HookEvery pixel of a Pixar film is a line-plane intersection
Every frame of a modern animated film is this section running at industrial scale. To colour one pixel, the renderer fires a ray from the camera into the scene — a straight line in three dimensions, a point plus a multiple of a direction vector — and asks three questions: which surface does this line hit first, where exactly, and at what angle? A 4K frame holds about 8.3 million pixels, films run at 24 frames per second, and production renderers fire dozens of rays per pixel to smooth the image, so a 90-minute feature settles literally trillions of line–plane intersections before anyone sees a single character move. The mathematics answering each one is the content of AQA's Section F, unchanged.
The section is a small toolkit with enormous reach: equations of lines and planes in 3D, the scalar product for every angle you will ever need, the vector product for manufacturing perpendicular directions and areas, and a family of intersection and shortest-distance problems that combine them. Nothing here is conceptually new if you hold one idea firmly: a line is a point plus one direction, a plane is a point plus one normal, and the two products are the instruments that measure how these objects sit relative to each other.
ModelA line in 3D: one point, one direction
The vector equation of a line is \(\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}\): a position vector \(\mathbf{a}\) to anchor it and a direction vector \(\mathbf{b}\) to point it, with the parameter \(\lambda\) acting as a dial that slides you along the line — every real value of \(\lambda\) is a point, every point is one value of \(\lambda\). Neither ingredient is unique: any point on the line can serve as \(\mathbf{a}\), and any non-zero multiple of \(\mathbf{b}\) gives the same line, which is why two correct answers to the same question can look completely different.
Eliminate \(\lambda\) and you get the Cartesian form: \(\frac{x-a_1}{b_1}=\frac{y-a_2}{b_2}=\frac{z-a_3}{b_3}\). Read it as a machine for recovering the vector form — the numbers under the fractions are the direction vector, the numbers being subtracted are a point. The double equality matters: in 3D a single linear equation describes a plane, not a line, so a line genuinely needs two equations, and the chained form is shorthand for exactly that pair. When a direction component is zero, say \(b_2=0\), that fraction cannot be written; the convention is to state \(y=a_2\) separately alongside the chain for the other two variables — a detail AQA is fond of testing.
ModelA plane: one point, one normal
A plane can be written parametrically as \(\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c}\) — a point and two non-parallel directions lying in it — but the form that does the real work is built on the normal vector \(\mathbf{n}\), the direction perpendicular to the whole plane. A point \(R\) with position vector \(\mathbf{r}\) lies in the plane through \(A\) exactly when the displacement \(\mathbf{r}-\mathbf{a}\) is perpendicular to \(\mathbf{n}\), and the scalar product turns that sentence into algebra: \((\mathbf{r}-\mathbf{a})\cdot\mathbf{n}=0\), usually rearranged to \(\mathbf{r}\cdot\mathbf{n}=\mathbf{a}\cdot\mathbf{n}=d\).
Write \(\mathbf{n}=\begin{pmatrix} n_1 \\ n_2 \\ n_3 \end{pmatrix}\) and the same statement becomes the Cartesian equation \(n_1x+n_2y+n_3z=d\). This is the single most useful decoding in the section: the coefficients of \(x\), \(y\), \(z\) in a plane's equation are its normal vector. Handed \(2x+y-2z=4\), you instantly know the normal is \(\begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}\), and almost every angle, distance or intersection question about that plane starts from exactly that observation. Given two directions in a plane instead of a normal, the vector product (two blocks down) manufactures the normal for you.
Find the Cartesian equation of the plane through \(A(1,3,-2)\) with normal \(\mathbf{n}=\begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}\).
The plane is \(\mathbf{r}\cdot\mathbf{n}=\mathbf{a}\cdot\mathbf{n}\). Compute the right side: \(\mathbf{a}\cdot\mathbf{n}=(1)(3)+(3)(-1)+(-2)(2)=3-3-4=-4\). So the plane is \(3x-y+2z=-4\). Check with the point: \(3(1)-3+2(-2)=-4\). One scalar product, one line of algebra — this sub-part is among the fastest marks on the paper.
MechanismThe scalar product: every angle in the section
The identity \(\mathbf{u}\cdot\mathbf{v}=|\mathbf{u}||\mathbf{v}|\cos\theta\) is the section's protractor, and its cleanest consequence is the perpendicularity test: non-zero vectors are perpendicular exactly when \(\mathbf{u}\cdot\mathbf{v}=0\), because \(\cos 90^\circ=0\). That one-line test underpins everything from checking a normal to finding the foot of a perpendicular.
For angles, the rule is: identify which two vectors actually carry the angle. Between two lines, use the two direction vectors: \(\cos\theta=\frac{|\mathbf{b}_1\cdot\mathbf{b}_2|}{|\mathbf{b}_1||\mathbf{b}_2|}\), with the modulus on top because the acute angle is conventionally reported. Between two planes, use the two normals — the angle between planes equals the angle between their normals. The trap is the mixed case. The angle between a line and a plane is measured to the plane's surface, but the vector you have for the plane is the normal, which stands \(90^\circ\) off the surface. The scalar product of \(\mathbf{b}\) and \(\mathbf{n}\) therefore delivers the complement of the angle you want, and the fix is one function swap: \(\sin\theta=\frac{|\mathbf{b}\cdot\mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}\). Writing \(\cos\) there is the most reliably lost mark in the whole topic — the renderer computing how light glances off a surface makes precisely this complement correction on every bounce.
MechanismThe vector product: perpendicularity on demand
The vector product \(\mathbf{a}\times\mathbf{b}\) takes two vectors and returns a third, perpendicular to both, with magnitude \(|\mathbf{a}||\mathbf{b}|\sin\theta\) and direction fixed by the right-hand rule. Computationally it is a determinant expansion: \[\mathbf{a}\times\mathbf{b}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix},\] and unlike the scalar product it is anticommutative: \(\mathbf{b}\times\mathbf{a}=-\mathbf{a}\times\mathbf{b}\), so order matters up to sign.
It earns its place three ways. First, it manufactures normals: two directions lying in a plane cross to give the plane's normal, unlocking the \(\mathbf{r}\cdot\mathbf{n}=d\) form. Second, it gives an alternative equation of a line: \((\mathbf{r}-\mathbf{a})\times\mathbf{b}=\mathbf{0}\) says the displacement from \(\mathbf{a}\) to \(\mathbf{r}\) is parallel to \(\mathbf{b}\) — parallel vectors have zero cross product, just as perpendicular ones have zero dot product. Third, its magnitude is an area: \(|\mathbf{a}\times\mathbf{b}|\) is the area of the parallelogram on \(\mathbf{a}\) and \(\mathbf{b}\), so a triangle with vertices \(A\), \(B\), \(C\) has area \(\frac{1}{2}\big|\overrightarrow{AB}\times\overrightarrow{AC}\big|\).
Find the area of the triangle with vertices \(A(1,0,0)\), \(B(0,2,0)\) and \(C(0,0,3)\).
\(\overrightarrow{AB}=\begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix}\), \(\overrightarrow{AC}=\begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix}\). The cross product: \(\mathbf{i}\)-component \((2)(3)-(0)(0)=6\); \(\mathbf{j}\)-component \(-\big((-1)(3)-(0)(-1)\big)=3\); \(\mathbf{k}\)-component \((-1)(0)-(2)(-1)=2\). So \(\overrightarrow{AB}\times\overrightarrow{AC}=\begin{pmatrix} 6 \\ 3 \\ 2 \end{pmatrix}\), with magnitude \(\sqrt{36+9+4}=\sqrt{49}=7\). Area \(=\frac{1}{2}\times 7=3.5\) square units — and as a free by-product, \(\begin{pmatrix} 6 \\ 3 \\ 2 \end{pmatrix}\) is a normal to the plane \(ABC\), giving its equation \(6x+3y+2z=6\) in one further scalar product.
CaseIntersections and shortest distances: the full pipeline
Two lines in 3D have three possible relationships, not two. Equate \(\mathbf{a}_1+\lambda\mathbf{b}_1=\mathbf{a}_2+\mu\mathbf{c}\) component by component, solve two of the three equations for \(\lambda\) and \(\mu\), then check the third: consistent means intersecting, inconsistent with non-parallel directions means skew — lines that never meet yet point different ways, impossible in 2D and everywhere in 3D. A line meets a plane even more cleanly: substitute the parametric coordinates \((a_1+\lambda b_1,\,a_2+\lambda b_2,\,a_3+\lambda b_3)\) into the plane's Cartesian equation, solve the resulting linear equation for \(\lambda\), and feed it back to get the point. If the \(\lambda\) terms cancel you have learned something structural: \(\mathbf{b}\cdot\mathbf{n}=0\), so the line is parallel to the plane — in it if the equation reduces to a truth, missing it entirely otherwise.
Every shortest distance in the section is a perpendicular distance. Point to plane has a formula worth memorising: from \(P(x_0,y_0,z_0)\) to \(n_1x+n_2y+n_3z=d\) the distance is \(\frac{|n_1x_0+n_2y_0+n_3z_0-d|}{|\mathbf{n}|}\). Point to line: write a general point \(Q(\lambda)\) on the line, demand \(\overrightarrow{PQ}\cdot\mathbf{b}=0\), solve for \(\lambda\), then take \(|\overrightarrow{PQ}|\) — the perpendicularity test doing honest work. Skew line to skew line: the common perpendicular direction is \(\mathbf{b}_1\times\mathbf{b}_2\), and the distance is the projection of any connecting vector onto it, \(\frac{|(\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{b}_1\times\mathbf{b}_2)|}{|\mathbf{b}_1\times\mathbf{b}_2|}\). Three problems, one idea: build the perpendicular, then measure along it.
The line \(L\): \(\mathbf{r}=\begin{pmatrix} 1 \\ 0 \\ 3 \end{pmatrix}+\lambda\begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix}\) and the plane \(\Pi\): \(2x+y-2z=4\). Find where \(L\) meets \(\Pi\), the angle between them, and the distance from \(A(1,0,3)\) to \(\Pi\).
Intersection: a general point of \(L\) is \((1+2\lambda,\,2\lambda,\,3+\lambda)\). Substitute: \(2(1+2\lambda)+2\lambda-2(3+\lambda)=4\), i.e. \(2+4\lambda+2\lambda-6-2\lambda=4\), so \(4\lambda-4=4\) and \(\lambda=2\). The point is \((5,4,5)\); check: \(10+4-10=4\). \(\checkmark\)
Angle: \(\mathbf{b}=\begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix}\), \(\mathbf{n}=\begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}\); \(\mathbf{b}\cdot\mathbf{n}=4+2-2=4\), \(|\mathbf{b}|=3\), \(|\mathbf{n}|=3\). Line-and-plane, so sine: \(\sin\theta=\frac{4}{9}\), giving \(\theta\approx 26.4^\circ\).
Distance: \(\frac{|2(1)+0-2(3)-4|}{\sqrt{4+1+4}}=\frac{|-8|}{3}=\frac{8}{3}\) units. Notice the sanity chain: \(A\) is off the plane (distance \(\frac{8}{3}\ne 0\)) but its line still lands in it at \(\lambda=2\) — exactly what a non-parallel line must do.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Vector questions on 7367 are long, multi-part and cumulative — a typical structure hands you a line and a plane, then asks for an intersection, an angle, and a perpendicular distance in sequence, each part feeding the next. Two habits protect the accuracy marks. First, name your vectors before computing: write down \(\mathbf{b}\), \(\mathbf{n}\), \(\overrightarrow{AB}\) explicitly, because a method mark survives an arithmetic slip only if the examiner can see the right vectors entering the right formula. Second, state which angle rule you are using and why — 'line and plane, so \(\sin\theta\)' is one short clause that both earns the method mark and stops you defaulting to cosine under time pressure.
Check everything that can be checked in five seconds: an intersection point must satisfy BOTH original equations; a computed normal must give zero when dotted with each direction it came from; a distance of zero should make you look for the point sitting in the plane. When two lines are involved, use different parameter letters — \(\lambda\) and \(\mu\) — because a shared letter silently forces the false assumption that the lines reach their meeting point at the same parameter value, and mark schemes explicitly penalise the collapse. Finally, leave exact surds until the last line: \(\frac{8}{3}\) and \(\arcsin\frac{4}{9}\) rounded early will drift off the mark scheme's tolerance by the time three parts have compounded.