow{AB}.
Show worked solution

Worked solution

  1. Understand what is being asked

    MN→=12AB→\overrightarrow{MN}=\tfrac{1}{2}\overrightarrow{AB}

    Before calculating, we identify the goal and choose the method: the midpoint (midline) theorem in vector form. Having a plan stops us getting lost part-way through.

  2. Recall the key result we will use

    MN→=12AB→\overrightarrow{MN}=\tfrac{1}{2}\overrightarrow{AB}

    The segment joining two midpoints is parallel to the third side and half its length.

  3. Position vector of MM

    OM→=12a=12(42)=(21)\overrightarrow{OM}=\tfrac{1}{2}\mathbf{a}=\tfrac{1}{2}\begin{pmatrix} 4 \\ 2 \end{pmatrix}=\begin{pmatrix} 2 \\ 1 \end{pmatrix}

    MM is the midpoint of OAOA, so halve a\mathbf{a}.

  4. Position vector of NN

    ON→=12b=12(26)=(13)\overrightarrow{ON}=\tfrac{1}{2}\mathbf{b}=\tfrac{1}{2}\begin{pmatrix} 2 \\ 6 \end{pmatrix}=\begin{pmatrix} 1 \\ 3 \end{pmatrix}

    NN is the midpoint of OBOB, so halve b\mathbf{b}.

  5. Find MN→\overrightarrow{MN}

    MN→=ON→−OM→=(13)−(21)=(−12)\overrightarrow{MN}=\overrightarrow{ON}-\overrightarrow{OM}=\begin{pmatrix} 1 \\ 3 \end{pmatrix}-\begin{pmatrix} 2 \\ 1 \end{pmatrix}=\begin{pmatrix} -1 \\ 2 \end{pmatrix}

    Subtract the position vectors of the two midpoints.

  6. Find AB→\overrightarrow{AB}

    AB→=b−a=(26)−(42)=(−24)\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix} 2 \\ 6 \end{pmatrix}-\begin{pmatrix} 4 \\ 2 \end{pmatrix}=\begin{pmatrix} -2 \\ 4 \end{pmatrix}

    The third side of the triangle.

  7. Compute 12AB→\tfrac{1}{2}\overrightarrow{AB}

    12(−24)=(−12)\tfrac{1}{2}\begin{pmatrix} -2 \\ 4 \end{pmatrix}=\begin{pmatrix} -1 \\ 2 \end{pmatrix}

    Halve AB→\overrightarrow{AB}.

  8. Compare

    MN→=(−12)=12AB→\overrightarrow{MN}=\begin{pmatrix} -1 \\ 2 \end{pmatrix}=\tfrac{1}{2}\overrightarrow{AB}

    The two results match, confirming the midpoint theorem numerically.

  9. State the answer

    MN→=(−12)\overrightarrow{MN}=\begin{pmatrix} -1 \\ 2 \end{pmatrix}

    So MN→\overrightarrow{MN} is parallel to AB→\overrightarrow{AB} and half its length.

  10. Why this method works

    MN→=12AB→\overrightarrow{MN}=\tfrac{1}{2}\overrightarrow{AB}

    Halving each side and subtracting factors out a clean one-half of the third side.

  11. Link to earlier work

    MN→=12AB→\overrightarrow{MN}=\tfrac{1}{2}\overrightarrow{AB}

    Uses midpoints and the parallel-vector condition together. Connecting new work to things you already know makes it far easier to remember.

  12. Write down what we are working with

    list the given vectors / points before substituting\text{list the given vectors / points before substituting}

    Carefully noting the given information first is a habit that prevents careless substitution errors.

  13. Check the answer

    (−12)\begin{pmatrix} -1 \\ 2 \end{pmatrix}

    Check the joining vector is exactly half the third-side vector. Checking your work is how you catch small mistakes before they cost marks.

  14. Interpret the result

    (−12)\begin{pmatrix} -1 \\ 2 \end{pmatrix}

    This proves the classic midline result with vectors.

  15. Watch out for a common slip

    MN→=12AB→\overrightarrow{MN}=\tfrac{1}{2}\overrightarrow{AB}

    A frequent mistake here is rushing the signs or the order of subtraction; going slowly on those parts keeps the work accurate.

  16. State the final answer clearly

    (−12)\begin{pmatrix} -1 \\ 2 \end{pmatrix}

    Always finish by writing the answer plainly so it is easy to read and mark.

Answer
(−12)\begin{pmatrix} -1 \\ 2 \end{pmatrix}
Question 5
8 markschallenging
PQRSPQRS is any quadrilateral with vertices at position vectors p,q,r,s\mathbf{p},\mathbf{q},\mathbf{r},\mathbf{s}. Let A,B,C,DA,B,C,D be the midpoints of PQ,QR,RS,SPPQ,QR,RS,SP respectively. Prove that ABCDABCD is a parallelogram.
Show worked solution

Worked solution

  1. Understand what is being asked

    AB→=DC→\overrightarrow{AB}=\overrightarrow{DC}

    Before calculating, we identify the goal and choose the method: the property that opposite sides are equal vectors. Having a plan stops us getting lost part-way through.

  2. Recall the key result we will use

    AB→=DC→\overrightarrow{AB}=\overrightarrow{DC}

    In a parallelogram opposite sides are equal and parallel as vectors.

  3. Midpoint AA of PQPQ

    A=12(p+q)\mathbf{A}=\tfrac{1}{2}(\mathbf{p}+\mathbf{q})

    The midpoint of a side is the average of its two endpoints.

  4. Midpoint BB of QRQR

    B=12(q+r)\mathbf{B}=\tfrac{1}{2}(\mathbf{q}+\mathbf{r})

    Same rule applied to side QRQR.

  5. Midpoint CC of RSRS

    C=12(r+s)\mathbf{C}=\tfrac{1}{2}(\mathbf{r}+\mathbf{s})

    And for side RSRS.

  6. Midpoint DD of SPSP

    D=12(s+p)\mathbf{D}=\tfrac{1}{2}(\mathbf{s}+\mathbf{p})

    And for side SPSP.

  7. Find AB→\overrightarrow{AB}

    AB→=B−A=12(r−p)\overrightarrow{AB}=\mathbf{B}-\mathbf{A}=\tfrac{1}{2}(\mathbf{r}-\mathbf{p})

    Subtract: the q\mathbf{q} terms cancel, leaving 12(r−p)\tfrac{1}{2}(\mathbf{r}-\mathbf{p}).

  8. Find DC→\overrightarrow{DC}

    DC→=C−D=12(r−p)\overrightarrow{DC}=\mathbf{C}-\mathbf{D}=\tfrac{1}{2}(\mathbf{r}-\mathbf{p})

    Here the s\mathbf{s} terms cancel, giving the same result.

  9. Compare

    AB→=DC→\overrightarrow{AB}=\overrightarrow{DC}

    Both equal 12(r−p)\tfrac{1}{2}(\mathbf{r}-\mathbf{p}), which is half the diagonal PRPR.

  10. Conclude

    ABCD is a parallelogramABCD\text{ is a parallelogram}

    One pair of opposite sides is equal and parallel, which is exactly the condition for a parallelogram.

  11. Why this method works

    AB→=DC→\overrightarrow{AB}=\overrightarrow{DC}

    Equal opposite sides let us reach a missing vertex by adding a known displacement.

  12. Link to earlier work

    AB→=DC→\overrightarrow{AB}=\overrightarrow{DC}

    This uses the vector addition and equality ideas from earlier. Connecting new work to things you already know makes it far easier to remember.

  13. Write down what we are working with

    list the given vectors / points before substituting\text{list the given vectors / points before substituting}

    Carefully noting the given information first is a habit that prevents careless substitution errors.

  14. Check the answer

    AB→=DC→=12(r−p)\overrightarrow{AB}=\overrightarrow{DC}=\tfrac{1}{2}(\mathbf{r}-\mathbf{p})

    Check both pairs of opposite sides come out equal. Checking your work is how you catch small mistakes before they cost marks.

  15. Interpret the result

    AB→=DC→=12(r−p)\overrightarrow{AB}=\overrightarrow{DC}=\tfrac{1}{2}(\mathbf{r}-\mathbf{p})

    The four points form a genuine parallelogram.

  16. Watch out for a common slip

    AB→=DC→\overrightarrow{AB}=\overrightarrow{DC}

    A frequent mistake here is rushing the signs or the order of subtraction; going slowly on those parts keeps the work accurate.

Answer
AB→=DC→=12(r−p)\overrightarrow{AB}=\overrightarrow{DC}=\tfrac{1}{2}(\mathbf{r}-\mathbf{p})

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