Vector geometry and proof Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Vector geometry and proof questions. See exactly how to solve problems on position vectors, triangle law, vectors in terms of a and b, midpoints.

position vectorstriangle lawvectors in terms of a and bmidpointsexact halvesparallelogram
GCSE Higher70 questionsStep-by-step solutions
Question 1
1 markeasy
OA=a\overrightarrow{OA} = \mathbf{a} and OB=b\overrightarrow{OB} = \mathbf{b}. Work out AB\overrightarrow{AB} in terms of a\mathbf{a} and b\mathbf{b}.

Worked solution

  1. Write down the two vectors the question gives you

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

    Every vector in the figure has to be written in terms of a\mathbf{a} and b\mathbf{b}, so start from the two the question hands you: OA=a\overrightarrow{OA} = \mathbf{a} and OB=b\overrightarrow{OB} = \mathbf{b}.

  2. Substitute the position vectors

    AB=(b)(a)\overrightarrow{AB} = \left(\mathbf{b}\right) - \left(\mathbf{a}\right)

    OB=b\overrightarrow{OB} = \mathbf{b} and OA=a\overrightarrow{OA} = \mathbf{a}. The brackets matter: the whole of the second vector is subtracted.

  3. State the answer

    AB=a+b\overrightarrow{AB} = -\mathbf{a} + \mathbf{b}

    So AB=a+b\overrightarrow{AB} = -\mathbf{a} + \mathbf{b}, written in terms of a\mathbf{a} and b\mathbf{b} as the question asked.

Answer
AB=a+b\overrightarrow{AB} = -\mathbf{a} + \mathbf{b}
Question 2
1 markeasy
OA=a\overrightarrow{OA} = \mathbf{a} and OB=b\overrightarrow{OB} = \mathbf{b}. Work out BA\overrightarrow{BA} in terms of a\mathbf{a} and b\mathbf{b}.

Worked solution

  1. Write down the two vectors the question gives you

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

    Every vector in the figure has to be written in terms of a\mathbf{a} and b\mathbf{b}, so start from the two the question hands you: OA=a\overrightarrow{OA} = \mathbf{a} and OB=b\overrightarrow{OB} = \mathbf{b}.

  2. Substitute the position vectors

    BA=(a)(b)\overrightarrow{BA} = \left(\mathbf{a}\right) - \left(\mathbf{b}\right)

    OA=a\overrightarrow{OA} = \mathbf{a} and OB=b\overrightarrow{OB} = \mathbf{b}. The brackets matter: the whole of the second vector is subtracted.

  3. State the answer

    BA=ab\overrightarrow{BA} = \mathbf{a} - \mathbf{b}

    So BA=ab\overrightarrow{BA} = \mathbf{a} - \mathbf{b}, written in terms of a\mathbf{a} and b\mathbf{b} as the question asked.

Answer
BA=ab\overrightarrow{BA} = \mathbf{a} - \mathbf{b}
Question 3
2 markseasy
OA=a\overrightarrow{OA} = \mathbf{a} and OB=b\overrightarrow{OB} = \mathbf{b}. MM is the midpoint of ABAB. Work out OM\overrightarrow{OM} in terms of a\mathbf{a} and b\mathbf{b}.

Worked solution

  1. Write down the two vectors the question gives you

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

    Every vector in the figure has to be written in terms of a\mathbf{a} and b\mathbf{b}, so start from the two the question hands you: OA=a\overrightarrow{OA} = \mathbf{a} and OB=b\overrightarrow{OB} = \mathbf{b}.

  2. Write down the position vector of M, the midpoint of AB

    OM=12(OA+OB)=12a+12b\overrightarrow{OM} = \frac{1}{2}\left(\overrightarrow{OA} + \overrightarrow{OB}\right) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}

    The midpoint of ABAB has position vector the average of the two ends, so OM=12a+12b\overrightarrow{OM} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}. Halving is exact — keep it as a fraction.

  3. State the answer

    OM=12a+12b\overrightarrow{OM} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}

    So OM=12a+12b\overrightarrow{OM} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}, written in terms of a\mathbf{a} and b\mathbf{b} as the question asked.

Answer
OM=12a+12b\overrightarrow{OM} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}
Question 4
2 markseasy
OA=a\overrightarrow{OA} = \mathbf{a} and OB=b\overrightarrow{OB} = \mathbf{b}. MM is the midpoint of ABAB. Work out AM\overrightarrow{AM} in terms of a\mathbf{a} and b\mathbf{b}.

Worked solution

  1. Write down the two vectors the question gives you

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

    Every vector in the figure has to be written in terms of a\mathbf{a} and b\mathbf{b}, so start from the two the question hands you: OA=a\overrightarrow{OA} = \mathbf{a} and OB=b\overrightarrow{OB} = \mathbf{b}.

  2. Write down the position vector of M, the midpoint of AB

    OM=12(OA+OB)=12a+12b\overrightarrow{OM} = \frac{1}{2}\left(\overrightarrow{OA} + \overrightarrow{OB}\right) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}

    The midpoint of ABAB has position vector the average of the two ends, so OM=12a+12b\overrightarrow{OM} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}. Halving is exact — keep it as a fraction.

  3. State the answer

    AM=(12a+12b)(a)=12a+12b\overrightarrow{AM} = \left(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) - \left(\mathbf{a}\right) = -\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}

    So AM=12a+12b\overrightarrow{AM} = -\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}, written in terms of a\mathbf{a} and b\mathbf{b} as the question asked.

Answer
AM=12a+12b\overrightarrow{AM} = -\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}
Question 5
1 markeasy
OA=a\overrightarrow{OA} = \mathbf{a} and OB=b\overrightarrow{OB} = \mathbf{b}. CC is the point such that OACBOACB is a parallelogram. Work out OC\overrightarrow{OC} in terms of a\mathbf{a} and b\mathbf{b}.

Worked solution

  1. Write down the two vectors the question gives you

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

    Every vector in the figure has to be written in terms of a\mathbf{a} and b\mathbf{b}, so start from the two the question hands you: OA=a\overrightarrow{OA} = \mathbf{a} and OB=b\overrightarrow{OB} = \mathbf{b}.

  2. Write down the position vector of C, the fourth vertex of the parallelogram OACB

    OC=OA+OBOC=a+b\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{OB} \Rightarrow \overrightarrow{OC} = \mathbf{a} + \mathbf{b}

    In a parallelogram OACBOACB the diagonals OCOC and ABAB bisect each other, so the two diagonals have the same midpoint and OC=OA+OB\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{OB}. That gives OC=a+b\overrightarrow{OC} = \mathbf{a} + \mathbf{b}.

  3. State the answer

    OC=a+b\overrightarrow{OC} = \mathbf{a} + \mathbf{b}

    So OC=a+b\overrightarrow{OC} = \mathbf{a} + \mathbf{b}, written in terms of a\mathbf{a} and b\mathbf{b} as the question asked.

Answer
OC=a+b\overrightarrow{OC} = \mathbf{a} + \mathbf{b}

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