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How do you represent quantities with both magnitude and direction, and use them in geometry and motion?

Vectors in two and three dimensions, magnitude and direction, addition and scalar multiplication, position vectors, unit vectors, and geometric applications.

A focused answer to the Edexcel A-Level Mathematics vectors content, covering vectors in two and three dimensions, magnitude and direction, addition and scalar multiplication, position vectors, unit vectors, and geometric applications such as collinearity and midpoints.

Generated by Claude Opus 4.89 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

Edexcel wants you to use vectors in two and three dimensions, write them in component or i\mathbf{i}, j\mathbf{j}, k\mathbf{k} form, find magnitudes and directions, add and subtract vectors and multiply by a scalar, use position vectors to find displacements, find unit vectors, and apply vectors to geometric problems such as proving points are collinear or finding a midpoint.

The answer

Components and magnitude

Adding, subtracting and scaling

Vectors add and subtract component by component: (x1i+y1j)+(x2i+y2j)=(x1+x2)i+(y1+y2)j(x_1\mathbf{i} + y_1\mathbf{j}) + (x_2\mathbf{i} + y_2\mathbf{j}) = (x_1 + x_2)\mathbf{i} + (y_1 + y_2)\mathbf{j}. Multiplying a vector by a scalar kk multiplies each component by kk, which stretches the vector by a factor kk and reverses it when kk is negative. Geometrically, addition is the triangle rule: place the tail of the second vector at the head of the first, and the resultant runs from the start to the finish.

Position vectors and displacement

The position vector of a point AA is OA=a\overrightarrow{OA} = \mathbf{a} from the origin. The displacement from AA to BB is AB=ba\overrightarrow{AB} = \mathbf{b} - \mathbf{a}, and the midpoint of ABAB has position vector 12(a+b)\tfrac{1}{2}(\mathbf{a} + \mathbf{b}).

Geometric applications

Examples in context

Try this

Q1. Find the magnitude of a=2i3j+6k\mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}. [2 marks]

  • Cue. a=4+9+36=49=7|\mathbf{a}| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7.

Q2. Points AA and BB have position vectors i+2j\mathbf{i} + 2\mathbf{j} and 5i2j5\mathbf{i} - 2\mathbf{j}. Find AB\overrightarrow{AB} and its magnitude. [3 marks]

  • Cue. AB=4i4j\overrightarrow{AB} = 4\mathbf{i} - 4\mathbf{j}, magnitude 32=42\sqrt{32} = 4\sqrt{2}.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20185 marksThe points AA and BB have position vectors a=3i+j2k\mathbf{a} = 3\mathbf{i} + \mathbf{j} - 2\mathbf{k} and b=7i2j+4k\mathbf{b} = 7\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}. Find AB\overrightarrow{AB}, its magnitude, and a unit vector in the direction of AB\overrightarrow{AB}.
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The displacement is AB=ba\overrightarrow{AB} = \mathbf{b} - \mathbf{a} (M1): AB=(73)i+(21)j+(4(2))k=4i3j+6k\overrightarrow{AB} = (7 - 3)\mathbf{i} + (-2 - 1)\mathbf{j} + (4 - (-2))\mathbf{k} = 4\mathbf{i} - 3\mathbf{j} + 6\mathbf{k} (A1).

Magnitude (M1): AB=42+(3)2+62=16+9+36=61|\overrightarrow{AB}| = \sqrt{4^2 + (-3)^2 + 6^2} = \sqrt{16 + 9 + 36} = \sqrt{61} (A1).

Unit vector: 161(4i3j+6k)\dfrac{1}{\sqrt{61}}(4\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) (A1).

Markers reward "end minus start", correct components, the magnitude, and dividing by the magnitude for the unit vector.

Edexcel 20224 marksPoints AA, BB and CC have position vectors i+2j\mathbf{i} + 2\mathbf{j}, 4i+8j4\mathbf{i} + 8\mathbf{j} and 6i+13j6\mathbf{i} + 13\mathbf{j}. Determine whether AA, BB and CC are collinear.
Show worked answer →

Find two displacements (M1): AB=(41)i+(82)j=3i+6j\overrightarrow{AB} = (4 - 1)\mathbf{i} + (8 - 2)\mathbf{j} = 3\mathbf{i} + 6\mathbf{j} and BC=(64)i+(138)j=2i+5j\overrightarrow{BC} = (6 - 4)\mathbf{i} + (13 - 8)\mathbf{j} = 2\mathbf{i} + 5\mathbf{j} (A1).

For collinearity BC\overrightarrow{BC} must be a scalar multiple of AB\overrightarrow{AB} (M1). If 3i+6j3\mathbf{i} + 6\mathbf{j} scaled by kk gave 2i+5j2\mathbf{i} + 5\mathbf{j}, then 3k=23k = 2 so k=23k = \dfrac{2}{3}, but then 6k=456k = 4 \ne 5.

The components are inconsistent, so the vectors are not parallel and AA, BB, CC are not collinear (A1).

Markers reward the two displacements, the scalar-multiple test, and the correct conclusion with reason.

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