Write the vector equation of a line by combining the position vector of any point on the line with a scalar multiple of a non-zero direction vector:
Here, varies over all real numbers, generating every point on the line.
| Symbol | Meaning |
|---|---|
| Position vector of a general point on the line | |
| Position vector of one known point on the line | |
| Non-zero vector parallel to the line | |
| Real scalar parameter |
The mechanism is translation and scaling. The vector moves from the fixed point represented by . Positive and negative values of move in opposite directions, while gives the known point itself. Because , changing the parameter traces the complete line rather than a single point.
Writing the equation from two points
Suppose the line passes through and .
First, choose one point to provide the position vector:
Next, calculate a direction vector by subtracting the position vectors:
Therefore,
To verify the result, substitute to recover , then solve for the parameter that gives . Here, produces , since adding to gives . This check also catches reversed subtraction: using is valid, but the corresponding parameter for becomes .
An equivalent equation may use as the position vector or any non-zero scalar multiple of as the direction vector. These describe the same geometric line.
A common misconception is to use the second point’s position vector directly as . The direction must instead be found by subtraction, such as .
IB exam technique
This is IB Mathematics AA HL, AHL 3.14, and the analytic method is suitable for Paper 1 as well as Papers 2 and 3. For a command such as find, show the subtraction used to obtain the direction vector before stating the final equation; a correct unsupported equation may not earn all method marks.
