Geometry and vectors
What this domain tests
Vector geometry in two and three dimensions: operations, scalar product, lines and planes, angles and distances. The exam rewards a clear coordinate set-up and correct use of the scalar product, including in spatial problems.
Core rules
- Vector operations: addition componentwise; scalar multiplication changes length, not direction (unless negative). The magnitude of (x, y, z) is sqrt(x^2 + y^2 + z^2); the unit vector is v/|v|.
- Scalar product: u.v = x1x2 + y1y2 + z1z2 = |u||v|cos(theta). Perpendicular vectors have u.v = 0. The angle between vectors follows from cos(theta) = (u.v)/(|u||v|).
- Cross product (3D): u x v is perpendicular to both, with magnitude |u||v|sin(theta); useful for finding a normal vector to a plane.
- Line: r = a + t d, where a is a point and d a direction vector.
- Plane: n.(r - a) = 0, or n.r = n.a, where n is a normal vector. The angle between planes is the angle between their normals.
- Distance from point P to a plane n.r = k: |n.p - k|/|n|.
Worked example
Find the plane through A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1), and the distance from the origin to it. Take AB = (-1, 1, 0) and AC = (-1, 0, 1). A normal is AB x AC = (1, 1, 1), since the first component is 11 - 00 = 1, the second is 0*(-1) - (-1)*1 = 1, and the third is (-1)0 - 1(-1) = 1. Using A, the plane is x + y + z = 1. For the origin, the distance is |0 + 0 + 0 - 1|/sqrt(1^2 + 1^2 + 1^2) = 1/sqrt(3) = sqrt(3)/3.
Traps that cost marks
- Confusing the scalar product (a number) with the cross product (a vector).
- Using direction vectors instead of normals when finding the angle between planes.
- Sign errors when expanding a determinant or subtracting position vectors.
- Forgetting to divide by |n| in distance formulas, or using a point not on the plane.