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"Show that the points are collinear" sounds like a five-second question, and with the right test it is. The trouble is that students compare the wrong vectors, stop after one component, or meet a question written purely in position vectors and do not know which test applies. This post gives four tests, says when each is quickest, and works through examples that also give you the ratio of division.
Where is this in the syllabus?
NCERT's Class 12 Vector Algebra chapter (2026-27 reprint) defines collinear vectors as vectors parallel to the same line, shows that two vectors are collinear exactly when one is a scalar multiple of the other, and works an example showing three points are collinear. The Class 12 Three Dimensional Geometry chapter returns to collinear points in its exercises. The CBSE Mathematics curriculum for 2026-27 lists collinear vectors, position vectors and the position vector of a point dividing a segment in a given ratio. The JEE (Main) 2026 syllabus covers vector algebra and 3D geometry. Check the documents for your exam year.
What are the four tests?
For points A, B and C with position vectors a, b and c:
Scalar multiple. AB = λ AC for some number λ. Since both vectors start at A, they can only be parallel if all three points lie on one line.
Cross product. AB × AC = 0. This is test 1 written in one line; it is also the statement that the triangle ABC has zero area.
Lengths. One distance equals the sum of the other two, such as AB + BC = AC. NCERT notes this as the case of equality in the triangle inequality. It is slow, because of square roots.
Position vector relation. If xa + yb + zc = 0 with x + y + z = 0, and x, y, z not all zero, the points are collinear.
Worked example 1: test and ratio together
Are A(1, −2, 3), B(3, 1, 1) and C(7, 7, −3) collinear?
AB = (3 − 1, 1 + 2, 1 − 3) = (2, 3, −2).
AC = (7 − 1, 7 + 2, −3 − 3) = (6, 9, −6).
AC = 3 AB in every component. So the points are collinear.
The multiple tells you more. AC = 3 AB means B is one third of the way from A to C, so B divides AC internally in the ratio 1 : 2. Check with lengths: |AB| = √17, |BC| = |(4, 6, −4)| = √68 = 2√17, and |AC| = 3√17. Indeed AB + BC = AC.
Do not stop at the first component. Compute λ from each component and confirm they agree.
Worked example 2: finding an unknown
For which k are P(1, 2, 3), Q(3, k, 7) and R(4, 8, 9) collinear?
PR = (3, 6, 6), with no unknown, so use it as the reference.
PQ = (2, k − 2, 4).
If PQ = λ PR, the x-components give 2 = 3λ, so λ = 2/3.
Then y: k − 2 = (2/3)(6) = 4, so k = 6. And z: 4 = (2/3)(6) = 4, which holds.
Because the z-component agrees, the answer k = 6 is genuine. Had z given a contradiction, no value of k would work. Always use the unknown-free vector as the reference; it saves algebra.
Worked example 3: the position vector relation
The position vectors of A, B and C satisfy 2a − 5b + 3c = 0. Are the points collinear, and how is C placed?
The coefficients add up to 2 − 5 + 3 = 0, so the points are collinear. To see why, solve for c:
c = (5b − 2a)/3 = (5b − 2a)/(5 − 2).
Compare this with the external section formula, (mb − na)/(m − n). With m = 5 and n = 2, C divides AB externally in the ratio 5 : 2.
Check with numbers. Take a = (1, 0, 2) and b = (3, 1, −1). Then c = (13/3, 5/3, −3). The distances are CA = √350/3 ≈ 6.24 and CB = √56/3 ≈ 2.49, and their ratio is 2.5, which is 5 : 2. Also c − a = (5/3)(b − a), so C lies beyond B on the line AB. Our post on the section formula in 3D explains why external division puts the point outside the segment.
If the coefficients did not add up to zero, you could not conclude anything about collinearity from the relation alone.
Worked example 4: when the points are not collinear
A(1, 0, 0), B(0, 1, 0), C(0, 0, 1). AB = (−1, 1, 0), AC = (−1, 0, 1).
AB × AC = (1 × 1 − 0 × 0, 0 × (−1) − (−1) × 1, (−1) × 0 − 1 × (−1)) = (1, 1, 1).
This is not the zero vector, so the points are not collinear. As a bonus, half its length, √3/2, is the area of triangle ABC. The cross product test is the one to choose when a question also asks for an area.
Which mistakes cost the most marks?
Comparing a with b instead of AB with AC.
Checking one component and declaring a multiple.
Reading λ = −2 as "not collinear". A negative multiple means the points are on one line, with A between B and C.
Applying the position vector test without checking x + y + z = 0.
Sign slips in the middle term of the cross product. Write the determinant out.
The vector habits in our post on vector algebra for JEE maths apply here too.
A practice routine
Day 1: Five collinearity checks by scalar multiple, each also giving the ratio.
Day 2: Five "find k" questions. Pick the unknown-free reference vector first.
Day 3: Five position vector relations. Check the coefficient sum, then find the ratio from the section formula.
Day 4: Redo three Day 1 questions by cross product and compare the time taken.
Mix these with other vector and 3D questions rather than doing one type all week. If the meaning of a cross product is the weak spot, the Fermi Concept Library covers every concept in mathematics with its own lesson, so you can revise that one idea and come back.
Tonight, find the value of k that makes (2, −1, k), (4, 3, 1) and (5, 5, −1) collinear, and state the ratio in which the middle point divides the other two.
