Mathematics RU

Practice · Chapter 22

Vectors

Vector coordinates and parallelograms, collinearity, the dot product and the angle between vectors, the distance from a point to a line and the area of a triangle.

How to solve it

A vector is a shift: how far right and how far up. The coordinates of $\overrightarrow{AB}$ are the end's coordinates minus the start's. The dot product joins algebra and geometry: it is computed from coordinates, and it tells you about the angle.

Step by step

  1. A vector from points: $\overrightarrow{AB} = (x_B - x_A,\ y_B - y_A)$. In a parallelogram $ABCD$ the vectors $\overrightarrow{AB}$ and $\overrightarrow{DC}$ are equal.
  2. Collinearity: the coordinates are proportional, $\vec b = k\vec a$.
  3. The dot product: by coordinates $x_1x_2 + y_1y_2$, by definition $|\vec a||\vec b|\cos\varphi$. Hence the angle: $\cos\varphi = \frac{\vec a \cdot \vec b}{|\vec a||\vec b|}$.
  4. The distance from $(x_0,\ y_0)$ to the line $ax + by + c = 0$: $\frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$.
  5. The area of a triangle from the vectors of two sides $(x_1,\ y_1)$ and $(x_2,\ y_2)$: $\frac{1}{2}|x_1y_2 - x_2y_1|$.
By coordinates. The product of the lengths. The angle between the vectors: at $\varphi = 90^\circ$ the product is zero. Example: $|\vec a| = 7$, $|\vec b| = 2$, the angle $120^\circ$: $\vec a \cdot \vec b = 7 \cdot 2 \cdot \left(-\frac{1}{2}\right) = -7$.

Common mistakes

  • Subtracting the coordinates the wrong way round: $\overrightarrow{AB}$ is “end minus start”.
  • Taking the vertices of a parallelogram out of order: the sides go round $A \to B \to C \to D$.
  • Forgetting the absolute value in the distance to a line — a distance is never negative.
  • Misreading a negative dot product: it means the angle is greater than $90^\circ$.

Example