Vector Equations of Lines (AI HL)
A vector equation describes every point on a line using a single known point plus a direction, scaled by a free parameter \(t\). It's the tool behind two of the most common HL vector questions - writing down the equation of a line through given points, and finding where two lines meet. This page covers both, with worked examples and the mistakes that lose marks. It's part of the broader Vectors topic.
12 questions on this sub-topic.
The line formula
Covered under IB syllabus reference AHL3.11: the vector equation of a line in two and three dimensions.
Vector equation of a line
\[\mathbf r = \mathbf a + t\mathbf d\]
\(\mathbf a\) is the position vector of any known point on the line, \(\mathbf d\) is a direction vector, and \(t\) is a scalar parameter. If two points \(A\) and \(B\) are given, take \(\mathbf a\) as \(A\)'s position vector and \(\mathbf d = \vec{AB} = B - A\).
✓ In the formula bookletIntersection of two lines
Give each line its own parameter, e.g. \(s\) for line 1 and \(t\) for line 2. Equate the \(x\)- and \(y\)- (and \(z\)-) components to get simultaneous equations, solve for \(s\) and \(t\), then check every equation is satisfied with the same pair before stating the point.
Need the full syllabus wording, the rest of the vector toolkit, or GDC keystrokes for solving simultaneous equations? See Vectors.
Worked examples
A line passes through \(A(1, 2, 3)\) and \(B(4, 0, 5).\)
(a) Find a direction vector.
(b) Write a vector equation of the line.
Worked solution
(a) \(\vec{AB} = (3, -2, 2).\) M1
\(\vec{AB} = (3, -2, 2).\) A1
(b) \(\mathbf r = (1, 2, 3) + t(3, -2, 2).\) M1
\(\mathbf r = (1, 2, 3) + t(3, -2, 2).\) A1
Line 1: \(\mathbf r = (1, 2) + s(2, 1).\) Line 2: \(\mathbf r = (7, -1) + t(-1, 2).\) Find their point of intersection.
Worked solution
Equate: \(1 + 2s = 7 - t\) and \(2 + s\) M1
\(= -1 + 2t.\) A1
Solving: M1
\(t = 2.4,\ s\) A1
\(= 1.8.\) A1
Point \(= (4.6, 3.8).\) A1
Common mistakes
- Using the same parameter letter for both lines. If both equations use \(t\), solving "\(t=t\)" only finds where the two parametrisations agree, not where the actual lines cross - use \(s\) for one line and \(t\) for the other so each can take a different value at the crossing point.
- Assuming crossing paths mean a collision. Solving the position equations only tells you the two lines meet somewhere - it does not mean two moving objects following those lines arrive there at the same time. A genuine collision needs the same time value \(t\) to satisfy both objects' equations simultaneously.
- Forgetting to verify with all coordinates. In three dimensions, solving two of the three component equations can give values of \(s\) and \(t\) that don't actually satisfy the third - always check the \(z\)-equation (or state that the lines are skew if it fails).
Ready to practise properly?
12 vector-line questions, marked instantly like the real exam.
Quick answers
What is the vector equation of a line?
\(\mathbf r = \mathbf a + t\mathbf d\), where \(\mathbf a\) is the position vector of a known point, \(\mathbf d\) is a direction vector, and \(t\) is a scalar parameter that generates every point on the line as it varies.
How do you find where two lines intersect?
Give each line its own parameter (commonly \(s\) and \(t\)), equate the corresponding \(x\), \(y\) (and \(z\)) components, and solve the resulting simultaneous equations. If a consistent solution exists, substitute it back into either line to get the point of intersection.