Vector arithmetic Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Vector arithmetic questions. See exactly how to solve problems on adding column vectors, negative components, subtracting column vectors, sign handling.

adding column vectorsnegative componentssubtracting column vectorssign handlingorder of subtractionscalar multiplication
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
a=(31)\mathbf{a} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} and b=(24)\mathbf{b} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}. Work out a+b\mathbf{a} + \mathbf{b} as a column vector.

Worked solution

  1. State the rule for combining column vectors

    (pq)+(rs)=(p+rq+s),(pq)−(rs)=(p−rq−s)\begin{pmatrix} p \\ q \end{pmatrix} + \begin{pmatrix} r \\ s \end{pmatrix} = \begin{pmatrix} p + r \\ q + s \end{pmatrix}, \quad \begin{pmatrix} p \\ q \end{pmatrix} - \begin{pmatrix} r \\ s \end{pmatrix} = \begin{pmatrix} p - r \\ q - s \end{pmatrix}

    Add or subtract the top components, then add or subtract the bottom components. A scalar multiplies both components. Nothing else is needed.

  2. Substitute and work down the two rows

    a+b=(31)+(24)=(3+21+4)=(55)\mathbf{a} + \mathbf{b} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} + \begin{pmatrix} 2 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 + 2 \\ 1 + 4 \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \end{pmatrix}

    Substituting the columns and combining the top row and the bottom row separately gives (55)\begin{pmatrix} 5 \\ 5 \end{pmatrix}.

  3. State the answer as a column vector

    a+b=(55)\mathbf{a} + \mathbf{b} = \begin{pmatrix} 5 \\ 5 \end{pmatrix}

    So a+b=(55)\mathbf{a} + \mathbf{b} = \begin{pmatrix} 5 \\ 5 \end{pmatrix}.

Answer
a+b=(55)\mathbf{a} + \mathbf{b} = \begin{pmatrix} 5 \\ 5 \end{pmatrix}
Question 2
2 markseasy
a=(5−2)\mathbf{a} = \begin{pmatrix} 5 \\ -2 \end{pmatrix} and b=(−13)\mathbf{b} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}. Work out a+b\mathbf{a} + \mathbf{b} as a column vector.

Worked solution

  1. State the rule for combining column vectors

    (pq)+(rs)=(p+rq+s),(pq)−(rs)=(p−rq−s)\begin{pmatrix} p \\ q \end{pmatrix} + \begin{pmatrix} r \\ s \end{pmatrix} = \begin{pmatrix} p + r \\ q + s \end{pmatrix}, \quad \begin{pmatrix} p \\ q \end{pmatrix} - \begin{pmatrix} r \\ s \end{pmatrix} = \begin{pmatrix} p - r \\ q - s \end{pmatrix}

    Add or subtract the top components, then add or subtract the bottom components. A scalar multiplies both components. Nothing else is needed.

  2. Substitute and work down the two rows

    a+b=(5−2)+(−13)=(5−1−2+3)=(41)\mathbf{a} + \mathbf{b} = \begin{pmatrix} 5 \\ -2 \end{pmatrix} + \begin{pmatrix} -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 5 - 1 \\ -2 + 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}

    Substituting the columns and combining the top row and the bottom row separately gives (41)\begin{pmatrix} 4 \\ 1 \end{pmatrix}.

  3. State the answer as a column vector

    a+b=(41)\mathbf{a} + \mathbf{b} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}

    So a+b=(41)\mathbf{a} + \mathbf{b} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}.

Answer
a+b=(41)\mathbf{a} + \mathbf{b} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}
Question 3
2 markseasy
a=(63)\mathbf{a} = \begin{pmatrix} 6 \\ 3 \end{pmatrix} and b=(25)\mathbf{b} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}. Work out a−b\mathbf{a} - \mathbf{b} as a column vector.

Worked solution

  1. State the rule for combining column vectors

    (pq)+(rs)=(p+rq+s),(pq)−(rs)=(p−rq−s)\begin{pmatrix} p \\ q \end{pmatrix} + \begin{pmatrix} r \\ s \end{pmatrix} = \begin{pmatrix} p + r \\ q + s \end{pmatrix}, \quad \begin{pmatrix} p \\ q \end{pmatrix} - \begin{pmatrix} r \\ s \end{pmatrix} = \begin{pmatrix} p - r \\ q - s \end{pmatrix}

    Add or subtract the top components, then add or subtract the bottom components. A scalar multiplies both components. Nothing else is needed.

  2. Substitute and work down the two rows

    a−b=(63)−(25)=(6−23−5)=(4−2)\mathbf{a} - \mathbf{b} = \begin{pmatrix} 6 \\ 3 \end{pmatrix} - \begin{pmatrix} 2 \\ 5 \end{pmatrix} = \begin{pmatrix} 6 - 2 \\ 3 - 5 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}

    Substituting the columns and combining the top row and the bottom row separately gives (4−2)\begin{pmatrix} 4 \\ -2 \end{pmatrix}.

  3. State the answer as a column vector

    a−b=(4−2)\mathbf{a} - \mathbf{b} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}

    So a−b=(4−2)\mathbf{a} - \mathbf{b} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}.

Answer
a−b=(4−2)\mathbf{a} - \mathbf{b} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}
Question 4
2 markseasy
a=(14)\mathbf{a} = \begin{pmatrix} 1 \\ 4 \end{pmatrix} and b=(3−2)\mathbf{b} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}. Work out a−b\mathbf{a} - \mathbf{b} as a column vector.

Worked solution

  1. State the rule for combining column vectors

    (pq)+(rs)=(p+rq+s),(pq)−(rs)=(p−rq−s)\begin{pmatrix} p \\ q \end{pmatrix} + \begin{pmatrix} r \\ s \end{pmatrix} = \begin{pmatrix} p + r \\ q + s \end{pmatrix}, \quad \begin{pmatrix} p \\ q \end{pmatrix} - \begin{pmatrix} r \\ s \end{pmatrix} = \begin{pmatrix} p - r \\ q - s \end{pmatrix}

    Add or subtract the top components, then add or subtract the bottom components. A scalar multiplies both components. Nothing else is needed.

  2. Substitute and work down the two rows

    a−b=(14)−(3−2)=(1−34+2)=(−26)\mathbf{a} - \mathbf{b} = \begin{pmatrix} 1 \\ 4 \end{pmatrix} - \begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 1 - 3 \\ 4 + 2 \end{pmatrix} = \begin{pmatrix} -2 \\ 6 \end{pmatrix}

    Substituting the columns and combining the top row and the bottom row separately gives (−26)\begin{pmatrix} -2 \\ 6 \end{pmatrix}.

  3. State the answer as a column vector

    a−b=(−26)\mathbf{a} - \mathbf{b} = \begin{pmatrix} -2 \\ 6 \end{pmatrix}

    So a−b=(−26)\mathbf{a} - \mathbf{b} = \begin{pmatrix} -2 \\ 6 \end{pmatrix}.

Answer
a−b=(−26)\mathbf{a} - \mathbf{b} = \begin{pmatrix} -2 \\ 6 \end{pmatrix}
Question 5
2 markseasy
a=(25)\mathbf{a} = \begin{pmatrix} 2 \\ 5 \end{pmatrix} and b=(41)\mathbf{b} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}. Work out b−a\mathbf{b} - \mathbf{a} as a column vector.

Worked solution

  1. State the rule for combining column vectors

    (pq)+(rs)=(p+rq+s),(pq)−(rs)=(p−rq−s)\begin{pmatrix} p \\ q \end{pmatrix} + \begin{pmatrix} r \\ s \end{pmatrix} = \begin{pmatrix} p + r \\ q + s \end{pmatrix}, \quad \begin{pmatrix} p \\ q \end{pmatrix} - \begin{pmatrix} r \\ s \end{pmatrix} = \begin{pmatrix} p - r \\ q - s \end{pmatrix}

    Add or subtract the top components, then add or subtract the bottom components. A scalar multiplies both components. Nothing else is needed.

  2. Substitute and work down the two rows

    b−a=(41)−(25)=(4−21−5)=(2−4)\mathbf{b} - \mathbf{a} = \begin{pmatrix} 4 \\ 1 \end{pmatrix} - \begin{pmatrix} 2 \\ 5 \end{pmatrix} = \begin{pmatrix} 4 - 2 \\ 1 - 5 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \end{pmatrix}

    Substituting the columns and combining the top row and the bottom row separately gives (2−4)\begin{pmatrix} 2 \\ -4 \end{pmatrix}.

  3. State the answer as a column vector

    b−a=(2−4)\mathbf{b} - \mathbf{a} = \begin{pmatrix} 2 \\ -4 \end{pmatrix}

    So b−a=(2−4)\mathbf{b} - \mathbf{a} = \begin{pmatrix} 2 \\ -4 \end{pmatrix}.

Answer
b−a=(2−4)\mathbf{b} - \mathbf{a} = \begin{pmatrix} 2 \\ -4 \end{pmatrix}

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