FP1-5 Further vectors (Further Pure 1) — revision question pack

3 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section FP1-5. 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.

How this checking works

FP1-5.1 · The vector product a x b of two vectors.

Explanation

The vector product of $\mathbf{a}$ and $\mathbf{b}$ is the vector $\mathbf{a}\times\mathbf{b}=|\mathbf{a}||\mathbf{b}|\sin\theta\,\hat{\mathbf{n}}$ , where $\theta$ is the angle between them and $\hat{\mathbf{n}}$ is the unit vector perpendicular to both in the right-handed sense.
In components it is the formal determinant with $\mathbf{i},\mathbf{j},\mathbf{k}$ in the first row and the components of $\mathbf{a}$ and $\mathbf{b}$ in the second and third.
Two consequences do most of the work in examinations: $|\mathbf{a}\times\mathbf{b}|$ is the area of the parallelogram with sides $\mathbf{a}$ and $\mathbf{b}$ , so half of it is the area of the corresponding triangle; and $\mathbf{a}\times\mathbf{b}=\mathbf{0}$ with both vectors non-zero means they are parallel.
The product is anticommutative, $\mathbf{a}\times\mathbf{b}=-\mathbf{b}\times\mathbf{a}$ , 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 $\mathbf{a}=2\mathbf{i}+3\mathbf{j}-\mathbf{k}$ and $\mathbf{b}=\mathbf{i}-\mathbf{j}+4\mathbf{k}$ , find $\mathbf{a}\times\mathbf{b}$ .

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

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

Common mistakes

Don't fall into the trap of forgetting the minus sign attached to the $\mathbf{j}$ component of the determinant expansion.
Don't fall into the trap of writing $\mathbf{a}\times\mathbf{b}=\mathbf{b}\times\mathbf{a}$ ; the product changes sign when the order is reversed.
Don't fall into the trap of quoting $|\mathbf{a}\times\mathbf{b}|$ 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

1.
Given $\mathbf{a}=3\mathbf{i}+\mathbf{j}-2\mathbf{k}$ and $\mathbf{b}=\mathbf{i}+4\mathbf{j}+\mathbf{k}$ , find $\mathbf{a}\times\mathbf{b}$ .
(3)
(Total for Question 1 is 3 marks)
2.
Find a unit vector perpendicular to both $\mathbf{i}+\mathbf{j}$ and $\mathbf{j}+\mathbf{k}$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
Find the area of the triangle with vertices $A(1,0,2)$ , $B(3,1,-1)$ and $C(2,4,1)$ , giving your answer in surd form.
(5)
(Total for Question 1 is 5 marks)
2.
Two vectors $\mathbf{a}$ and $\mathbf{b}$ have magnitudes $3$ and $5$ and the angle between them is $30^\circ$ . Find $\left|\mathbf{a}\times\mathbf{b}\right|$ and state what it measures.
(3)
(Total for Question 2 is 3 marks)
3.
Find the area of the parallelogram whose adjacent sides are $\mathbf{a}=3\mathbf{i}-\mathbf{j}+2\mathbf{k}$ and $\mathbf{b}=\mathbf{i}+2\mathbf{j}-\mathbf{k}$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Prove that $\left|\mathbf{a}\times\mathbf{b}\right|^2+\left(\mathbf{a}\cdot\mathbf{b}\right)^2=|\mathbf{a}|^2|\mathbf{b}|^2$ , and use it to find $\left|\mathbf{a}\times\mathbf{b}\right|$ when $|\mathbf{a}|=4$ , $|\mathbf{b}|=7$ and $\mathbf{a}\cdot\mathbf{b}=-10$ .
(6)
(Total for Question 1 is 6 marks)
2.
Find the perpendicular distance from the point $C(1,2,2)$ to the line through $A(2,-1,3)$ and $B(4,0,1)$ .
(6)
(Total for Question 2 is 6 marks)
3.
Use a vector product to find the Cartesian equation of the plane through the points $A(1,1,0)$ , $B(2,0,3)$ and $C(0,2,1)$ .
(6)
(Total for Question 3 is 6 marks)
4.
Prove that if $\mathbf{a}\times\mathbf{b}=\mathbf{0}$ and neither $\mathbf{a}$ nor $\mathbf{b}$ is the zero vector, then $\mathbf{a}$ and $\mathbf{b}$ are parallel.
(5)
(Total for Question 4 is 5 marks)
5.
Given $\mathbf{a}=2\mathbf{i}-\mathbf{j}+\mathbf{k}$ and $\mathbf{b}=\mathbf{i}+\lambda\mathbf{j}-2\mathbf{k}$ , find the values of $\lambda$ for which $\left|\mathbf{a}\times\mathbf{b}\right|=\sqrt{35}$ .
(6)
(Total for Question 5 is 6 marks)

FP1-5.2 · The scalar triple product a.b x c.

Explanation

The scalar triple product $\mathbf{a}\cdot\left(\mathbf{b}\times\mathbf{c}\right)$ is the $3\times3$ determinant whose rows are the components of $\mathbf{a}$ , $\mathbf{b}$ and $\mathbf{c}$ .
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, $\mathbf{a}\cdot\left(\mathbf{b}\times\mathbf{c}\right)=\mathbf{b}\cdot\left(\mathbf{c}\times\mathbf{a}\right)=\mathbf{c}\cdot\left(\mathbf{a}\times\mathbf{b}\right)$ , 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 $O$ , $A(1,0,2)$ , $B(2,3,0)$ and $C(0,1,4)$ .

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

Answer: The volume is $\dfrac83$ .

Common mistakes

Don't fall into the trap of dividing by $3$ rather than $6$ 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

1.
Given $\mathbf{a}=\mathbf{i}+2\mathbf{j}+3\mathbf{k}$ , $\mathbf{b}=\mathbf{j}+4\mathbf{k}$ and $\mathbf{c}=2\mathbf{i}+\mathbf{k}$ , find $\mathbf{a}\cdot\left(\mathbf{b}\times\mathbf{c}\right)$ .
(3)
(Total for Question 1 is 3 marks)
2.
Find the volume of the parallelepiped with edges $2\mathbf{i}$ , $3\mathbf{j}$ and $\mathbf{i}+\mathbf{j}+4\mathbf{k}$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Find the volume of the tetrahedron with vertices $O(0,0,0)$ , $A(1,0,2)$ , $B(2,3,0)$ and $C(0,1,4)$ .
(4)
(Total for Question 1 is 4 marks)
2.
Determine whether the vectors $\mathbf{a}=\mathbf{i}+2\mathbf{j}+\mathbf{k}$ , $\mathbf{b}=2\mathbf{i}+\mathbf{j}+3\mathbf{k}$ and $\mathbf{c}=4\mathbf{i}+5\mathbf{j}+5\mathbf{k}$ are coplanar.
(4)
(Total for Question 2 is 4 marks)
3.
Explain, using the row properties of determinants, why $\mathbf{a}\cdot\left(\mathbf{b}\times\mathbf{c}\right)=\mathbf{b}\cdot\left(\mathbf{c}\times\mathbf{a}\right)$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Find the volume of the tetrahedron with vertices $A(1,1,1)$ , $B(2,3,1)$ , $C(0,2,4)$ and $D(3,0,2)$ .
(6)
(Total for Question 1 is 6 marks)
2.
Find the value of $\lambda$ for which the vectors $\mathbf{i}+2\mathbf{j}+3\mathbf{k}$ , $2\mathbf{i}+\lambda\mathbf{j}+\mathbf{k}$ and $3\mathbf{i}+\mathbf{j}+2\mathbf{k}$ are coplanar.
(6)
(Total for Question 2 is 6 marks)
3.
Show that $\mathbf{a}\cdot\left(\mathbf{b}\times\mathbf{c}\right)=\left(\mathbf{a}\times\mathbf{b}\right)\cdot\mathbf{c}$ , so that the dot and the cross may be interchanged in a scalar triple product.
(5)
(Total for Question 3 is 5 marks)
4.
The tetrahedron $OABC$ has $\overrightarrow{OA}=\mathbf{i}+\mathbf{k}$ , $\overrightarrow{OB}=2\mathbf{j}+\mathbf{k}$ and $\overrightarrow{OC}=\lambda\mathbf{i}+\mathbf{j}+\mathbf{k}$ , and its volume is $5$ . Find the two possible values of $\lambda$ .
(7)
(Total for Question 4 is 7 marks)
5.
Show that the shortest distance between the skew lines $\mathbf{r}=\mathbf{a}+\lambda\mathbf{u}$ and $\mathbf{r}=\mathbf{b}+\mu\mathbf{v}$ is $\dfrac{\left|\left(\mathbf{b}-\mathbf{a}\right)\cdot\left(\mathbf{u}\times\mathbf{v}\right)\right|}{\left|\mathbf{u}\times\mathbf{v}\right|}$ , and apply it to the lines through $(1,0,0)$ with direction $\mathbf{i}+\mathbf{j}$ and through $(0,1,0)$ with direction $\mathbf{j}+\mathbf{k}$ .
(8)
(Total for Question 5 is 8 marks)

FP1-5.3 · Applications of vectors to three dimensional geometry involving points, lines and planes.

Explanation

A line through $\mathbf{a}$ with direction $\mathbf{b}$ can be written as $\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}$ , in Cartesian form $\dfrac{x-a_1}{b_1}=\dfrac{y-a_2}{b_2}=\dfrac{z-a_3}{b_3}$ , or in the vector-product form $\left(\mathbf{r}-\mathbf{a}\right)\times\mathbf{b}=\mathbf{0}$ , which simply says that $\mathbf{r}-\mathbf{a}$ is parallel to $\mathbf{b}$ .
The components $b_1:b_2:b_3$ are the direction ratios, and dividing by $|\mathbf{b}|$ gives the direction cosines, whose squares sum to $1$ .
A plane is $\mathbf{r}\cdot\mathbf{n}=d$ , with $\mathbf{n}$ 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 $\sin\theta=\dfrac{|\mathbf{b}\cdot\mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}$ ; the distance from a point to a plane is $\dfrac{|\mathbf{p}\cdot\mathbf{n}-d|}{|\mathbf{n}|}$ ; and the shortest distance between skew lines uses the scalar triple product with the common perpendicular $\mathbf{u}\times\mathbf{v}$ .

Worked example

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

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

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

Common mistakes

Don't fall into the trap of using $\cos\theta$ rather than $\sin\theta$ 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

1.
Write the line through $(1,2,3)$ with direction $2\mathbf{i}-\mathbf{j}+4\mathbf{k}$ in the form $\left(\mathbf{r}-\mathbf{a}\right)\times\mathbf{b}=\mathbf{0}$ and in Cartesian form.
(3)
(Total for Question 1 is 3 marks)
2.
Find the direction cosines of the line with direction vector $3\mathbf{i}-4\mathbf{j}+12\mathbf{k}$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Find the Cartesian equation of the plane through the point $(1,2,-1)$ with normal vector $2\mathbf{i}-\mathbf{j}+3\mathbf{k}$ .
(3)
(Total for Question 1 is 3 marks)
2.
Find the acute angle between the line with direction $2\mathbf{i}+\mathbf{j}-2\mathbf{k}$ and the plane $3x+4y=12$ , giving your answer to $1$ decimal place.
(5)
(Total for Question 2 is 5 marks)
3.
Find the perpendicular distance from the point $(2,3,-1)$ to the plane $x+2y-2z=5$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Find the coordinates of the point where the line $\mathbf{r}=\left(\mathbf{i}+2\mathbf{j}+3\mathbf{k}\right)+\lambda\left(2\mathbf{i}-\mathbf{j}+\mathbf{k}\right)$ meets the plane $3x+y-z=4$ .
(5)
(Total for Question 1 is 5 marks)
2.
Show that the equation $\mathbf{r}\times\left(\mathbf{i}+2\mathbf{j}-\mathbf{k}\right)=3\mathbf{i}+\mathbf{j}+5\mathbf{k}$ represents a straight line, and find its Cartesian equation.
(7)
(Total for Question 2 is 7 marks)
3.
Find the shortest distance between the skew lines $\mathbf{r}=\left(\mathbf{i}+2\mathbf{j}\right)+\lambda\left(\mathbf{i}+\mathbf{k}\right)$ and $\mathbf{r}=\left(\mathbf{j}+\mathbf{k}\right)+\mu\left(\mathbf{i}+\mathbf{j}\right)$ .
(6)
(Total for Question 3 is 6 marks)
4.
Find the Cartesian equation of the plane that contains the line $\mathbf{r}=\left(\mathbf{i}+2\mathbf{k}\right)+\lambda\left(2\mathbf{i}+\mathbf{j}-\mathbf{k}\right)$ and passes through the point $(3,2,1)$ .
(6)
(Total for Question 4 is 6 marks)
5.
The planes $x+y+z=3$ and $2x-y+z=1$ meet in a line. Find an equation of that line in the form $\left(\mathbf{r}-\mathbf{a}\right)\times\mathbf{b}=\mathbf{0}$ .
(7)
(Total for Question 5 is 7 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

FP1-5.1 · The vector product a x b of two vectors.

Tier 1 · Easy

Mark scheme for FP1-5.1 Tier 1 · Easy
Question	Scheme	Marks
1	Expanding the determinant gives $\mathbf{i}(1+8)-\mathbf{j}(3+2)+\mathbf{k}(12-1)$ $9\mathbf{i}-5\mathbf{j}+11\mathbf{k}$	3
(3 marks)		3
Notes
Expanding the determinant gives $\mathbf{i}(1+8)-\mathbf{j}(3+2)+\mathbf{k}(12-1)$ . Independent check by scalar products: $(9,-5,11)\cdot(3,1,-2)=27-5-22=0$ and $(9,-5,11)\cdot(1,4,1)=9-20+11=0$ , so the result is perpendicular to both.
2	$(\mathbf{i}+\mathbf{j})\times(\mathbf{j}+\mathbf{k})=\mathbf{i}-\mathbf{j}+\mathbf{k}$ Its magnitude is $\sqrt3$ A unit vector is $\dfrac{1}{\sqrt3}\left(\mathbf{i}-\mathbf{j}+\mathbf{k}\right)$	4
(4 marks)		4
Notes
The determinant gives $\mathbf{i}(1-0)-\mathbf{j}(1-0)+\mathbf{k}(1-0)$ , whose magnitude is $\sqrt{1+1+1}$ . Independent check: $(1,-1,1)\cdot(1,1,0)=0$ and $(1,-1,1)\cdot(0,1,1)=0$ , so the vector is perpendicular to both; the negative of this unit vector is also acceptable.

Tier 2 · Standard

Mark scheme for FP1-5.1 Tier 2 · Standard
Question	Scheme	Marks
1	$\overrightarrow{AB}=(2,1,-3)$ and $\overrightarrow{AC}=(1,4,-1)$ $\overrightarrow{AB}\times\overrightarrow{AC}=(11,-1,7)$ $\left\|\overrightarrow{AB}\times\overrightarrow{AC}\right\|=\sqrt{171}=3\sqrt{19}$ The area is $\dfrac{3\sqrt{19}}{2}$	5
(5 marks)		5
Notes
Form two edge vectors from the same vertex, take their vector product and halve its magnitude. Independent check by the scalar-product route: $\left\|\overrightarrow{AB}\right\|^2=14$ , $\left\|\overrightarrow{AC}\right\|^2=18$ and $\overrightarrow{AB}\cdot\overrightarrow{AC}=9$ , so $\|\mathbf{u}\|^2\|\mathbf{v}\|^2-(\mathbf{u}\cdot\mathbf{v})^2=252-81=171$ , the same value.
2	$\left\|\mathbf{a}\times\mathbf{b}\right\|=3\times5\times\sin30^\circ$ $=7.5$ It is the area of the parallelogram with sides $\mathbf{a}$ and $\mathbf{b}$	3
(3 marks)		3
Notes
Apply $\left\|\mathbf{a}\times\mathbf{b}\right\|=\|\mathbf{a}\|\|\mathbf{b}\|\sin\theta$ with $\sin30^\circ=\tfrac12$ . Independent check by Lagrange's identity: $\mathbf{a}\cdot\mathbf{b}=15\cos30^\circ=\tfrac{15\sqrt3}{2}$ , so $\left\|\mathbf{a}\times\mathbf{b}\right\|^2=225-\tfrac{675}{4}=\tfrac{225}{4}$ , giving $7.5$ .
3	$\mathbf{a}\times\mathbf{b}=-3\mathbf{i}+5\mathbf{j}+7\mathbf{k}$ The area is $\sqrt{9+25+49}=\sqrt{83}$	4
(4 marks)		4
Notes
Expanding the determinant gives $\mathbf{i}(1-4)-\mathbf{j}(-3-2)+\mathbf{k}(6+1)$ , and the area is the magnitude. Independent check: $\|\mathbf{a}\|^2=14$ , $\|\mathbf{b}\|^2=6$ and $\mathbf{a}\cdot\mathbf{b}=3-2-2=-1$ , so $14\times6-1=83$ .

Tier 3 · Hard

Mark scheme for FP1-5.1 Tier 3 · Hard
Question	Scheme	Marks
1	$\left\|\mathbf{a}\times\mathbf{b}\right\|=\|\mathbf{a}\|\|\mathbf{b}\|\sin\theta$ and $\mathbf{a}\cdot\mathbf{b}=\|\mathbf{a}\|\|\mathbf{b}\|\cos\theta$ Squaring and adding gives $\|\mathbf{a}\|^2\|\mathbf{b}\|^2\left(\sin^2\theta+\cos^2\theta\right)$ $=\|\mathbf{a}\|^2\|\mathbf{b}\|^2$ $\left\|\mathbf{a}\times\mathbf{b}\right\|^2=16\times49-100=684$ $\left\|\mathbf{a}\times\mathbf{b}\right\|=6\sqrt{19}$	6
(6 marks)		6
Notes
The identity follows immediately from the two definitions and the Pythagorean identity. Substituting the given magnitudes gives $784-100=684=36\times19$ . Independent check componentwise with $\mathbf{a}=(4,0,0)$ and $\mathbf{b}=\left(-\tfrac52,\tfrac{\sqrt{171}}{2},0\right)$ , which satisfy the given data: $\mathbf{a}\times\mathbf{b}=\left(0,0,2\sqrt{171}\right)$ , of magnitude $2\sqrt{171}=6\sqrt{19}$ .
2	$\overrightarrow{AB}=(2,1,-2)$ with $\left\|\overrightarrow{AB}\right\|=3$ $\overrightarrow{AC}=(-1,3,-1)$ $\overrightarrow{AC}\times\overrightarrow{AB}=(-5,-4,-7)$ $\left\|\overrightarrow{AC}\times\overrightarrow{AB}\right\|=\sqrt{90}=3\sqrt{10}$ The distance is $\dfrac{3\sqrt{10}}{3}=\sqrt{10}$	6
(6 marks)		6
Notes
The distance from a point to a line equals the area of the parallelogram on $\overrightarrow{AC}$ and $\overrightarrow{AB}$ divided by the base length $\left\|\overrightarrow{AB}\right\|$ . Independent check by minimising: the foot of the perpendicular is $A+\lambda\overrightarrow{AB}$ with $\lambda=\dfrac{\overrightarrow{AC}\cdot\overrightarrow{AB}}{9}=\dfrac{-2+3+2}{9}=\dfrac13$ , giving the foot $\left(\tfrac83,-\tfrac23,\tfrac73\right)$ , whose distance from $C$ is $\sqrt{\tfrac{25}{9}+\tfrac{64}{9}+\tfrac{1}{9}}=\sqrt{10}$ .
3	$\overrightarrow{AB}=(1,-1,3)$ and $\overrightarrow{AC}=(-1,1,1)$ $\overrightarrow{AB}\times\overrightarrow{AC}=(-4,-4,0)$ A normal is $\mathbf{n}=(1,1,0)$ The plane is $x+y=d$ with $d=1+1=2$ $x+y=2$	6
(6 marks)		6
Notes
The vector product of two edge vectors is normal to the plane; dividing by the common factor $-4$ simplifies it, and substituting $A$ fixes the constant. Independent check: $B$ gives $2+0=2$ and $C$ gives $0+2=2$ , so all three points satisfy the equation.
4	$\left\|\mathbf{a}\times\mathbf{b}\right\|=\|\mathbf{a}\|\|\mathbf{b}\|\sin\theta=0$ Since $\|\mathbf{a}\|\ne0$ and $\|\mathbf{b}\|\ne0$ , it follows that $\sin\theta=0$ Hence $\theta=0$ or $\theta=\pi$ In both cases $\mathbf{a}$ and $\mathbf{b}$ are parallel, with $\mathbf{b}=k\mathbf{a}$ for some non-zero scalar $k$	5
(5 marks)		5
Notes
Taking magnitudes reduces the vector equation to $\|\mathbf{a}\|\|\mathbf{b}\|\sin\theta=0$ , and the non-zero hypotheses force $\sin\theta=0$ . Independent check componentwise: $\mathbf{a}\times\mathbf{b}=\mathbf{0}$ gives $a_2b_3=a_3b_2$ , $a_3b_1=a_1b_3$ and $a_1b_2=a_2b_1$ , which are exactly the conditions for the two component triples to be proportional.
5	$\mathbf{a}\times\mathbf{b}=(2-\lambda)\mathbf{i}+5\mathbf{j}+(2\lambda+1)\mathbf{k}$ $\left\|\mathbf{a}\times\mathbf{b}\right\|^2=(2-\lambda)^2+25+(2\lambda+1)^2=5\lambda^2+30$ $5\lambda^2+30=35$ gives $\lambda^2=1$ $\lambda=1$ or $\lambda=-1$	6
(6 marks)		6
Notes
Expanding the determinant gives the three components in terms of $\lambda$ ; squaring and collecting terms produces $5\lambda^2+30$ , with the $\lambda$ terms cancelling. Independent check with $\lambda=1$ : $\mathbf{a}\times\mathbf{b}=(1,5,3)$ , whose magnitude squared is $1+25+9=35$ ; with $\lambda=-1$ it is $(3,5,-1)$ , again giving $35$ .

FP1-5.2 · The scalar triple product a.b x c.

Tier 1 · Easy

Mark scheme for FP1-5.2 Tier 1 · Easy
Question	Scheme	Marks
1	$\mathbf{b}\times\mathbf{c}=(1,8,-2)$ $\mathbf{a}\cdot(1,8,-2)=1+16-6=11$	3
(3 marks)		3
Notes
Compute the vector product first, then take the scalar product with $\mathbf{a}$ . Independent check by direct determinant expansion along the first row: $1(1\times1-4\times0)-2(0\times1-4\times2)+3(0\times0-1\times2)=1+16-6=11$ .
2	The determinant is $2\times3\times4=24$ The volume is $24$	3
(3 marks)		3
Notes
Expanding the determinant with two rows already aligned to the axes leaves only the product of the diagonal entries. Independent check geometrically: the base parallelogram on $2\mathbf{i}$ and $3\mathbf{j}$ has area $6$ , and the perpendicular height is the $\mathbf{k}$ component $4$ , giving $24$ .

Tier 2 · Standard

Mark scheme for FP1-5.2 Tier 2 · Standard
Question	Scheme	Marks
1	The determinant of the three position vectors is $16$ The volume is $\dfrac{16}{6}=\dfrac83$	4
(4 marks)		4
Notes
With a vertex at the origin the position vectors are already the edge vectors, so the volume is one sixth of the determinant. Independent check by base and height: the triangle $OAB$ has area $\tfrac12\left\|(1,0,2)\times(2,3,0)\right\|=\tfrac12\left\|(-6,4,3)\right\|=\tfrac{\sqrt{61}}{2}$ , and the perpendicular height of $C$ above the plane $OAB$ is $\dfrac{16}{\sqrt{61}}$ , so the volume is $\tfrac13\times\tfrac{\sqrt{61}}{2}\times\dfrac{16}{\sqrt{61}}=\dfrac83$ .
2	$\mathbf{a}\cdot\left(\mathbf{b}\times\mathbf{c}\right)=1(5-15)-2(10-12)+1(10-4)$ $=-10+4+6=0$ The scalar triple product is zero, so the three vectors are coplanar	4
(4 marks)		4
Notes
Expand the determinant along the first row; a zero value means the parallelepiped is flat. Independent check by finding the linear relation: $\mathbf{c}=2\mathbf{a}+\mathbf{b}$ , since $2(1,2,1)+(2,1,3)=(4,5,5)$ , so $\mathbf{c}$ lies in the plane of $\mathbf{a}$ and $\mathbf{b}$ .
3	Each product is the determinant with rows $\mathbf{a},\mathbf{b},\mathbf{c}$ and $\mathbf{b},\mathbf{c},\mathbf{a}$ respectively Interchanging two rows of a determinant changes its sign Moving $\mathbf{a}$ from the first row to the last requires two interchanges Two sign changes leave the value unaltered, so the two products are equal	4
(4 marks)		4
Notes
A cyclic permutation of three rows is an even permutation, so the determinant is unchanged. Independent check with $\mathbf{a}=(1,2,3)$ , $\mathbf{b}=(0,1,4)$ , $\mathbf{c}=(2,0,1)$ : both products evaluate to $11$ .

Tier 3 · Hard

Mark scheme for FP1-5.2 Tier 3 · Hard
Question	Scheme	Marks
1	$\overrightarrow{AB}=(1,2,0)$ , $\overrightarrow{AC}=(-1,1,3)$ and $\overrightarrow{AD}=(2,-1,1)$ The determinant is $1(1+3)-2(-1-6)+0(1-2)$ $=4+14=18$ The volume is $\dfrac{18}{6}=3$	6
(6 marks)		6
Notes
Because no vertex is at the origin, form the three edge vectors from $A$ before taking the determinant, then divide by $6$ . Independent check by a different base vertex: from $B$ the edges are $(-1,-2,0)$ , $(-2,-1,3)$ and $(1,-3,1)$ , whose determinant is $-18$ , and the modulus divided by $6$ is again $3$ .
2	The scalar triple product is $1(2\lambda-1)-2(4-3)+3(2-3\lambda)$ $=2\lambda-1-2+6-9\lambda$ $=3-7\lambda$ Setting this to zero gives $\lambda=\dfrac37$	6
(6 marks)		6
Notes
Expand the determinant along the first row, collect the terms in $\lambda$ and set the result to zero. Independent check by substituting $\lambda=\tfrac37$ : the determinant becomes $3-7\times\tfrac37=0$ , and the three vectors then satisfy a linear relation, confirming coplanarity.
3	$\mathbf{a}\cdot\left(\mathbf{b}\times\mathbf{c}\right)$ is the determinant with rows $\mathbf{a},\mathbf{b},\mathbf{c}$ $\left(\mathbf{a}\times\mathbf{b}\right)\cdot\mathbf{c}=\mathbf{c}\cdot\left(\mathbf{a}\times\mathbf{b}\right)$ is the determinant with rows $\mathbf{c},\mathbf{a},\mathbf{b}$ The second is a cyclic permutation of the first, an even permutation of rows Hence the two determinants are equal	5
(5 marks)		5
Notes
Both expressions are determinants of the same three rows in cyclically permuted order, and a cyclic permutation of three rows leaves a determinant unchanged. Independent check with $\mathbf{a}=(1,0,2)$ , $\mathbf{b}=(2,3,0)$ , $\mathbf{c}=(0,1,4)$ : $\mathbf{b}\times\mathbf{c}=(12,-8,2)$ so $\mathbf{a}\cdot\left(\mathbf{b}\times\mathbf{c}\right)=12+4=16$ , while $\mathbf{a}\times\mathbf{b}=(-6,4,3)$ so $\left(\mathbf{a}\times\mathbf{b}\right)\cdot\mathbf{c}=0+4+12=16$ .
4	The determinant is $1(2-1)-0(0-\lambda)+1(0-2\lambda)=1-2\lambda$ The volume is $\dfrac{\|1-2\lambda\|}{6}=5$ $\|1-2\lambda\|=30$ $1-2\lambda=30$ gives $\lambda=-\dfrac{29}{2}$ $1-2\lambda=-30$ gives $\lambda=\dfrac{31}{2}$	7
(7 marks)		7
Notes
Expand the determinant of the three edge vectors, equate one sixth of its modulus to $5$ and solve the resulting modulus equation, which has two roots. Independent check with $\lambda=\tfrac{31}{2}$ : the determinant is $1-31=-30$ , whose modulus divided by $6$ is $5$ as required.
5	$\mathbf{u}\times\mathbf{v}$ is perpendicular to both lines, so the shortest distance is the projection of $\mathbf{b}-\mathbf{a}$ onto that common perpendicular That projection has length $\dfrac{\left\|\left(\mathbf{b}-\mathbf{a}\right)\cdot\left(\mathbf{u}\times\mathbf{v}\right)\right\|}{\left\|\mathbf{u}\times\mathbf{v}\right\|}$ Here $\mathbf{u}\times\mathbf{v}=(1,-1,1)$ with $\left\|\mathbf{u}\times\mathbf{v}\right\|=\sqrt3$ $\mathbf{b}-\mathbf{a}=(-1,1,0)$ , so the scalar triple product is $-1-1+0=-2$ The shortest distance is $\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3}$	8
(8 marks)		8
Notes
Any vector joining the lines has the same component along the common perpendicular $\mathbf{u}\times\mathbf{v}$ , and that component is the shortest distance; dividing the scalar triple product by $\left\|\mathbf{u}\times\mathbf{v}\right\|$ computes it. Independent check by minimisation: the squared distance between $(1+\lambda,\lambda,0)$ and $(0,1+\mu,\mu)$ is $(1+\lambda)^2+(\lambda-1-\mu)^2+\mu^2$ , whose stationary point at $\lambda=-\tfrac13$ , $\mu=-\tfrac23$ gives squared distance $\tfrac49+\tfrac49+\tfrac49=\tfrac43$ , and $\sqrt{\tfrac43}=\tfrac{2\sqrt3}{3}$ .

FP1-5.3 · Applications of vectors to three dimensional geometry involving points, lines and planes.

Tier 1 · Easy

Mark scheme for FP1-5.3 Tier 1 · Easy
Question	Scheme	Marks
1	$\left(\mathbf{r}-\left(\mathbf{i}+2\mathbf{j}+3\mathbf{k}\right)\right)\times\left(2\mathbf{i}-\mathbf{j}+4\mathbf{k}\right)=\mathbf{0}$ $\dfrac{x-1}{2}=\dfrac{y-2}{-1}=\dfrac{z-3}{4}$	3
(3 marks)		3
Notes
The vector-product form states that $\mathbf{r}-\mathbf{a}$ is parallel to $\mathbf{b}$ ; equating the three ratios of components gives the Cartesian form. Independent check: at $\lambda=1$ the point is $(3,1,7)$ , and $\tfrac{3-1}{2}=\tfrac{1-2}{-1}=\tfrac{7-3}{4}=1$ .
2	The magnitude is $\sqrt{9+16+144}=13$ The direction cosines are $\dfrac{3}{13}$ , $-\dfrac{4}{13}$ and $\dfrac{12}{13}$	3
(3 marks)		3
Notes
Divide each component by the magnitude of the direction vector. Independent check: $\left(\tfrac3{13}\right)^2+\left(\tfrac4{13}\right)^2+\left(\tfrac{12}{13}\right)^2=\tfrac{9+16+144}{169}=1$ , as direction cosines must satisfy.

Tier 2 · Standard

Mark scheme for FP1-5.3 Tier 2 · Standard
Question	Scheme	Marks
1	$\mathbf{r}\cdot(2,-1,3)=d$ with $d=(1,2,-1)\cdot(2,-1,3)$ $d=2-2-3=-3$ $2x-y+3z=-3$	3
(3 marks)		3
Notes
Substitute the given point into $\mathbf{r}\cdot\mathbf{n}$ to find the constant. Independent check: the point $(0,3,0)$ satisfies $0-3+0=-3$ and the vector from $(1,2,-1)$ to it, namely $(-1,1,1)$ , has zero scalar product with the normal, so it lies in the plane.
2	The normal to the plane is $(3,4,0)$ $\sin\theta=\dfrac{\|(2,1,-2)\cdot(3,4,0)\|}{3\times5}$ $=\dfrac{10}{15}=\dfrac23$ $\theta=41.8^\circ$	5
(5 marks)		5
Notes
The angle between a line and a plane is the complement of the angle between the direction and the normal, so a sine rather than a cosine appears. Independent check: the angle between the direction and the normal is $\arccos\tfrac23=48.2^\circ$ , and $90^\circ-48.2^\circ=41.8^\circ$ .
3	$\mathbf{p}\cdot\mathbf{n}=2+6+2=10$ $\|\mathbf{n}\|=\sqrt{1+4+4}=3$ The distance is $\dfrac{\|10-5\|}{3}=\dfrac53$	4
(4 marks)		4
Notes
Apply $\dfrac{\|\mathbf{p}\cdot\mathbf{n}-d\|}{\|\mathbf{n}\|}$ with $\mathbf{n}=(1,2,-2)$ and $d=5$ . Independent check by projection: the foot of the perpendicular is $(2,3,-1)-\tfrac{5}{9}(1,2,-2)=\left(\tfrac{13}{9},\tfrac{17}{9},\tfrac{1}{9}\right)$ , which satisfies $\tfrac{13}{9}+\tfrac{34}{9}-\tfrac{2}{9}=\tfrac{45}{9}=5$ , and its distance from the point is $\tfrac59\times3=\tfrac53$ .

Tier 3 · Hard

Mark scheme for FP1-5.3 Tier 3 · Hard
Question	Scheme	Marks
1	A general point is $\left(1+2\lambda,\ 2-\lambda,\ 3+\lambda\right)$ $3(1+2\lambda)+(2-\lambda)-(3+\lambda)=4$ $2+4\lambda=4$ , so $\lambda=\dfrac12$ The point is $\left(2,\ \dfrac32,\ \dfrac72\right)$	5
(5 marks)		5
Notes
Substitute the parametric coordinates into the plane equation and solve the resulting linear equation for $\lambda$ . Independent check: $3(2)+\tfrac32-\tfrac72=6-2=4$ , so the point lies on the plane, and it also lies on the line at $\lambda=\tfrac12$ .
2	With $\mathbf{r}=(x,y,z)$ the left side is $\left(-y-2z,\ x+z,\ 2x-y\right)$ Equating components gives $-y-2z=3$ , $x+z=1$ and $2x-y=5$ The third equation is the sum of the first and twice the second, so only two are independent Setting $x=t$ gives $z=1-t$ and $y=2t-5$ $\dfrac{x}{1}=\dfrac{y+5}{2}=\dfrac{z-1}{-1}$	7
(7 marks)		7
Notes
Expanding the vector product gives three linear equations, one of which is dependent, so the solution set is a line with direction $\mathbf{i}+2\mathbf{j}-\mathbf{k}$ , as the form $\left(\mathbf{r}-\mathbf{a}\right)\times\mathbf{b}=\mathbf{0}$ predicts. Independent check: the point $(0,-5,1)$ gives $\mathbf{r}\times\mathbf{b}=(0,-5,1)\times(1,2,-1)=(5-2,\ 1-0,\ 0+5)=(3,1,5)$ as required.
3	$\mathbf{u}\times\mathbf{v}=(1,0,1)\times(1,1,0)=(-1,1,1)$ $\left\|\mathbf{u}\times\mathbf{v}\right\|=\sqrt3$ $\mathbf{b}-\mathbf{a}=(-1,-1,1)$ $\left(\mathbf{b}-\mathbf{a}\right)\cdot\left(\mathbf{u}\times\mathbf{v}\right)=1-1+1=1$ The shortest distance is $\dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{3}$	6
(6 marks)		6
Notes
Compute the common perpendicular as the vector product of the two directions, then project the joining vector onto it. Independent check: the two lines are indeed skew because $\mathbf{u}$ and $\mathbf{v}$ are not parallel and the scalar triple product is non-zero, and a direct minimisation of the squared distance between general points gives the same value $\tfrac13$ for the square.
4	A second direction in the plane is $(3,2,1)-(1,0,2)=(2,2,-1)$ $\mathbf{n}=(2,1,-1)\times(2,2,-1)=(1,0,2)$ $\mathbf{r}\cdot(1,0,2)=d$ with $d=(1,0,2)\cdot(1,0,2)=5$ $x+2z=5$	6
(6 marks)		6
Notes
Take the vector product of the line's direction with the vector from a point of the line to the given point; that is normal to the plane, and substituting either point fixes the constant. Independent check: the given point gives $3+2=5$ , and a general point of the line gives $(1+2\lambda)+2(2-\lambda)=5$ for every $\lambda$ , so the whole line lies in the plane.
5	The direction is $\mathbf{b}=(1,1,1)\times(2,-1,1)=(2,1,-3)$ Setting $z=0$ gives $x+y=3$ and $2x-y=1$ Solving gives $x=\dfrac43$ and $y=\dfrac53$ $\mathbf{a}=\dfrac43\mathbf{i}+\dfrac53\mathbf{j}$ $\left(\mathbf{r}-\dfrac43\mathbf{i}-\dfrac53\mathbf{j}\right)\times\left(2\mathbf{i}+\mathbf{j}-3\mathbf{k}\right)=\mathbf{0}$	7
(7 marks)		7
Notes
The line of intersection is perpendicular to both normals, so its direction is their vector product; any one point common to the planes then serves as $\mathbf{a}$ . Independent check: $\left(\tfrac43,\tfrac53,0\right)$ satisfies $\tfrac43+\tfrac53=3$ and $\tfrac83-\tfrac53=1$ ; adding the direction gives $\left(\tfrac{10}{3},\tfrac83,-3\right)$ , and $\tfrac{10}{3}+\tfrac83-3=3$ with $\tfrac{20}{3}-\tfrac83-3=1$ .