M35 · Vectors | Edexcel GCSE Maths | Mr Kidman's Classroom

Revise the key ideas

A vector has magnitude and direction. A column vector $(\begin{matrix} 4 \\ −3 \end{matrix})$ represents 4 units right and 3 down. Its components are signed displacements, not a coordinate pair identifying one fixed point.
Equal vectors have equal components even when drawn in different positions. The negative vector reverses direction: − $(\begin{matrix} 4 \\ −3 \end{matrix})$ = $(\begin{matrix} −4 \\ 3 \end{matrix})$ .
Add component by component: $(\begin{matrix} 2 \\ 3 \end{matrix})$ + $(\begin{matrix} 4 \\ −1 \end{matrix})$ = $(\begin{matrix} 6 \\ 2 \end{matrix})$ . Geometrically, place the second arrow's tail at the first arrow's head.
Head-to-tail vector addition
Enlarge diagram: Head-to-tail vector addition
Subtract using the reverse vector: $(\begin{matrix} 5 \\ 2 \end{matrix})$ − $(\begin{matrix} 1 \\ 4 \end{matrix})$ = $(\begin{matrix} 4 \\ −2 \end{matrix})$ . Order matters.
Multiplying by a scalar changes length and may reverse direction. 3 $(\begin{matrix} 2 \\ −1 \end{matrix})$ = $(\begin{matrix} 6 \\ −3 \end{matrix})$ , while −2 $(\begin{matrix} 2 \\ −1 \end{matrix})$ = $(\begin{matrix} −4 \\ 2 \end{matrix})$ .
If A = (1, 2) and B = (5, −1), vector AB is B − A = $(\begin{matrix} 4 \\ −3 \end{matrix})$ . Vector BA is its negative.
A vector's magnitude follows Pythagoras: | $(\begin{matrix} 3 \\ 4 \end{matrix})$ | = √(3² + 4²) = 5. A translation's vector is measured in the coordinate system's units.

Use bold letters such as a or directed labels such as $\vec{AB}$ for vectors. If OA = a and OB = b, then AB = b − a.
The midpoint M of AB has OM = 12(a + b). If M divides AB in ratio 1 : 2 from A, OM = a + 13(b − a).
Vectors that are non-zero scalar multiples are parallel. To prove three points are collinear, show vectors along two joining segments are scalar multiples and share a point.
Worked example: In triangle OAB, midpoints M of OA and N of OB have OM = 12a and ON = 12b. Therefore MN = 12(b − a) = 12AB, proving MN is parallel to AB and half its length.
For a parallelogram OABC in order, if OA = a and OC = b, then OB = a + b. Diagonals can be expressed in more than one route to locate their intersection.
Direction matters in each route: AB + BC = AC, but AB + CB is different. Draw arrows and keep start/end labels consistent.
For an intersection problem, express its position along each line with parameters, equate coefficients of independent vectors and solve the resulting simultaneous equations.

Practice questions with explained answers will be added here.

For now, cover the worked answers, try the calculations yourself, then compare each step. Include units and reasons where needed.