Edexcel A Level Mathematics

Conducting Body: Pearson Edexcel
Who it is for: Students aged 16-18 in their final school years
Structure: Three written exam papers taken at the end of the course
Content: Pure Mathematics, Statistics, and Mechanics
Duration: Three 2-hour papers
Question Types: A mix of standard procedures and multi-step problems
When held: Annually in the summer exam series
Qualifies you for: University entry, especially for STEM and Economics degrees

Edexcel A Level Mathematics is an advanced qualification for students in their final two years of secondary school. It is one of the most popular A Level subjects and is a common prerequisite for university degrees in science, technology, engineering, computing, finance, and economics. The course builds significantly on GCSE knowledge, introducing more abstract concepts and a greater emphasis on mathematical reasoning and modelling.

The final assessment consists of three written papers taken at the end of the two-year course. Two of these papers cover Pure Mathematics, which includes topics like advanced algebra, calculus, and trigonometry. The third paper assesses applied mathematics, with separate sections on Statistics and Mechanics. A calculator is permitted in all examinations, and a formula booklet is provided.

The main difficulty for many students is the shift towards unstructured problem solving. While some questions test standard procedures, many require students to interpret a situation, select the appropriate mathematical techniques from across the syllabus, and construct a logical solution. Success requires a deep understanding of the core principles, not just memorisation, and the ability to apply them accurately under time pressure.

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From a mathematics paper rather than Edexcel A Level Mathematics, worked the way every question on this site is. Edexcel A Level Mathematics questions are being added.

JEE Advanced 2021

Let $O$ be the origin and $\overrightarrow{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$ , $\overrightarrow{OB} = \hat{i} - 2\hat{j} + 2\hat{k}$ and $\overrightarrow{OC} = \frac{1}{2}(\overrightarrow{OB} - \lambda\overrightarrow{OA})$ for some $\lambda > 0$ . If $|\overrightarrow{OB} \times \overrightarrow{OA}| = \frac{9}{2}$ , then which of the following statements is (are) TRUE?

Vector Algebra Let

O

be the origin and

\overrightarrow{OA} = 2\hat{i} + 2\hat{j} + \hat{k}

\overrightarrow{OB} = \hat{i} - 2\hat{j} + 2\hat{k}

and

\overrightarrow{OC} = \frac{1}{2}(\overrightarrow{OB} - \lambda\overrightarrow{OA})

for some

\lambda > 0

. If

|\overrightarrow{OB} \times \overrightarrow{OA}| = \frac{9}{2}

, then which of the following statements is (are) TRUE?

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$\text{Given: } \overrightarrow{OA} = 2\hat{i} + 2\hat{j} + \hat{k},\quad \overrightarrow{OB} = \hat{i} - 2\hat{j} + 2\hat{k}$ $\overrightarrow{OC} = \dfrac{1}{2}\big(\overrightarrow{OB} - \lambda\overrightarrow{OA}\big),\quad \lambda > 0$ $|\overrightarrow{OB} \times \overrightarrow{OA}| = \dfrac{9}{2}$ $\text{To find: which statements are TRUE?}$ $(1)\ \text{proj}_{OA}\,\overrightarrow{OC} = -\dfrac{3}{2}\qquad (2)\ [OAB] = \dfrac{9}{2}$ $(3)\ [ABC] = \dfrac{9}{2}\qquad (4)\ \text{diagonals of } OA,OC \text{ parallelogram} = \dfrac{\pi}{3}$ $\overrightarrow{OC} = \dfrac{1}{2}(\overrightarrow{OB} - \lambda\overrightarrow{OA}),\quad \lambda > 0$ $\overrightarrow{OA}\cdot\overrightarrow{OB} = (2)(1) + (2)(-2) + (1)(2)$ $\overrightarrow{OA}\cdot\overrightarrow{OB} = 2 - 4 + 2 = 0 \Rightarrow \overrightarrow{OA}\perp\overrightarrow{OB}$ $|\overrightarrow{OA}| = \sqrt{4+4+1} = 3,\qquad |\overrightarrow{OB}| = \sqrt{1+4+4} = 3$ $\overrightarrow{OB}\times\overrightarrow{OA} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 2 & 2 & 1 \end{vmatrix}$ $\overrightarrow{OB}\times\overrightarrow{OA} = (-6,\ 3,\ 6)$ $|\overrightarrow{OB}\times\overrightarrow{OA}| = \sqrt{36+9+36} = 9$ $[OAB] = \dfrac{1}{2}|\overrightarrow{OA}\times\overrightarrow{OB}| = \dfrac{9}{2}$ $\overrightarrow{OB}\times\overrightarrow{OC} = \dfrac{1}{2}\big(\overrightarrow{OB}\times\overrightarrow{OB} - \lambda(\overrightarrow{OB}\times\overrightarrow{OA})\big) = -\dfrac{\lambda}{2}(\overrightarrow{OB}\times\overrightarrow{OA})$ $|\overrightarrow{OB}\times\overrightarrow{OC}| = \dfrac{\lambda}{2}\cdot 9 = \dfrac{9\lambda}{2} = \dfrac{9}{2} \Rightarrow \lambda = 1$ $\lambda = 1:\quad \overrightarrow{OC} = \dfrac{1}{2}(\overrightarrow{OB} - \overrightarrow{OA}) = \dfrac{1}{2}(-1,\ -4,\ 1)$ $\overrightarrow{OC} = \left(-\dfrac{1}{2},\ -2,\ \dfrac{1}{2}\right)$ $\text{Plan: } 1.\ \text{proj}_{OA}\,\overrightarrow{OC}$ $2.\ [OAB];\qquad 3.\ [ABC] = \dfrac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|$ $4.\ \text{angle between diagonals of the } OA,OC \text{ parallelogram}$ $\text{proj}_{OA}\overrightarrow{OC} = \dfrac{\overrightarrow{OC}\cdot\overrightarrow{OA}}{|\overrightarrow{OA}|}$ $\overrightarrow{OC}\cdot\overrightarrow{OA} = -\dfrac{1}{2}(2) + (-2)(2) + \dfrac{1}{2}(1) = -\dfrac{9}{2}$ $\text{proj} = \dfrac{-9/2}{3} = -\dfrac{3}{2}\ \checkmark$ $[OAB] = \dfrac{1}{2}|\overrightarrow{OA}\times\overrightarrow{OB}| = \dfrac{1}{2}(3)(3) = \dfrac{9}{2}\ \checkmark$ $\overrightarrow{AB} = B - A = (1-2,\ -2-2,\ 2-1) = (-1,\ -4,\ 1)$ $\overrightarrow{AC} = C - A = \left(-\dfrac{1}{2}-2,\ -2-2,\ \dfrac{1}{2}-1\right) = \left(-\dfrac{5}{2},\ -4,\ -\dfrac{1}{2}\right)$ $\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -4 & 1 \\ -\frac{5}{2} & -4 & -\frac{1}{2} \end{vmatrix}$ $= \hat{i}(2+4) - \hat{j}\left(\dfrac{1}{2}+\dfrac{5}{2}\right) + \hat{k}(4-10)$ $\overrightarrow{AB}\times\overrightarrow{AC} = (6,\ -3,\ -6)$ $|\overrightarrow{AB}\times\overrightarrow{AC}| = \sqrt{36+9+36} = 9$ $[ABC] = \dfrac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}| = \dfrac{9}{2}\ \checkmark$ $\vec{d_1} = \overrightarrow{OA} + \overrightarrow{OC} = \left(\dfrac{3}{2},\ 0,\ \dfrac{3}{2}\right)$ $\vec{d_2} = \overrightarrow{OA} - \overrightarrow{OC} = \left(\dfrac{5}{2},\ 4,\ \dfrac{1}{2}\right)$ $\vec{d_1}\cdot\vec{d_2} = \left(\dfrac{3}{2}\right)\left(\dfrac{5}{2}\right) + 0 + \left(\dfrac{3}{2}\right)\left(\dfrac{1}{2}\right) = \dfrac{9}{2}$ $|\vec{d_1}| = \dfrac{3\sqrt{2}}{2},\qquad |\vec{d_2}| = \dfrac{3\sqrt{10}}{2}$ $\cos\theta = \dfrac{9/2}{\frac{3\sqrt2}{2}\cdot\frac{3\sqrt{10}}{2}} = \dfrac{1}{\sqrt{5}} \Rightarrow \theta \approx 63.4^\circ \neq \dfrac{\pi}{3}\ \text{✗}$ $\boxed{\text{True: } (1),\ (2),\ (3)}$

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