Skip to main content
WASpecialist MathematicsSyllabus dot point

How do vectors describe lines and planes and the relationships between them?

Write vector and parametric equations of lines and planes and find intersections, distances and angles

WACE Specialist Unit 4 vector geometry: vector and parametric equations of lines and planes, the cartesian and scalar (normal) forms, intersections, the angle between lines and planes, parallel and skew lines, and distance calculations in three dimensions.

Generated by Claude Opus 4.79 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. Vector and parametric equations of a line
  3. Equations of a plane
  4. Angles
  5. Intersections
  6. Distances

What this dot point is asking

SCSA Unit 4 uses three-dimensional vectors to describe lines and planes. You must write each object in vector, parametric and cartesian form, decide whether lines intersect, are parallel or are skew, find points of intersection, and compute the angles between lines, between planes, and between a line and a plane.

Vector and parametric equations of a line

A line through point AA with position vector a\mathbf{a} and direction vector d\mathbf{d} is

r=a+λd,λR.\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}, \qquad \lambda \in \mathbb{R}.

Splitting into components gives the parametric form x=a1+λd1x = a_1 + \lambda d_1, y=a2+λd2y = a_2 + \lambda d_2, z=a3+λd3z = a_3 + \lambda d_3.

Two lines are parallel if their direction vectors are scalar multiples. If they are not parallel, they either intersect (a common point exists) or are skew (no common point and not parallel).

Equations of a plane

A plane with normal vector n\mathbf{n} passing through a point with position vector a\mathbf{a} satisfies

If a plane is given by three points, find two direction vectors in the plane and take their cross product to obtain the normal n\mathbf{n}.

Angles

Use the dot product, uv=uvcosϕ\mathbf{u}\cdot\mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos\phi, to extract angles. Take the acute angle by using the modulus of the dot product.

  • Between two lines: angle between their direction vectors, cosθ=d1d2d1d2\cos\theta = \dfrac{|\mathbf{d_1}\cdot\mathbf{d_2}|}{|\mathbf{d_1}||\mathbf{d_2}|}.
  • Between two planes: angle between their normals.
  • Between a line and a plane: if α\alpha is the angle between the direction d\mathbf{d} and the normal n\mathbf{n}, the line-plane angle is 90α90^\circ - \alpha, so sinθ=dndn\sin\theta = \dfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}.

Intersections

To intersect a line with a plane, substitute the parametric coordinates into the plane equation and solve for λ\lambda, then back-substitute. To intersect two lines, equate components and solve; if the resulting equations are inconsistent the lines are skew (assuming they are not parallel).

Distances

The distance from a point with position vector p\mathbf{p} to a plane rn=k\mathbf{r}\cdot\mathbf{n} = k is pnkn\dfrac{|\mathbf{p}\cdot\mathbf{n} - k|}{|\mathbf{n}|}. The shortest distance from a point to a line uses the perpendicular component: project the point-to-line vector onto d\mathbf{d} and subtract.