Vector and Cartesian equations of lines and planes, the scalar and vector products, angles between lines and planes, intersections, and shortest distances including between skew lines.

What this dot point is asking

Edexcel wants you to write lines and planes in vector, parametric and Cartesian form, use the scalar product for angles and perpendicularity and the vector product for normals, find where lines and planes intersect, and compute shortest distances from a point to a line or plane and between two skew lines. Vectors are a guaranteed Core Pure topic with a long multi-part question, so accuracy in component arithmetic is everything.

Lines, planes and the two products

A line in three dimensions needs a point on it, $\mathbf{a}$, and a direction $\mathbf{d}$, giving $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$. Splitting into components gives the parametric form, and eliminating $\lambda$ gives the symmetric Cartesian form $\frac{x - a_1}{d_1} = \frac{y - a_2}{d_2} = \frac{z - a_3}{d_3}$. A plane needs a point and a normal $\mathbf{n}$; its equation is $\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}$, which expands to $n_1 x + n_2 y + n_3 z = d$.

The two products do completely different jobs, and choosing the wrong one is the most common error.

Angles and intersections

The scalar product gives the angle between two lines directly from their directions. For the angle between a line and a plane, the dot product of the direction with the normal gives the angle $\alpha$ between the line and the normal, so the angle to the plane is $90^\circ - \alpha$, which is why you use $\sin$ rather than $\cos$ in the line-plane formula.

To find where a line meets a plane, substitute the parametric line into the plane equation and solve for $\lambda$, then back-substitute to find the point. If the line is parallel to the plane ($\mathbf{d}\cdot\mathbf{n} = 0$) there is either no intersection or the whole line lies in the plane.

Shortest distances

The shortest distance from a point to a plane projects the vector from any point on the plane to the given point onto the unit normal. The shortest distance between two skew lines uses the common perpendicular $\mathbf{d}_1\times\mathbf{d}_2$: project the vector joining the two base points onto this direction.

Examples in context

Vectors connect to several other Core Pure threads. The cross product reappears as the area of a parallelogram (and half of it as a triangle's area), tying vectors to the determinant: $\mathbf{a}\times\mathbf{b}$ has components that are exactly $2 \times 2$ determinants. Transformation matrices act on position vectors, so the matrices dot point and this one share notation. In further coordinate systems, the same dot-product machinery finds angles for conic tangents and normals. The scalar triple product $(\mathbf{a}\times\mathbf{b})\cdot\mathbf{c}$, the volume of a parallelepiped, is the determinant of the matrix with those three vectors as rows, linking volume to singularity (a zero triple product means the vectors are coplanar).

Try this

Q1. Find a normal to the plane through $\mathbf{a} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ containing directions $\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$. [3 marks]

• Cue. Take the vector product: $\mathbf{n} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$.

Q2. Are the vectors $\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$ perpendicular? [2 marks]

• Cue. Scalar product $= 2 - 1 - 1 = 0$, so yes, they are perpendicular.

Q3. Find the point where the line $\mathbf{r} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ meets the plane $x + y + z = 6$. [3 marks]

• Cue. Substitute: $3\lambda = 6$, so $\lambda = 2$ and the point is $(2, 2, 2)$.