Vectors and Their Applications
The Vectors and Their Applications domain covers vector algebra in two and three dimensions, the scalar and vector products, and their use in line, plane and triangle geometry. The exam rewards candidates who keep the scalar/vector distinction clear and who can translate a geometric condition - perpendicular, parallel, coplanar - into an algebraic one.
Core rules
- If a = a1i + a2j + a3k then |a| = sqrt(a1^2 + a2^2 + a3^2) and the unit vector is a/|a|.
- Scalar product: a·b = |a||b| cos θ = a1b1 + a2b2 + a3b3, and the result is a scalar. a is perpendicular to b exactly when a·b = 0.
- Vector product: |a × b| = |a||b| sin θ with direction from the right-hand rule; i × j = k, j × k = i, k × i = j, and b × a = -(a × b). a is parallel to b exactly when a × b = 0.
- Area of the triangle with sides a and b from a common vertex is (1/2)|a × b|; the parallelogram on the same sides has area |a × b|.
- Line through a with direction d: r = a + λd. Plane with normal n through a: (r - a)·n = 0.
- The scalar triple product a·(b × c) gives the parallelepiped volume; zero means the three vectors are coplanar.
Worked example
With a = 2i - j + k and b = i + j + 2k: a·b = (2)(1) + (-1)(1) + (1)(2) = 3, while |a| = |b| = sqrt(4 + 1 + 1) = sqrt 6. Hence cos θ = 3/(sqrt6 · sqrt6) = 3/6 = 1/2, so the angle between a and b is 60°.
Traps
- Reporting a·b as a vector, or quoting a × b without its modulus when an area is wanted.
- Reversing the cyclic order in i × j = k, which flips the sign of the product.
- Forgetting that a line's direction vector may be replaced by any non-zero multiple of it.
- Using degrees in the angle formula when the question works in radians.