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Edexcel A-level Further Maths revision notes

Further vectors (Further Pure 1)

Section 5
Years 1 and 2
Years 1 and 2: this holds AS subject content and content the exam board adds beyond it for the full A-level.
3 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9FM0 section 5

Checked against Edexcel 9FM0 section 5. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

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In the exam: Formulae booklet provided · calculator allowed in every paper

Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

The vector product a x b of two vectors.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The vector product of a\mathbf{a} and b\mathbf{b} is the vector a×b=∣a∣∣b∣sin⁡θ n^\mathbf{a}\times\mathbf{b}=|\mathbf{a}||\mathbf{b}|\sin\theta\,\hat{\mathbf{n}}, where θ\theta is the angle between them and n^\hat{\mathbf{n}} is the unit vector perpendicular to both in the right-handed sense.
  • In components it is the formal determinant with i,j,k\mathbf{i},\mathbf{j},\mathbf{k} in the first row and the components of a\mathbf{a} and b\mathbf{b} in the second and third.
  • Two consequences do most of the work in examinations: ∣a×b∣|\mathbf{a}\times\mathbf{b}| is the area of the parallelogram with sides a\mathbf{a} and b\mathbf{b}, so half of it is the area of the corresponding triangle; and a×b=0\mathbf{a}\times\mathbf{b}=\mathbf{0} with both vectors non-zero means they are parallel.
  • The product is anticommutative, a×b=−b×a\mathbf{a}\times\mathbf{b}=-\mathbf{b}\times\mathbf{a}, and it is not associative, so bracket every triple product carefully.
The parallelogram spanned by a and b has area |a × b|, and the product itself points along the perpendicular.
Worked example

Given a=2i+3j−k\mathbf{a}=2\mathbf{i}+3\mathbf{j}-\mathbf{k} and b=i−j+4k\mathbf{b}=\mathbf{i}-\mathbf{j}+4\mathbf{k}, find a×b\mathbf{a}\times\mathbf{b}.

  1. 1.The i\mathbf{i} component is 3×4−(−1)×(−1)=12−1=113\times4-(-1)\times(-1)=12-1=11.
  2. 2.The j\mathbf{j} component is −(2×4−(−1)×1)=−(8+1)=−9-\left(2\times4-(-1)\times1\right)=-(8+1)=-9.
  3. 3.The k\mathbf{k} component is 2×(−1)−3×1=−52\times(-1)-3\times1=-5.

Answer: a×b=11i−9j−5k\mathbf{a}\times\mathbf{b}=11\mathbf{i}-9\mathbf{j}-5\mathbf{k}.

Common mistakes

  • Don't fall into the trap of forgetting the minus sign attached to the j\mathbf{j} component of the determinant expansion.
  • Don't fall into the trap of writing a×b=b×a\mathbf{a}\times\mathbf{b}=\mathbf{b}\times\mathbf{a}; the product changes sign when the order is reversed.
  • Don't fall into the trap of quoting ∣a×b∣|\mathbf{a}\times\mathbf{b}| as the area of the triangle rather than of the parallelogram.

Exam tip

Check your answer by taking the scalar product with each original vector; both must be zero, and that costs ten seconds.

Tier 1 · Easy

ORIGINAL

1.

Given a=3i+j−2k\mathbf{a}=3\mathbf{i}+\mathbf{j}-2\mathbf{k} and b=i+4j+k\mathbf{b}=\mathbf{i}+4\mathbf{j}+\mathbf{k}, find a×b\mathbf{a}\times\mathbf{b}.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Find the area of the triangle with vertices A(1,0,2)A(1,0,2), B(3,1,−1)B(3,1,-1) and C(2,4,1)C(2,4,1), giving your answer in surd form.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

Prove that ∣a×b∣2+(a⋅b)2=∣a∣2∣b∣2\left|\mathbf{a}\times\mathbf{b}\right|^2+\left(\mathbf{a}\cdot\mathbf{b}\right)^2=|\mathbf{a}|^2|\mathbf{b}|^2, and use it to find ∣a×b∣\left|\mathbf{a}\times\mathbf{b}\right| when ∣a∣=4|\mathbf{a}|=4, ∣b∣=7|\mathbf{b}|=7 and a⋅b=−10\mathbf{a}\cdot\mathbf{b}=-10.

(6)

(Total for Question 1 is 6 marks)

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Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

The scalar triple product a.b x c.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The scalar triple product a⋅(b×c)\mathbf{a}\cdot\left(\mathbf{b}\times\mathbf{c}\right) is the 3×33\times3 determinant whose rows are the components of a\mathbf{a}, b\mathbf{b} and c\mathbf{c}.
  • Its modulus is the volume of the parallelepiped with those three edges, and one sixth of its modulus is the volume of the tetrahedron on the same three edges.
  • Because swapping two rows of a determinant changes only its sign, the product is unchanged by a cyclic rotation, a⋅(b×c)=b⋅(c×a)=c⋅(a×b)\mathbf{a}\cdot\left(\mathbf{b}\times\mathbf{c}\right)=\mathbf{b}\cdot\left(\mathbf{c}\times\mathbf{a}\right)=\mathbf{c}\cdot\left(\mathbf{a}\times\mathbf{b}\right), and the dot and the cross may be interchanged.
  • The product is zero exactly when the three vectors are coplanar, which is the standard test for coplanarity and the standard route to finding an unknown component that makes three vectors lie in one plane.
  • Always take the modulus before quoting a volume; a negative determinant only records the handedness of the three edges.
Worked example

Find the volume of the tetrahedron with vertices OO, A(1,0,2)A(1,0,2), B(2,3,0)B(2,3,0) and C(0,1,4)C(0,1,4).

  1. 1.The three edge vectors from OO are (1,0,2)(1,0,2), (2,3,0)(2,3,0) and (0,1,4)(0,1,4).
  2. 2.The determinant is 1(12−0)−0(8−0)+2(2−0)=12+4=161(12-0)-0(8-0)+2(2-0)=12+4=16.
  3. 3.The volume is 16×16\dfrac{1}{6}\times16.

Answer: The volume is 83\dfrac83.

Common mistakes

  • Don't fall into the trap of dividing by 33 rather than 66 for a tetrahedron.
  • Don't fall into the trap of quoting a negative volume instead of taking the modulus of the determinant.
  • Don't fall into the trap of using position vectors of the vertices instead of edge vectors when the tetrahedron does not have a vertex at the origin.

Exam tip

For a tetrahedron with no vertex at the origin, subtract one vertex from the other three first; the triple product needs edge vectors, not position vectors.

Tier 1 · Easy

ORIGINAL

1.

Given a=i+2j+3k\mathbf{a}=\mathbf{i}+2\mathbf{j}+3\mathbf{k}, b=j+4k\mathbf{b}=\mathbf{j}+4\mathbf{k} and c=2i+k\mathbf{c}=2\mathbf{i}+\mathbf{k}, find a⋅(b×c)\mathbf{a}\cdot\left(\mathbf{b}\times\mathbf{c}\right).

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Find the volume of the tetrahedron with vertices O(0,0,0)O(0,0,0), A(1,0,2)A(1,0,2), B(2,3,0)B(2,3,0) and C(0,1,4)C(0,1,4).

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Find the volume of the tetrahedron with vertices A(1,1,1)A(1,1,1), B(2,3,1)B(2,3,1), C(0,2,4)C(0,2,4) and D(3,0,2)D(3,0,2).

(6)

(Total for Question 1 is 6 marks)

Years 1 and 2
Years 1 and 2: this holds AS subject content and content the exam board adds beyond it for the full A-level.

Applications of vectors to three dimensional geometry involving points, lines and planes.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A line through a\mathbf{a} with direction b\mathbf{b} can be written as r=a+λb\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}, in Cartesian form x−a1b1=y−a2b2=z−a3b3\dfrac{x-a_1}{b_1}=\dfrac{y-a_2}{b_2}=\dfrac{z-a_3}{b_3}, or in the vector-product form (r−a)×b=0\left(\mathbf{r}-\mathbf{a}\right)\times\mathbf{b}=\mathbf{0}, which simply says that r−a\mathbf{r}-\mathbf{a} is parallel to b\mathbf{b}.
  • The components b1:b2:b3b_1:b_2:b_3 are the direction ratios, and dividing by ∣b∣|\mathbf{b}| gives the direction cosines, whose squares sum to 11.
  • A plane is r⋅n=d\mathbf{r}\cdot\mathbf{n}=d, with n\mathbf{n} obtained from a vector product when three points or a line and a point are given.
  • The standard results follow from these: the angle between a line and a plane satisfies sin⁡θ=∣b⋅n∣∣b∣∣n∣\sin\theta=\dfrac{|\mathbf{b}\cdot\mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}; the distance from a point to a plane is ∣p⋅n−d∣∣n∣\dfrac{|\mathbf{p}\cdot\mathbf{n}-d|}{|\mathbf{n}|}; and the shortest distance between skew lines uses the scalar triple product with the common perpendicular u×v\mathbf{u}\times\mathbf{v}.
Worked example

Find the point of intersection of the line r=(i+2j+3k)+λ(2i−j+k)\mathbf{r}=\left(\mathbf{i}+2\mathbf{j}+3\mathbf{k}\right)+\lambda\left(2\mathbf{i}-\mathbf{j}+\mathbf{k}\right) with the plane 3x+y−z=43x+y-z=4.

  1. 1.A general point on the line is (1+2λ, 2−λ, 3+λ)\left(1+2\lambda,\ 2-\lambda,\ 3+\lambda\right).
  2. 2.Substituting: 3(1+2λ)+(2−λ)−(3+λ)=43(1+2\lambda)+(2-\lambda)-(3+\lambda)=4.
  3. 3.This simplifies to 2+4λ=42+4\lambda=4, so λ=12\lambda=\dfrac12.

Answer: The point of intersection is (2, 32, 72)\left(2,\ \dfrac32,\ \dfrac72\right).

Common mistakes

  • Don't fall into the trap of using cos⁡θ\cos\theta rather than sin⁡θ\sin\theta for the angle between a line and a plane, since the normal is perpendicular to the plane.
  • Don't fall into the trap of writing direction cosines without dividing by the magnitude of the direction vector.
  • Don't fall into the trap of forgetting the modulus in the point-to-plane distance formula.

Exam tip

Write the general point of the line in one bracket before substituting into the plane; almost every error here is an arithmetic slip made while substituting three expressions at once.

Tier 1 · Easy

ORIGINAL

1.

Write the line through (1,2,3)(1,2,3) with direction 2i−j+4k2\mathbf{i}-\mathbf{j}+4\mathbf{k} in the form (r−a)×b=0\left(\mathbf{r}-\mathbf{a}\right)\times\mathbf{b}=\mathbf{0} and in Cartesian form.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Find the Cartesian equation of the plane through the point (1,2,−1)(1,2,-1) with normal vector 2i−j+3k2\mathbf{i}-\mathbf{j}+3\mathbf{k}.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

Find the coordinates of the point where the line r=(i+2j+3k)+λ(2i−j+k)\mathbf{r}=\left(\mathbf{i}+2\mathbf{j}+3\mathbf{k}\right)+\lambda\left(2\mathbf{i}-\mathbf{j}+\mathbf{k}\right) meets the plane 3x+y−z=43x+y-z=4.

(5)

(Total for Question 1 is 5 marks)

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