Lines in Vector Form (AA HL)
A line in three dimensions can't be written as \(y=mx+c\) any more - instead you fix one point on it and give a direction to travel along. This page covers setting up that vector equation, converting to Cartesian form, and deciding whether two lines meet, run parallel, or pass each other as skew lines. It's part of the broader Vectors, Lines & Planes topic.
23 questions on this sub-topic.
Key facts you'll need
Covered under IB syllabus reference AHL3.14: the vector equation of a line in two and three dimensions, \(\mathbf r=\mathbf a+\lambda\mathbf b\), including its parametric and Cartesian forms. Setting up and comparing these equations also leans on two prior-knowledge vector tools.
Magnitude of a vector
\(|\mathbf v|=\sqrt{x^2+y^2+z^2}\)
Not in the formula booklet - this comes from prior knowledge. You'll use it to find a unit direction vector or the distance between two points on a line.
Vector (cross) product
\(\mathbf a\times\mathbf b=|\mathbf a||\mathbf b|\sin\theta\,\hat{\mathbf n}\)
In the formula booklet. Two direction vectors are parallel exactly when \(\mathbf a\times\mathbf b=\mathbf 0\) - a quick way to rule parallel lines in or out before testing for skewness.
Need the full syllabus wording or the GDC methods for solving simultaneous equations? See Vectors, Lines & Planes.
Worked examples
Find a vector equation of the line through \(A(1,2,3)\) and \(B(4,0,5).\)
Worked solution
Compute \(\vec{AB} = B-A.\) M1 This evaluates to \(\vec{AB} = (3, -2, 2).\) A1
Use point \(A=(1,2,3).\) M1
Step 2b - assemble the vector equation. \(\mathbf r = (1,2,3) + t(3,-2,2).\) A1
Determine whether \(\mathbf r_1=(1,0,0)+s(1,1,0)\) and \(\mathbf r_2=(0,0,2)+t(0,1,1)\) intersect, are parallel, or are skew.
(a) State \(s.\)
(b) State \(t.\)
Worked solution
\((1,1,0)\) and \((0,1,1)\) are not scalar multiples, so the lines are not parallel. M1 A1
\(x: 1+s = 0 \Rightarrow s = -1.\) \(z: 0 = 2+t \Rightarrow t = -2.\) M1 A1 A1
\(y: s = t \Rightarrow -1 = -2\), false. M1 No common point and not parallel, so the lines are skew. A1
Common mistakes
- Assuming non-intersecting lines must be parallel. In three dimensions, two lines can fail to meet without being parallel - that's exactly what a skew line is.
- Mixing up the point and the direction vector. In \(\mathbf r=\mathbf a+\lambda\mathbf b\), \(\mathbf a\) is a fixed position and \(\mathbf b\) is the direction you travel in - swapping them, or using two points where a point and a direction are needed, gives the wrong line entirely.
- Stopping after solving only two of the three equations. With two unknown parameters and three coordinate equations, you must check the third equation is also satisfied before concluding the lines intersect - skip it and you can wrongly rule a skew pair as meeting.
Ready to practise properly?
23 vector-line questions, marked instantly like the real exam.
Quick answers
What is the vector equation of a line?
\(\mathbf r=\mathbf a+t\mathbf b\), where \(\mathbf a\) is the position vector of a known point on the line and \(\mathbf b\) is a direction vector. \(t\) is a scalar parameter that generates every point on the line as it varies.
What does it mean for two lines to be skew?
Skew lines are lines in three dimensions that never meet but also are not parallel - their direction vectors are not scalar multiples of each other, and no values of the two parameters satisfy all three coordinate equations at once.