Overview
This chapter applies vectors to geometry — triple products, and the equations of lines and planes with angle and distance calculations.
1. Triple Products
Scalar triple product: [a b c] = a·(b×c) = volume of parallelepiped
[a b c] = 0 ⇒ a, b, c are coplanar
Vector triple product: a×(b×c) = (a·c)b − (a·b)c
2. Line
Vector form: r = a + t b
Cartesian form: (x−x₁)/l = (y−y₁)/m = (z−z₁)/n (direction ratios l, m, n)
Angle between two lines: cos θ = (b₁·b₂)/(|b₁||b₂|)
3. Plane
Vector form: r·n = d
Cartesian: ax + by + cz = d (normal n = (a, b, c))
Distance of point (x₁,y₁,z₁) from plane = |ax₁+by₁+cz₁−d| / √(a²+b²+c²)
Angle between planes: cos θ = (n₁·n₂)/(|n₁||n₂|)
Worked Examples
Example 1 — Scalar triple product / coplanarity
Are a = i+j, b = j+k, c = i+k coplanar?
[a b c] = |1 1 0; 0 1 1; 1 0 1|
= 1(1−0) − 1(0−1) + 0 = 1 + 1 = 2 ≠ 0
→ not coplanar; volume = 2.
Example 2 — Cross product (area)
Area of triangle with adjacent sides a = 2i+j, b = i+3j.
a×b = (2)(3) − (1)(1) = 5 (k-component) → |a×b| = 5
Area of triangle = ½|a×b| = 5/2 sq.units
Example 3 — Angle between lines
Direction ratios (1, 2, 2) and (2, 2, 1). Find the angle.
cos θ = (1·2 + 2·2 + 2·1)/(3 × 3) = (2+4+2)/9 = 8/9
θ = cos⁻¹(8/9)
Example 4 — Distance of a point from a plane
Distance of (1, 2, 3) from 2x − y + 2z − 6 = 0.
= |2(1) − 2 + 2(3) − 6| / √(4+1+4) = |2 − 2 + 6 − 6|/3 = 0/3 = 0
→ the point lies on the plane.
Common Mistakes
- Forgetting [a b c] = 0 means coplanar.
- Using a·b (dot) when the cross product is needed for area.
- Dropping the modulus in the distance formula.
- Using direction ratios without normalising in angle formulas.
Exam Tips & Revision Checklist
- Scalar triple product = determinant of components.
- Area of triangle = ½|a×b|; volume = |[a b c]|.
- Practised: triple product/coplanarity, cross-product area, angle between lines, point–plane distance.
Frequently Asked Questions
[a b c] = a·(b×c) gives the volume of the parallelepiped formed by the three vectors. If it is 0, the vectors are coplanar.