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The position vector of a point \(A\) is \(\overrightarrow{OA} = \mathbf{a}\), its vector from the origin. The vector from \(A\) to \(B\) is \(\mathbf{b} - \mathbf{a}\), and the distance between the points is its magnitude. You use this to find distances, midpoints, points that divide a line segment in a ratio, and side lengths and angles of triangles.
Key facts
\(\overrightarrow{OA} = \mathbf{a}\) (position vector of \(A\))
Relative to a fixed origin \(O\), the position vector of \(A(x, y, z)\) is \(\mathbf{a} = \overrightarrow{OA} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\)
The distance \(AB\) is \(|\mathbf{b} - \mathbf{a}|\). For \(A(1, 2, 3)\) and \(B(2, 4, 5)\):\(\mathbf{b} - \mathbf{a} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\), so \(AB = \sqrt{1 + 4 + 4} = 3\)
Leave distances as exact surds unless asked for a decimal.
Special points
diagram
Midpoint: \(\overrightarrow{OM} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})\), which is the average of the coordinates.
\(P\) divides \(AB\) in the ratio \(m : n\): start at \(A\) and go \(\dfrac{m}{m + n}\) of the way to \(B\)
\(\overrightarrow{OP} = \mathbf{a} + \dfrac{m}{m + n}(\mathbf{b} - \mathbf{a})\): start at \(A\) and go \(\dfrac{m}{m + n}\) of the way to \(B\)
Parallelogram \(ABCD\) (vertices in order): \(\overrightarrow{AB} = \overrightarrow{DC}\) and \(\overrightarrow{AD} = \overrightarrow{BC}\)
In parallelogram \(ABCD\), opposite sides are equal vectors: \(\overrightarrow{AB} = \overrightarrow{DC}\) and \(\overrightarrow{AD} = \overrightarrow{BC}\)
Triangles
Find the three side lengths with the distance formula. Two equal sides show the triangle is isosceles; \(a^2 + b^2 = c^2\) shows a right angle.
Find an angle with the cosine rule, \(\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}\), and the area with \(\dfrac{1}{2}ab\sin C\)
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Find the distance between two points
3 marksGrade C
Find the vector between the points: end minus start.
Square the components, add and take the square root.
Example. The points \(A\) and \(B\) have coordinates \((2, -1, 5)\) and \((-4, 1, 2)\). Find \(\overrightarrow{AB}\) and the distance \(AB\)
Turn the ratio into a fraction of the whole segment: \(1 : 2\) means \(\dfrac{1}{3}\) of the way.
Add that fraction of \(\overrightarrow{AB}\) to \(\mathbf{a}\)
Example. \(A\) and \(B\) have position vectors \(\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\) and \(7\mathbf{i} - 2\mathbf{j} + 10\mathbf{k}\). The point \(P\) lies on \(AB\) with \(AP : PB = 1 : 2\). Find the position vector of \(P\)
Use the order of the letters: in \(ABCD\), \(\overrightarrow{AD} = \overrightarrow{BC}\)
Add the known side vector to the position vector of the starting vertex.
Example. \(ABCD\) is a parallelogram, with its vertices in that order. \(A\), \(B\) and \(C\) have coordinates \((1, 2, 0)\), \((4, 3, -1)\) and \((6, 0, 2)\). Find the coordinates of \(D\)
Find the three vectors along the sides of the triangle.
Find the three side lengths.
Use the cosine rule for the angle required (the side opposite it goes after the minus sign).
Example. The points \(P\), \(Q\) and \(R\) have coordinates \((1, 0, 2)\), \((3, 4, 1)\) and \((-1, 2, 5)\). Find the size of angle \(QPR\), giving your answer to 1 decimal place.
Finding \(\mathbf{a} - \mathbf{b}\) for \(\overrightarrow{AB}\): the distance is the same, but the vector points the wrong way.
Writing the midpoint as \(\dfrac{1}{2}(\mathbf{b} - \mathbf{a})\): that is half of \(\overrightarrow{AB}\), not a position vector.
Going \(\dfrac{1}{2}\) of the way for a ratio \(1 : 2\) instead of \(\dfrac{1}{3}\)
Squaring negative differences wrongly in the distance formula.
More tips
Memory tricks
'End minus start' for any vector between two points.
Midpoint: average each coordinate.
Parallelogram check: after finding \(D\), confirm that \(\overrightarrow{DC} = \overrightarrow{AB}\)
The order of the letters matters: \(ABCD\) means \(A\) to \(B\) to \(C\) to \(D\) round the shape.
Exam technique
Show the vector between the points before its magnitude: it is usually a method mark.
Give coordinates when asked for a point, and a vector when asked for a position vector.
The scalar (dot) product is not part of A level Mathematics: angles come from side lengths and the cosine rule.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Write the position vector of a pointThe point \((3, -1, 2)\) has position vector \(3\mathbf{i} - \mathbf{j} + 2\mathbf{k}\)
Grade D
Find the vector between two points\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\): subtract the start from the end.
Grade C
Find the distance between two points\[AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\]
Grade C
Find a midpoint or dividing point\(\overrightarrow{OM} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})\); for \(AP : PB = m : n\), \(\overrightarrow{OP} = \mathbf{a} + \dfrac{m}{m + n}(\mathbf{b} - \mathbf{a})\)
Grade B
Find a missing vertex of a parallelogramUse equal opposite sides: in \(ABCD\),\(\overrightarrow{AD} = \overrightarrow{BC}\), so \(\mathbf{d} = \mathbf{a} + \mathbf{c} - \mathbf{b}\)
Grade A
Find angles in a triangle from coordinatesFind all three side lengths, then use the cosine rule.
Grade A
Find unknown coordinates from a distanceForm an equation from the distance squared and solve it, usually a quadratic.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy6 marks
Relative to a fixed origin \(O\), the points \(A\) and \(B\) have position vectors\[\overrightarrow{OA} = -6\mathbf{i} + 5\mathbf{j}, \qquad \overrightarrow{OB} = 4\mathbf{i} + 4\mathbf{j}\]respectively.
(a) Find \(\overrightarrow{AB}\).[2]
(b) Find the exact distance \(AB\), giving your answer as a simplified surd where appropriate.[2]
(c) Find the position vector of the midpoint of \(AB\).[2]
Relative to a fixed origin \(O\), the points \(A\), \(B\) and \(C\) have position vectors \(-6\mathbf{i}\), \(-8\mathbf{i} - 4\mathbf{j}\) and \(-3\mathbf{i} - 9\mathbf{j}\) respectively, where \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors. The point \(P\), with position vector \(x\mathbf{i} + y\mathbf{j}\), is the same distance from each of \(A\), \(B\) and \(C\).
(a) Use \(PA = PB\) to show that \(x + 2y = -11\).[3]