MathsCBSE Class 12 · Vector Algebra

Section Formula and Collinearity of Points Using Vectors (Class 12)

Updated 15 October 2026 · Fact-checked

If A and B have position vectors a and b, the point dividing AB internally in the ratio m : n has position vector (mb + na)/(m + n); externally it is (mb − na)/(m − n). Three points A, B, C are collinear when AB is a scalar multiple of AC, that is AB = λ AC.

Notice the cross-over: m, the part next to A, multiplies b, and n, the part next to B, multiplies a.

Illustration drawn for QuestionBuddy.

Understand Section Formula and Collinearity of Vectors

Position vectors. Fix an origin O. The position vector of a point $P (x, y, z)$ is $O P = x \hat{i} + y \hat{j} + z \hat{k}$ . If $A$ and $B$ have position vectors $a$ and $b$ , then $A B = b - a$ ('head minus tail').

Section formula (internal). The point $R$ dividing $A B$ internally in the ratio $m : n$ (so $A R : R B = m : n$ ) has position vector
$r = \frac{m b + n a}{m + n}$ .
Notice the cross-over: $m$ (the part next to A) multiplies $b$ . The midpoint is the case $m = n$ : $\frac{a + b}{2}$ . The centroid of triangle ABC is $\frac{a + b + c}{3}$ .

Section formula (external). If $R$ lies on line AB extended with $A R : R B = m : n$ ( $m \neq = n$ ), then
$r = \frac{m b - n a}{m - n}$ .
An equivalent way to think: every point on line AB is $a + t (b - a)$ ; internal division has $0 < t < 1$ , external has $t > 1$ or $t < 0$ .

Collinear vectors and points. Two vectors are collinear (parallel) if one is a scalar multiple of the other: $b = λ a$ ; equivalently, their components are proportional. Three points A, B, C are collinear if $A B$ and $A C$ (or $A B$ and $B C$ ) are collinear vectors, since they share the point A. The scalar $λ$ also gives the ratio: if $A C = \frac{3}{2} A B$ , then $A B : B C = 2 : 1$ .

Finding an unknown. If one coordinate is unknown, write the proportionality of components and solve. If a ratio is unknown, write $r = \frac{k b + a}{k + 1}$ (ratio $k : 1$ ) and compare components.

Vector proofs in geometry. Results like the midpoint theorem follow in two lines: if D and E are midpoints of AB and AC, $D E = \frac{a + c}{2} - \frac{a + b}{2} = \frac{1}{2} (c - b) = \frac{1}{2} B C$ .

This is NCERT Section 10.5 (vector joining two points, section formula) and appears as 1- to 3-mark questions and within case studies.

Section Formula and Collinearity of Vectors: formulas to remember

Vector joining two points: $A B = b - a$; Head minus tail.
Internal division: $r = \frac{m b + n a}{m + n}$; $A R : R B = m : n$ .
External division: $r = \frac{m b - n a}{m - n}$; $m \neq = n$ .
Midpoint and centroid: $\frac{a + b}{2}, \frac{a + b + c}{3}$; Special cases of the section formula.
Collinear vectors: $b = λ a ⟺ \frac{b _{1}}{a _{1}} = \frac{b _{2}}{a _{2}} = \frac{b _{3}}{a _{3}}$; Components proportional.
Collinear points: $A C = λ A B$; Then $A B : B C = 1 : (λ - 1)$ when $λ > 1$ .

How to solve Section Formula and Collinearity of Vectors questions

Use this for section-formula, ratio and collinearity questions.

1Write all position vectors in component form.
2For a division point, decide internal or external, then apply $\frac{m b \pm n a}{m \pm n}$ with $m$ next to $b$ .
3For collinearity, compute $A B$ and $A C$ (head minus tail) and check whether one is a multiple of the other.
4For an unknown coordinate, set the component ratios equal and solve.
5For an unknown ratio, use $k : 1$ and compare one convenient component; check the others.
6State the conclusion in words: 'Hence A, B, C are collinear and B divides AC in the ratio 2 : 1.'

Section Formula and Collinearity of Vectors in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Dividing a segment in a given ratio

Suppose a zip-line runs straight from a platform at $A (0, 0, 20)$ to a landing point $B (60, 80, 0)$ , in metres. A safety stop is placed one quarter of the way from A to B.

Find the position of the safety stop and the length of the zip-line.

Try it first, then show the solutionSolution

One quarter of the way means $A R : R B = 1 : 3$ .
$r = \frac{1 \cdot b + 3 a}{4} = \frac{( 60 , 80 , 0 ) + ( 0 , 0 , 60 )}{4} = (15, 20, 15)$ .
$∣ A B ∣ = 6 0^{2} + 8 0^{2} + 2 0^{2} = 10400 \approx 102$ m.

Answer: Safety stop at $(15, 20, 15)$ ; zip-line about 102 m long.

Memory hook: m next to b: the bigger share goes to the far end.

Collinearity of points

Imagine an electrician wants to run one straight cable through three lamp posts at $P (1, 2, 0)$ , $Q (3, 5, 0)$ and $R (7, 11, 0)$ (in metres).

Check whether a single straight cable can pass through all three posts.

Try it first, then show the solutionSolution

$P Q = (2, 3, 0)$ and $P R = (6, 9, 0)$ .
$P R = 3 P Q$ , so the vectors are collinear and share P.
Hence P, Q, R lie on one line; $P Q : QR = 1 : 2$ .

Answer: Yes; the posts are collinear, with Q one third of the way from P to R.

Memory hook: Collinear points: one direction vector is a multiple of the other.

Tricks to remember Section Formula and Collinearity of Vectors

Short memory aids. Say each one aloud once, then use it on a question the same day.

Cross-multiply the ratio
In $\frac{m b + n a}{m + n}$ , m pairs with b and n pairs with a: the numbers cross over.
Use it when: Avoiding the reversed section formula.
External means minus
Change both plus signs to minus for external division.
Use it when: External division questions.
Head minus tail
$A B = b - a$ : where you arrive minus where you start.
Use it when: Every displacement vector.
Proportional components, parallel vectors
Divide matching components; if all ratios agree, the vectors are parallel.
Use it when: Collinearity tests and finding unknowns.

Section Formula and Collinearity of Vectors: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Internal division · 2 marks · Very short answerBoard level

Find the position vector of the point R dividing the line joining $P (\hat{i} + 2 \hat{j} - \hat{k})$ and $Q (4 \hat{i} - \hat{j} + 5 \hat{k})$ internally in the ratio $2 : 1$ .

Try it first, then show the solutionSolution

$r = \frac{2 q + 1 p}{3} = \frac{( 8 i ^ - 2 j ^ + 10 k ^ ) + ( i ^ + 2 j ^ - k ^ )}{3}$ . (1 mark)
$= \frac{9 i ^ + 0 j ^ + 9 k ^}{3} = 3 \hat{i} + 3 \hat{k}$ . (1 mark)

Answer: $3 \hat{i} + 3 \hat{k}$

Example 2: External division · 2 marks · Very short answerBoard level

For the same P and Q, find the point dividing PQ externally in the ratio $2 : 1$ .

Try it first, then show the solutionSolution

$r = \frac{2 q - p}{2 - 1} = (8 \hat{i} - 2 \hat{j} + 10 \hat{k}) - (\hat{i} + 2 \hat{j} - \hat{k})$ . (1 mark)
$= 7 \hat{i} - 4 \hat{j} + 11 \hat{k}$ . (1 mark)

Answer: $7 \hat{i} - 4 \hat{j} + 11 \hat{k}$

Example 3: Proving collinearity and finding the ratio · 3 marks · Short answerHarder board level

Show that $A (2, - 1, 3)$ , $B (4, 3, 1)$ and $C (5, 5, 0)$ are collinear, and find the ratio in which B divides AC.

Try it first, then show the solutionSolution

$A B = 2 \hat{i} + 4 \hat{j} - 2 \hat{k}$ , $A C = 3 \hat{i} + 6 \hat{j} - 3 \hat{k}$ . (1 mark)
$A C = \frac{3}{2} A B$ , so the vectors are parallel with common point A: A, B, C are collinear. (1 mark)
$B C = A C - A B = \frac{1}{2} A B$ , so $A B : B C = 2 : 1$ (internal). Check: $\frac{2 c + a}{3} = \frac{( 10 , 10 , 0 ) + ( 2 , - 1 , 3 )}{3} = (4, 3, 1)$ . (1 mark)

Answer: Collinear; B divides AC internally in the ratio $2 : 1$ .

Example 4: An unknown coordinate · 3 marks · Short answerHarder board level

Find $λ$ if the points $A (1, λ, 2)$ , $B (2, 3, 1)$ and $C (4, 7, - 1)$ are collinear.

Try it first, then show the solutionSolution

$A B = (1, 3 - λ, - 1)$ , $B C = (2, 4, - 2)$ . (1 mark)
Collinear points need $\frac{1}{2} = \frac{3 - λ}{4} = \frac{- 1}{- 2}$ . (1 mark)
$\frac{3 - λ}{4} = \frac{1}{2} \Rightarrow λ = 1$ . (1 mark)

Answer: $λ = 1$

Example 5: A vector proof of the midpoint theorem · 3 marks · Short answerHOTS / competency

Using vectors, prove that the segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

Try it first, then show the solutionSolution

Let A, B, C have position vectors $a, b, c$ , and let D, E be the midpoints of AB and AC: $d = \frac{a + b}{2}$ , $e = \frac{a + c}{2}$ . (1 mark)
$D E = e - d = \frac{c - b}{2} = \frac{1}{2} B C$ . (1 mark)
So $D E$ is a scalar multiple of $B C$ (parallel), and $∣ D E ∣ = \frac{1}{2} ∣ B C ∣$ . (1 mark)

Answer: $D E = \frac{1}{2} B C$ : parallel and half as long.

Example 6: Case study: checkpoints on a drone route · 4 marks · Case-basedHOTS / competency

Suppose a delivery drone flies in a straight line from $A (2, 1, 0)$ to $B (8, 4, 6)$ (in km).
(i) Find the two checkpoints that divide AB into three equal parts. (2 marks)
(ii) Find the length of the route AB. (1 mark)
(iii) The drone continues straight on to $C (11, 5.5, 9)$ . In what ratio does C divide AB externally? (1 mark)

Try it first, then show the solutionSolution

(i) Ratio $1 : 2$ : $\frac{b + 2 a}{3} = \frac{( 8 , 4 , 6 ) + ( 4 , 2 , 0 )}{3} = (4, 2, 2)$ . Ratio $2 : 1$ : $\frac{2 b + a}{3} = \frac{( 16 , 8 , 12 ) + ( 2 , 1 , 0 )}{3} = (6, 3, 4)$ . (2 marks)
(ii) $∣ A B ∣ = 6^{2} + 3^{2} + 6^{2} = 81 = 9$ km. (1 mark)
(iii) $A C = (9, 4.5, 9) = \frac{3}{2} A B$ , so $A C = 1.5 A B$ and $B C = 0.5 A B$ ; $A C : C B = 3 : 1$ externally. Check: $\frac{3 b - a}{2} = (11, 5.5, 9)$ . (1 mark)

Answer: (i) $(4, 2, 2)$ and $(6, 3, 4)$ (ii) 9 km (iii) $3 : 1$ externally.

Where students lose marks in Section Formula and Collinearity of Vectors

The slips examiners see most often, and what to write instead.

Mistake: Reversing m and n
m must multiply the position vector of the far end B.
Wrong working
Ratio $2 : 1$ : $r = \frac{2 p + q}{3}$ .
Right working
$r = \frac{2 q + p}{3}$ .
Do this: Write $A R : R B = m : n$ and remember the cross-over.
Usually costs: The answer.
Mistake: Using the internal formula for external division
The external point lies outside the segment and needs minus signs.
Wrong working
External $2 : 1$ : $\frac{2 q + p}{3}$ .
Right working
$\frac{2 q - p}{2 - 1}$ .
Do this: Read 'internally' or 'externally' twice.
Usually costs: 1 to 2 marks.
Mistake: Checking only two components for collinearity
All three component ratios must be equal.
Wrong working
$\frac{1}{2} = \frac{- 1}{- 2}$ , so collinear.
Right working
Also check the middle component: $\frac{3 - λ}{4} = \frac{1}{2}$ .
Do this: Write all three ratios.
Usually costs: 1 mark.
Mistake: Writing AB as a − b
The displacement from A to B is head minus tail.
Wrong working
$A B = a - b$ .
Right working
$A B = b - a$ .
Do this: Say 'B minus A' aloud.
Usually costs: Sign errors in every later step.

Exam tips

Section-formula and collinearity questions appeared in the 2023, 2024, 2025 and 2026 papers as 1- to 3-mark items; show the formula with your values substituted.
Leave answers in $\hat{i}, \hat{j}, \hat{k}$ form unless coordinates are asked.
For collinearity, a single line ' $A C = λ A B$ with common point A' is the key reason.
Check a division point by verifying that it lies between A and B for internal division.
In case studies, express every location as a position vector first.

Practice questions on Section Formula and Collinearity of Vectors

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The position vector of the midpoint of the segment joining points with position vectors $a$ and $b$ is:
(a) $a + b$
(b) $\frac{a - b}{2}$
(c) $\frac{a + b}{2}$
(d) $b - a$
Hint
Internal division in the ratio 1 : 1.
Answer
(c) $\frac{a + b}{2}$ .
Q2 · 2 marks · Very short answerBoard level
Find the position vector of the point dividing the join of $A (2, 3, 4)$ and $B (- 1, 0, 5)$ internally in the ratio $1 : 2$ .
Hint
$\frac{1 \cdot b + 2 a}{3}$ .
Answer
$\frac{( - 1 , 0 , 5 ) + ( 4 , 6 , 8 )}{3} = (1, 2, \frac{13}{3})$ , i.e. $\hat{i} + 2 \hat{j} + \frac{13}{3} \hat{k}$ .
Q3 · 2 marks · Very short answerBoard level
Show that $a = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $b = - 4 \hat{i} + 6 \hat{j} - 8 \hat{k}$ are collinear.
Hint
Compare components.
Answer
$b = - 2 a$ , so they are collinear (opposite in direction).
Q4 · 2 marks · Very short answerHarder board level
Find the centroid of the triangle with vertices $A (1, 2, 3)$ , $B (3, 0, 1)$ and $C (2, 4, 5)$ .
Hint
$\frac{a + b + c}{3}$ .
Answer
$\frac{( 6 , 6 , 9 )}{3} = (2, 2, 3)$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The points with position vectors $a$ , $b$ and $3 b - 2 a$ are collinear.
Reason (R): For these points, $A C = 3 A B$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
$A C = (3 b - 2 a) - a$ .
Answer
(a). $A C = 3 (b - a) = 3 A B$ , which proves collinearity.
Q6 · 3 marks · Short answerHarder board level
Find the ratio in which $C (5, 4, 1)$ divides the segment joining $A (1, 2, - 1)$ and $B (7, 5, 2)$ .
Hint
Compare $A C$ and $C B$ .
Answer
$A C = (4, 2, 2)$ and $C B = (2, 1, 1)$ , so $A C = 2 C B$ : C divides AB internally in the ratio $2 : 1$ .
Q7 · 3 marks · Short answerHOTS / competency
If $a, b, c$ are the position vectors of A, B, C and $2 a + 3 b - 5 c = 0$ , show that A, B, C are collinear and find the ratio in which C divides AB.
Hint
Solve for $c$ and compare with the section formula.
Answer
$c = \frac{3 b + 2 a}{5} = \frac{3 b + 2 a}{3 + 2}$ , which is the section formula with $m = 3$ , $n = 2$ . So C lies on AB and divides it internally in the ratio $3 : 2$ ; hence A, B, C are collinear.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Internal section formula?Show answer
$\frac{m b + n a}{m + n}$ .
2.External section formula?Show answer
$\frac{m b - n a}{m - n}$ .
3.Position vector of the centroid of ABC?Show answer
$\frac{a + b + c}{3}$ .
4.When are $a$ and $b$ collinear?Show answer
When $b = λ a$ for some scalar λ.
5.What is $A B$ in terms of position vectors?Show answer
$b - a$ .
6.Midpoint of $(2, 4, 6)$ and $(0, 2, - 2)$ ?Show answer
$(1, 3, 2)$ .

Section Formula and Collinearity of Vectors: frequently asked questions

What is the section formula in vectors?

If R divides AB internally in the ratio m : n, its position vector is (m b + n a)/(m + n). For external division, it is (m b − n a)/(m − n).

How do I prove three points are collinear using vectors?

Find AB and AC (or AB and BC) and show that one is a scalar multiple of the other. Since they share a point, the three points lie on one line.

What is the difference between collinear vectors and collinear points?

Collinear vectors are parallel vectors (one is a multiple of the other) and may lie on different lines. Collinear points lie on the same line; you test them by checking that the vectors joining them are collinear and share a point.

Which NCERT section covers this?

Section 10.5 (Multiplication of a Vector by a Scalar), including 10.5.2 (Vector joining two points) and 10.5.3 (Section formula), in Chapter 10 of NCERT Mathematics Part II for Class 12.

Is coplanarity of vectors in the syllabus?

No. The scalar triple product and coplanarity were dropped from the 2026-27 syllabus.