Vectors in the Plane and in Space
What the exam tests
Vectors in the Plane and in Space tests components, dot and cross products, angles, lines, and planes. Kyoto problems often combine a perpendicularity condition with a distance or area computation.
Core rules
- For a = (a_1, a_2, a_3) and b = (b_1, b_2, b_3), the dot product is a·b = a_1b_1 + a_2b_2 + a_3b_3 = |a||b|cos θ.
- a and b are perpendicular exactly when a·b = 0. The angle satisfies cos θ = (a·b)/(|a||b|).
- The cross product a × b is perpendicular to both a and b; its magnitude |a × b| equals the area of the parallelogram spanned by a and b. In components, a × b = (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1).
- A plane with normal n = (A, B, C) through P has equation A(x - x_0) + B(y - y_0) + C(z - z_0) = 0.
Worked example
Let a = (1, 2, -1) and b = (3, 0, 2). Then a·b = 1·3 + 2·0 + (-1)·2 = 1. Also a × b = (2·2 - (-1)·0, (-1)·3 - 1·2, 1·0 - 2·3) = (4, -5, -6). The cosine of the angle between a and b is 1/(√6√13), since |a| = √6 and |b| = √13.
Traps
Do not reverse the order in a cross product: b × a = -(a × b). Do not use the dot product to find a perpendicular vector in space; it gives a scalar condition, while the cross product gives a vector. Check signs carefully in the component formula.