Teach Me Maths

Vector Proof

Question 11 mark

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. Express \overrightarrow{AB} in terms of \mathbf{a} and \mathbf{b}.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Travel from A to O and then from O to B.

Worked solution

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. Express \overrightarrow{AB} in terms of \mathbf{a} and \mathbf{b}.

  1. \overrightarrow{AO}=-\mathbf{a}.
  2. Therefore \overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=\mathbf{b}-\mathbf{a}.

Question 22 marks

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. M is the midpoint of AB. Express \overrightarrow{OM} in terms of \mathbf{a} and \mathbf{b}.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Use half of the vector from A to B.

Worked solution

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. M is the midpoint of AB. Express \overrightarrow{OM} in terms of \mathbf{a} and \mathbf{b}.

  1. Go from A to B via O: \overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=-\mathbf{a}+\mathbf{b}=\mathbf{b}-\mathbf{a}.
  2. M is the midpoint of AB, so \overrightarrow{AM}=\frac12\overrightarrow{AB}=\frac12(\mathbf{b}-\mathbf{a}).
  3. \overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM}=\mathbf{a}+\frac12\mathbf{b}-\frac12\mathbf{a}=\frac12\mathbf{a}+\frac12\mathbf{b}

Question 32 marks

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. P lies on AB with AP:PB=2:3. Express \overrightarrow{OP} in terms of \mathbf{a} and \mathbf{b}.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

AP is two fifths of the complete vector AB.

Worked solution

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. P lies on AB with AP:PB=2:3. Express \overrightarrow{OP} in terms of \mathbf{a} and \mathbf{b}.

  1. \overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=\mathbf{b}-\mathbf{a}.
  2. AP:PB=2:3 splits AB into 5 equal parts, and AP is 2 of them, so \overrightarrow{AP}=\frac25(\mathbf{b}-\mathbf{a}).
  3. \overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}=\mathbf{a}+\frac25\mathbf{b}-\frac25\mathbf{a}=\frac35\mathbf{a}+\frac25\mathbf{b}

Question 42 marks

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. P lies on OA with OP:PA=3:1. Express \overrightarrow{PB} in terms of \mathbf{a} and \mathbf{b}.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Reverse OP to travel from P back to O.

Worked solution

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. P lies on OA with OP:PA=3:1. Express \overrightarrow{PB} in terms of \mathbf{a} and \mathbf{b}.

  1. \overrightarrow{OP}=\frac34\mathbf{a}, so \overrightarrow{PO}=-\frac34\mathbf{a}.
  2. \overrightarrow{PB}=\overrightarrow{PO}+\overrightarrow{OB}=\mathbf{b}-\frac34\mathbf{a}.

Question 51 mark

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. Which expression equals \overrightarrow{BA}?

Hint

The order of the two endpoint letters sets the direction.

Worked solution

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. Which expression equals \overrightarrow{BA}?

  1. \mathbf{a}-\mathbf{b}
  2. \mathbf{b}-\mathbf{a}
  3. \mathbf{a}+\mathbf{b}
  4. -\mathbf{a}-\mathbf{b}
  1. Reversing a vector changes its sign, so \overrightarrow{BO}=-\mathbf{b}.
  2. Go from B to A via O: \overrightarrow{BA}=\overrightarrow{BO}+\overrightarrow{OA}=-\mathbf{b}+\mathbf{a}=\mathbf{a}-\mathbf{b}.
  3. \mathbf{b}-\mathbf{a} is \overrightarrow{AB}: the same line, but in the opposite direction.

Question 63 marks

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. P and Q are the midpoints of OA and OB. Express \overrightarrow{PQ} in terms of \mathbf{a} and \mathbf{b}.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Subtract the position vector of P from the position vector of Q.

Worked solution

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. P and Q are the midpoints of OA and OB. Express \overrightarrow{PQ} in terms of \mathbf{a} and \mathbf{b}.

  1. \overrightarrow{OP}=\frac12\mathbf{a} and \overrightarrow{OQ}=\frac12\mathbf{b}.
  2. \overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}=\frac12(\mathbf{b}-\mathbf{a}).

Question 72 marks

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. P and Q are the midpoints of OA and OB. Which vector argument proves PQ\parallel AB?

Hint

A non-zero scalar multiple gives parallel vectors.

Worked solution

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. P and Q are the midpoints of OA and OB. Which vector argument proves PQ\parallel AB?

  1. \overrightarrow{PQ}=\frac12(\mathbf{a}+\mathbf{b}), so the lines are parallel.
  2. \overrightarrow{PQ}=\frac12(\mathbf{b}-\mathbf{a}) and \overrightarrow{AB}=\mathbf{a}-\mathbf{b}, so PQ is not parallel to AB.
  3. \overrightarrow{PQ}=\frac12(\mathbf{b}-\mathbf{a})=\frac12\overrightarrow{AB}, so the vectors have the same direction.
  4. \overrightarrow{PQ}=\overrightarrow{OA}, so the lines are parallel.
  1. P and Q are midpoints, so \overrightarrow{OP}=\frac12\mathbf{a} and \overrightarrow{OQ}=\frac12\mathbf{b}.
  2. \overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}=\frac12\mathbf{b}-\frac12\mathbf{a}=\frac12(\mathbf{b}-\mathbf{a}).
  3. \overrightarrow{AB}=\mathbf{b}-\mathbf{a}, so \overrightarrow{PQ}=\frac12\overrightarrow{AB}. One vector is a multiple of the other, so PQ is parallel to AB.
  4. \overrightarrow{AB} is \mathbf{b}-\mathbf{a}, not \mathbf{a}-\mathbf{b}. Even a negative multiple would still mean the lines are parallel.

Question 83 marks

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. P lies on OA with OP:PA=1:2. Q lies on OB with OQ:QB=2:1. Express \overrightarrow{PQ} in terms of \mathbf{a} and \mathbf{b}.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Convert each ratio into a fraction of its complete side.

Worked solution

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. P lies on OA with OP:PA=1:2. Q lies on OB with OQ:QB=2:1. Express \overrightarrow{PQ} in terms of \mathbf{a} and \mathbf{b}.

  1. \overrightarrow{OP}=\frac13\mathbf{a} and \overrightarrow{OQ}=\frac23\mathbf{b}.
  2. \overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}=\frac23\mathbf{b}-\frac13\mathbf{a}.

Question 92 marks

Relative to O, the position vectors of P and Q are 2\mathbf{a}+\mathbf{b} and 5\mathbf{a}+4\mathbf{b}. Express \overrightarrow{PQ} in terms of \mathbf{a} and \mathbf{b}.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Use final position minus initial position.

Worked solution

Relative to O, the position vectors of P and Q are 2\mathbf{a}+\mathbf{b} and 5\mathbf{a}+4\mathbf{b}. Express \overrightarrow{PQ} in terms of \mathbf{a} and \mathbf{b}.

  1. \overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}=(5\mathbf{a}+4\mathbf{b})-(2\mathbf{a}+\mathbf{b}).
  2. Collecting the coefficients gives 3\mathbf{a}+3\mathbf{b}, or 3(\mathbf{a}+\mathbf{b}).

Question 103 marks

Relative to O, P, Q and R have position vectors \mathbf{a}+2\mathbf{b}, 3\mathbf{a}+5\mathbf{b} and 5\mathbf{a}+8\mathbf{b}, where \mathbf{a} and \mathbf{b} are non-parallel. Which argument proves that P, Q and R are collinear?

Hint

Compare vectors with the same starting point.

Worked solution

Relative to O, P, Q and R have position vectors \mathbf{a}+2\mathbf{b}, 3\mathbf{a}+5\mathbf{b} and 5\mathbf{a}+8\mathbf{b}, where \mathbf{a} and \mathbf{b} are non-parallel. Which argument proves that P, Q and R are collinear?

  1. \overrightarrow{PR}=2\overrightarrow{PQ}, and these vectors have the common starting point P.
  2. \overrightarrow{OR}=5\mathbf{a}+8\mathbf{b} is not a multiple of \overrightarrow{OP}=\mathbf{a}+2\mathbf{b}, so P, Q and R are not collinear.
  3. \overrightarrow{PQ} and \overrightarrow{QR} are both 2\mathbf{a}+3\mathbf{b}, so PQ is parallel to QR; parallel lines never meet, so the points are not collinear.
  4. \overrightarrow{OQ}=\overrightarrow{OP}+\overrightarrow{OR}, so Q is between P and R.
  1. \overrightarrow{PQ}=(3\mathbf{a}+5\mathbf{b})-(\mathbf{a}+2\mathbf{b})=2\mathbf{a}+3\mathbf{b}
  2. \overrightarrow{PR}=(5\mathbf{a}+8\mathbf{b})-(\mathbf{a}+2\mathbf{b})=4\mathbf{a}+6\mathbf{b}=2(2\mathbf{a}+3\mathbf{b})=2\overrightarrow{PQ}
  3. So PR is parallel to PQ. Both lines go through P, so they are the same straight line: P, Q and R are collinear.
  4. Parallel vectors with a point in common lie on one line, so PQ\parallel QR shows that the points are collinear, not that they are not. Comparing position vectors from O doesn't test whether the points are in line, and \overrightarrow{OP}+\overrightarrow{OR}=6\mathbf{a}+10\mathbf{b}, not \overrightarrow{OQ}.

Question 113 marks

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. P lies on OA with OP:PA=2:3. Q lies on AB with AQ:QB=1:2. Express \overrightarrow{PQ} in terms of \mathbf{a} and \mathbf{b}.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Find OQ by travelling through A, then subtract OP.

Worked solution

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. P lies on OA with OP:PA=2:3. Q lies on AB with AQ:QB=1:2. Express \overrightarrow{PQ} in terms of \mathbf{a} and \mathbf{b}.

  1. \overrightarrow{OQ}=\mathbf{a}+\frac13(\mathbf{b}-\mathbf{a})=\frac23\mathbf{a}+\frac13\mathbf{b}.
  2. \overrightarrow{OP}=\frac25\mathbf{a}.
  3. Subtracting gives \overrightarrow{PQ}=(\frac23-\frac25)\mathbf{a}+\frac13\mathbf{b}=\frac4{15}\mathbf{a}+\frac13\mathbf{b}.

Question 123 marks

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. M is the midpoint of AB. N lies on OB with ON:NB=2:1. Express \overrightarrow{MN} in terms of \mathbf{a} and \mathbf{b}.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Subtract the midpoint position vector from ON.

Worked solution

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. M is the midpoint of AB. N lies on OB with ON:NB=2:1. Express \overrightarrow{MN} in terms of \mathbf{a} and \mathbf{b}.

  1. \overrightarrow{OM}=\frac12\mathbf{a}+\frac12\mathbf{b} and \overrightarrow{ON}=\frac23\mathbf{b}.
  2. \overrightarrow{MN}=\overrightarrow{ON}-\overrightarrow{OM}=-\frac12\mathbf{a}+\frac16\mathbf{b}.

Question 133 marks

OACB is a parallelogram with \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. P is the midpoint of AC and Q is the midpoint of BC. Express \overrightarrow{PQ} in terms of \mathbf{a} and \mathbf{b}.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Find both midpoint position vectors using the parallel sides.

Worked solution

OACB is a parallelogram with \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. P is the midpoint of AC and Q is the midpoint of BC. Express \overrightarrow{PQ} in terms of \mathbf{a} and \mathbf{b}.

  1. \overrightarrow{AC}=\mathbf{b} and \overrightarrow{BC}=\mathbf{a} in the parallelogram.
  2. \overrightarrow{OP}=\mathbf{a}+\frac12\mathbf{b} and \overrightarrow{OQ}=\mathbf{b}+\frac12\mathbf{a}.
  3. Subtracting gives \overrightarrow{PQ}=\frac12\mathbf{b}-\frac12\mathbf{a}=\frac12(\mathbf{b}-\mathbf{a}).

Question 143 marks

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. OP:PA=3:2 and OQ:QB=3:2, with P on OA and Q on OB. Given \overrightarrow{PQ}=k\overrightarrow{AB}, find k. Give your answer as a fraction.

Hint

Write each position vector as the same fraction of its complete side.

Worked solution

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. OP:PA=3:2 and OQ:QB=3:2, with P on OA and Q on OB. Given \overrightarrow{PQ}=k\overrightarrow{AB}, find k. Give your answer as a fraction.

  1. \overrightarrow{OP}=\frac35\mathbf{a} and \overrightarrow{OQ}=\frac35\mathbf{b}.
  2. \overrightarrow{PQ}=\frac35\mathbf{b}-\frac35\mathbf{a}=\frac35(\mathbf{b}-\mathbf{a})=\frac35\overrightarrow{AB}.
  3. Therefore k=\frac35.

Question 153 marks

OACB is a parallelogram. \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. X is the point on the diagonal OC such that OX:XC=1:3.

Express \overrightarrow{AX} in terms of \mathbf{a} and \mathbf{b}. Give your answer in its simplest form.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Find \overrightarrow{OC} first, then \overrightarrow{OX}, then go from A to X via O.

Worked solution

OACB is a parallelogram. \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. X is the point on the diagonal OC such that OX:XC=1:3.

Express \overrightarrow{AX} in terms of \mathbf{a} and \mathbf{b}. Give your answer in its simplest form.

  1. In the parallelogram, \overrightarrow{AC}=\overrightarrow{OB}=\mathbf{b}, so \overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{AC}=\mathbf{a}+\mathbf{b}.
  2. OX:XC=1:3, so X is \frac14 of the way along OC: \overrightarrow{OX}=\frac14(\mathbf{a}+\mathbf{b}).
  3. \overrightarrow{AX}=\overrightarrow{AO}+\overrightarrow{OX}=-\mathbf{a}+\frac14\mathbf{a}+\frac14\mathbf{b}
  4. \overrightarrow{AX}=\frac14\mathbf{b}-\frac34\mathbf{a}

Question 16Challenge3 marks

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. P lies on AB with AP:PB=1:2. Q lies on OA with OQ:QA=2:1.

Express \overrightarrow{QP} in terms of \mathbf{a} and \mathbf{b}. Give your answer in its simplest form.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Find \overrightarrow{OP} by going through A, then subtract \overrightarrow{OQ}.

Worked solution

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. P lies on AB with AP:PB=1:2. Q lies on OA with OQ:QA=2:1.

Express \overrightarrow{QP} in terms of \mathbf{a} and \mathbf{b}. Give your answer in its simplest form.

  1. \overrightarrow{AB}=\mathbf{b}-\mathbf{a}, and AP is \frac13 of AB, so \overrightarrow{OP}=\mathbf{a}+\frac13(\mathbf{b}-\mathbf{a})=\frac23\mathbf{a}+\frac13\mathbf{b}.
  2. OQ is \frac23 of OA, so \overrightarrow{OQ}=\frac23\mathbf{a}.
  3. \overrightarrow{QP}=\overrightarrow{OP}-\overrightarrow{OQ}=\frac23\mathbf{a}+\frac13\mathbf{b}-\frac23\mathbf{a}=\frac13\mathbf{b}
  4. \overrightarrow{QP}=\frac13\overrightarrow{OB} is a multiple of \overrightarrow{OB}, which proves that QP is parallel to OB.

Question 17Challenge4 marks

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. P lies on OA with OP:PA=1:3, and Q lies on OB with OQ:QB=1:3. R lies on AB with AR:RB=2:1. S lies on PQ with PS:SQ=2:1. Given \overrightarrow{OS}=k\overrightarrow{OR}, find k. Give your answer as a fraction.

Hint

Find OR and OS, then compare both coefficients.

Worked solution

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. P lies on OA with OP:PA=1:3, and Q lies on OB with OQ:QB=1:3. R lies on AB with AR:RB=2:1. S lies on PQ with PS:SQ=2:1. Given \overrightarrow{OS}=k\overrightarrow{OR}, find k. Give your answer as a fraction.

  1. \overrightarrow{OR}=\overrightarrow{OA}+\frac23\overrightarrow{AB}=\mathbf{a}+\frac23(\mathbf{b}-\mathbf{a})=\frac13\mathbf{a}+\frac23\mathbf{b}
  2. \overrightarrow{OP}=\frac14\mathbf{a} and \overrightarrow{OQ}=\frac14\mathbf{b}, so \overrightarrow{PQ}=\frac14\mathbf{b}-\frac14\mathbf{a}.
  3. PS:SQ=2:1, so \overrightarrow{PS}=\frac23\overrightarrow{PQ}=\frac16\mathbf{b}-\frac16\mathbf{a}.
  4. \overrightarrow{OS}=\overrightarrow{OP}+\overrightarrow{PS}=\frac14\mathbf{a}-\frac16\mathbf{a}+\frac16\mathbf{b}=\frac1{12}\mathbf{a}+\frac16\mathbf{b}
  5. The red route O\to P\to S has one quarter of each coefficient in the blue route O\to A\to R. This gives \overrightarrow{OS}=\tfrac14\overrightarrow{OR}.
    Vector routes to R and STriangle OAB as in the question, with P and Q one quarter along OA and OB. R divides AB in ratio 2 to 1, and S divides PQ in ratio 2 to 1. The red route is O to P to S. The blue route is O to A to R. Dashed OSR shows the resulting common line.Compare the directed routesOABPQRSabOS = ¼ ORSame direction; quarter the length
    Same triangle as the question. The red route O\to P\to S is a quarter of the blue route O\to A\to R, leg by leg, so O, S and R lie on one straight line (dashed).
  6. Each coefficient in \overrightarrow{OS} is \frac14 of the matching one in \overrightarrow{OR}=\frac13\mathbf{a}+\frac23\mathbf{b}, so \overrightarrow{OS}=\frac14\overrightarrow{OR} and k=\frac14.

Question 18Challenge4 marks

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. Q lies on OA. P lies on the straight line BQ, with \overrightarrow{OP}=\frac15\mathbf{a}+\frac25\mathbf{b}. Write \overrightarrow{OQ}=k\mathbf{a} and find k. Give your answer as a fraction.

Hint

At Q the coefficient of \mathbf{b} must be zero.

Worked solution

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. Q lies on OA. P lies on the straight line BQ, with \overrightarrow{OP}=\frac15\mathbf{a}+\frac25\mathbf{b}. Write \overrightarrow{OQ}=k\mathbf{a} and find k. Give your answer as a fraction.

  1. \overrightarrow{BP}=\overrightarrow{OP}-\overrightarrow{OB}=\frac15\mathbf{a}+\frac25\mathbf{b}-\mathbf{b}=\frac15\mathbf{a}-\frac35\mathbf{b}
  2. Q is on the line BP, so \overrightarrow{BQ} is a multiple of \overrightarrow{BP}: \overrightarrow{BQ}=t\left(\frac15\mathbf{a}-\frac35\mathbf{b}\right) for some number t.
  3. \overrightarrow{OQ}=\overrightarrow{OB}+\overrightarrow{BQ}=\frac t5\mathbf{a}+\left(1-\frac{3t}5\right)\mathbf{b}
  4. Q is on OA, so \overrightarrow{OQ} has no \mathbf{b} part: 1-\frac{3t}5=0, so t=\frac53.
  5. The route O\to B\to Q ends on OA, so the \mathbf b contribution must cancel. The directed segment BQ is \tfrac53 times BP.
    Finding Q on OA using the line BPTriangle OAB as in the question. Q lies on OA and on the continuation of the line from B through P. The directed route O to B to Q ends at Q. P is three fifths along BQ.Extend the route through POABPQbBQ = ⁵⁄₃ BPOQ = ⅓ a
    Same triangle as the question. The line from B through P meets OA at Q, with BQ=\tfrac53BP.
  6. Then \overrightarrow{OQ}=\frac15\times\frac53\mathbf{a}=\frac13\mathbf{a}, so k=\frac13.

Question 19Challenge4 marks

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. M lies on AB with AM:MB=3:2. N lies on the line OA beyond O, and BN\parallel OM. Write \overrightarrow{ON}=k\mathbf{a} and find k. Give your answer as a fraction.

Hint

Use the coefficient of \mathbf{b} to find the scalar relating BN and OM.

Worked solution

In triangle OAB, \overrightarrow{OA}=\mathbf{a} and \overrightarrow{OB}=\mathbf{b}. M lies on AB with AM:MB=3:2. N lies on the line OA beyond O, and BN\parallel OM. Write \overrightarrow{ON}=k\mathbf{a} and find k. Give your answer as a fraction.

  1. \overrightarrow{AB}=\mathbf{b}-\mathbf{a} and AM is \frac35 of AB, so \overrightarrow{OM}=\mathbf{a}+\frac35(\mathbf{b}-\mathbf{a})=\frac25\mathbf{a}+\frac35\mathbf{b}.
  2. \overrightarrow{BN}=\overrightarrow{BO}+\overrightarrow{ON}=k\mathbf{a}-\mathbf{b}
  3. BN is parallel to OM, so \overrightarrow{BN} is a multiple of \overrightarrow{OM}: k\mathbf{a}-\mathbf{b}=t\left(\frac25\mathbf{a}+\frac35\mathbf{b}\right) for some number t.
  4. Compare the \mathbf{b} parts: -1=\frac35t, so t=-\frac53.
  5. The negative multiple makes \overrightarrow{BN} point opposite to \overrightarrow{OM}. The endpoint N lies beyond O, so its \mathbf a coefficient is negative.
    Negative multiple and opposite vector directionTriangle OAB as in the question, with N on the line AO extended beyond O. OM and BN are parallel with opposite directed arrows. The directed vector ON is minus two thirds of OA.A negative multiple reverses directionOABMNaBN = −⁵⁄₃ OMON = −⅔ a
    Same triangle as the question. \overrightarrow{OM} and \overrightarrow{BN} are parallel but point in opposite directions, and N is on the other side of O from A.
  6. Compare the \mathbf{a} parts: k=\frac25\times\left(-\frac53\right)=-\frac23. It is negative because N is on the other side of O from A.

Question 20Challenge4 marks

Relative to O, the position vectors of A, B, C and D are \mathbf{a}, \mathbf{b}, 2\mathbf{a}+3\mathbf{b} and 4\mathbf{a}+\mathbf{b}, where \mathbf{a} and \mathbf{b} are not parallel.

(a)

\overrightarrow{CD}=k\overrightarrow{AB}. Find the value of k.

2 marks

(b)

Which statement about the quadrilateral ABCD is correct?

2 marks

Hint

Compare the vectors for opposite sides: AB with CD, and AD with BC.

Worked solution

Relative to O, the position vectors of A, B, C and D are \mathbf{a}, \mathbf{b}, 2\mathbf{a}+3\mathbf{b} and 4\mathbf{a}+\mathbf{b}, where \mathbf{a} and \mathbf{b} are not parallel.

Part (a)

\overrightarrow{CD}=k\overrightarrow{AB}. Find the value of k.

  1. \overrightarrow{AB}=\mathbf{b}-\mathbf{a}
  2. \overrightarrow{CD}=\overrightarrow{OD}-\overrightarrow{OC}=(4\mathbf{a}+\mathbf{b})-(2\mathbf{a}+3\mathbf{b})=2\mathbf{a}-2\mathbf{b}
  3. The arrow from C to D points opposite to the arrow from A to B, and has twice its length.
    Parallel, with opposite directionsQuadrilateral ABCD as in the question, with A=a, B=b, C=2a+3b and D=4a+b. The arrows AB and CD are parallel with opposite directions, and CD is twice as long.Parallel, with opposite directionsABCDCD = −2 ABNegative sign reverses direction
    The quadrilateral from the question: \overrightarrow{CD} is parallel to \overrightarrow{AB}, twice as long and in the opposite direction.
  4. 2\mathbf{a}-2\mathbf{b}=-2(\mathbf{b}-\mathbf{a}), so \overrightarrow{CD}=-2\overrightarrow{AB} and k=-2.

Part (b)

Which statement about the quadrilateral ABCD is correct?

  1. ABCD is a parallelogram, because \overrightarrow{CD} is a multiple of \overrightarrow{AB}.
  2. ABCD is a trapezium: AB is parallel to CD, but AD is not parallel to BC.
  3. ABCD is a trapezium: AD is parallel to BC, but AB is not parallel to CD.
  4. ABCD has no parallel sides, because \overrightarrow{CD} and \overrightarrow{AB} point in opposite directions.
  1. From part (a), \overrightarrow{CD} is a multiple of \overrightarrow{AB}, so AB is parallel to CD. A negative multiple just means the opposite direction; the lines are still parallel.
  2. \overrightarrow{AD}=(4\mathbf{a}+\mathbf{b})-\mathbf{a}=3\mathbf{a}+\mathbf{b} and \overrightarrow{BC}=(2\mathbf{a}+3\mathbf{b})-\mathbf{b}=2\mathbf{a}+2\mathbf{b}.
  3. 3\mathbf{a}+\mathbf{b} is not a multiple of 2\mathbf{a}+2\mathbf{b} (the \mathbf{a} and \mathbf{b} parts are in different ratios), so AD is not parallel to BC.
  4. The remaining opposite sides have different coefficient ratios, so they are not parallel. Together with part (a), this establishes exactly one pair of parallel sides.
    Compare the other pair of sidesQuadrilateral ABCD as in the question. The arrows AD = 3a + b and BC = 2a + 2b point in different directions, so AD and BC are not parallel.Compare the other pair of sidesABCD3a + b2a + 2bCoefficient ratios differ
    The same quadrilateral: \overrightarrow{AD}=3\mathbf{a}+\mathbf{b} and \overrightarrow{BC}=2\mathbf{a}+2\mathbf{b} point in different directions, so AD is not parallel to BC.
  5. Exactly one pair of parallel sides makes ABCD a trapezium. A parallelogram needs both pairs of opposite sides parallel.

More on this topic: Vector Proof worksheet with full solutions

All Edexcel GCSE Higher practice questions