100% FREE Updated: Apr 2026 Vectors, Matrices and 3D Geometry Vectors

Vector operations

Comprehensive study notes on Vector operations for CMI BS Hons preparation. This chapter covers key concepts, formulas, and examples needed for your exam.

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Vector operations

This chapter introduces the fundamental operations on vectors, including addition and scalar multiplication. Mastering these concepts, alongside understanding position vectors and the section formula, is crucial as they form the bedrock for advanced topics in 3D geometry and linear algebra, frequently assessed in the CMI BS Hons examination.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Vector addition | | 2 | Scalar multiplication | | 3 | Position vectors | | 4 | Section formula |

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We begin with Vector addition.

Part 1: Vector addition

Vector Addition

Overview

Vector addition combines two or more vectors to produce a resultant vector. It is one of the most basic vector operations and is the starting point for displacement, force, geometry, and coordinate reasoning. In exam problems, vector addition is tested both algebraically through components and geometrically through triangle and parallelogram laws. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

  • Add vectors in component form.

  • Understand vector addition geometrically.

  • Use the triangle law and parallelogram law of addition.

  • Compute resultants of multiple vectors.

  • Use vector addition in coordinate and geometry problems.

---

Core Idea

📖 Vector Addition

If
a⃗=⟨x1,y1⟩\qquad \vec{a}=\langle x_1,y_1\rangle
and
b⃗=⟨x2,y2⟩\qquad \vec{b}=\langle x_2,y_2\rangle,
then

a⃗+b⃗=⟨x1+x2, y1+y2⟩\qquad \vec{a}+\vec{b}=\langle x_1+x_2,\ y_1+y_2\rangle

In three dimensions: ⟨x1,y1,z1⟩+⟨x2,y2,z2⟩=⟨x1+x2, y1+y2, z1+z2⟩\qquad \langle x_1,y_1,z_1\rangle + \langle x_2,y_2,z_2\rangle = \langle x_1+x_2,\ y_1+y_2,\ z_1+z_2\rangle ::: ---

Geometric Interpretation

❗ Triangle Law

To add a⃗\vec{a} and b⃗\vec{b} geometrically:

  • draw a⃗\vec{a}
    • place the tail of b⃗\vec{b} at the head of a⃗\vec{a}
      • the vector from the starting point of a⃗\vec{a} to the final point of b⃗\vec{b} is a⃗+b⃗\vec{a}+\vec{b}

❗ Parallelogram Law

If two vectors start from the same point, then their sum is represented by the diagonal of the parallelogram formed on them.

These two laws describe the same algebraic operation geometrically. ---

Algebraic Properties

📐 Vector Addition Rules

For vectors a⃗,b⃗,c⃗\vec{a},\vec{b},\vec{c}:

  • a⃗+b⃗=b⃗+a⃗\qquad \vec{a}+\vec{b}=\vec{b}+\vec{a}

    • (a⃗+b⃗)+c⃗=a⃗+(b⃗+c⃗)\qquad (\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})

      • a⃗+0⃗=a⃗\qquad \vec{a}+\vec{0}=\vec{a}

        • a⃗+(−a⃗)=0⃗\qquad \vec{a}+(-\vec{a})=\vec{0}

So vector addition is commutative and associative. ---

Resultant Vector

📖 Resultant

The sum of several vectors is called their resultant.

If
R⃗=a⃗+b⃗+c⃗\qquad \vec{R}=\vec{a}+\vec{b}+\vec{c},
then R⃗\vec{R} is the single vector having the same overall effect as the three vectors together.

---

Minimal Worked Examples

Example 1 If a⃗=⟨2,3⟩\qquad \vec{a}=\langle 2,3\rangle and b⃗=⟨−1,4⟩\qquad \vec{b}=\langle -1,4\rangle, then a⃗+b⃗=⟨2+(−1),3+4⟩=⟨1,7⟩\qquad \vec{a}+\vec{b}=\langle 2+(-1),3+4\rangle = \langle 1,7\rangle So a⃗+b⃗=⟨1,7⟩\qquad \boxed{\vec{a}+\vec{b}=\langle 1,7\rangle} --- Example 2 If u⃗=⟨1,0,2⟩\qquad \vec{u}=\langle 1,0,2\rangle and v⃗=⟨3,−2,5⟩\qquad \vec{v}=\langle 3,-2,5\rangle, then u⃗+v⃗=⟨4,−2,7⟩\qquad \vec{u}+\vec{v}=\langle 4,-2,7\rangle ---

Zero Vector and Additive Inverse

📐 Important Special Cases

The zero vector is
0⃗\qquad \vec{0}

If
a⃗=⟨x,y⟩\qquad \vec{a}=\langle x,y\rangle,
then
−a⃗=⟨−x,−y⟩\qquad -\vec{a}=\langle -x,-y\rangle

So
a⃗+(−a⃗)=0⃗\qquad \vec{a}+(-\vec{a})=\vec{0}

---

Common Patterns

💡 Typical Exam Patterns

  • Add two vectors in component form.

  • Find a resultant of several vectors.

  • Use triangle law in geometry.

  • Use

AB⃗+BC⃗=AC⃗\qquad \vec{AB}+\vec{BC}=\vec{AC}
  • Express one unknown vector from a vector equation.

---

Important Geometry Identity

📐 Path Addition

For points A,B,CA,B,C,

AB⃗+BC⃗=AC⃗\qquad \vec{AB}+\vec{BC}=\vec{AC}

This is one of the most important vector identities in geometry. It expresses the idea that going from AA to BB and then from BB to CC is equivalent to going directly from AA to CC. ::: ---

Common Mistakes

⚠️ Avoid These Errors
    • ❌ adding magnitudes instead of components
✅ vector addition is componentwise
    • ❌ mixing vector addition with scalar multiplication
✅ keep operations separate
    • ❌ forgetting sign when adding negative components
✅ use brackets carefully
    • ❌ thinking order matters for the sum
✅ vector addition is commutative
---

CMI Strategy

💡 How to Attack Vector-Addition Problems

  • Convert vectors to components if possible.

  • Add corresponding components directly.

  • In geometry, translate movement into path addition.

  • Use the zero vector to simplify vector equations.

  • Interpret the final vector geometrically if needed.

---

Practice Questions

:::question type="MCQ" question="If a⃗=⟨3,2⟩\vec{a}=\langle 3,2\rangle and b⃗=⟨−1,5⟩\vec{b}=\langle -1,5\rangle, then a⃗+b⃗\vec{a}+\vec{b} is" options=["⟨2,7⟩\langle 2,7\rangle","⟨4,−3⟩\langle 4,-3\rangle","⟨−2,7⟩\langle -2,7\rangle","⟨2,−7⟩\langle 2,-7\rangle"] answer="A" hint="Add components." solution="We have a⃗+b⃗=⟨3+(−1), 2+5⟩=⟨2,7⟩\qquad \vec{a}+\vec{b}=\langle 3+(-1),\ 2+5\rangle = \langle 2,7\rangle Hence the correct option is A\boxed{A}." ::: :::question type="NAT" question="If u⃗=⟨2,−3,1⟩\vec{u}=\langle 2,-3,1\rangle and v⃗=⟨4,1,−2⟩\vec{v}=\langle 4,1,-2\rangle, find u⃗+v⃗\vec{u}+\vec{v}." answer="6,-2,-1" hint="Add corresponding coordinates." solution="We add componentwise: u⃗+v⃗=⟨2+4, −3+1, 1+(−2)⟩=⟨6,−2,−1⟩\qquad \vec{u}+\vec{v} = \langle 2+4,\ -3+1,\ 1+(-2)\rangle = \langle 6,-2,-1\rangle Hence the answer is ⟨6,−2,−1⟩\boxed{\langle 6,-2,-1\rangle}." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["a⃗+b⃗=b⃗+a⃗\vec{a}+\vec{b}=\vec{b}+\vec{a}","(a⃗+b⃗)+c⃗=a⃗+(b⃗+c⃗)(\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})","a⃗+0⃗=a⃗\vec{a}+\vec{0}=\vec{a}","a⃗+b⃗\vec{a}+\vec{b} is found by multiplying corresponding components"] answer="A,B,C" hint="Recall the algebraic properties of vector addition." solution="1. True.
  • True.
  • True.
  • False, because vector addition is done by adding corresponding components, not multiplying them.
    • Hence the correct answer is A,B,C\boxed{A,B,C}." ::: :::question type="SUB" question="Using vector addition, prove that for any points A,B,CA,B,C, we have AB⃗+BC⃗=AC⃗\vec{AB}+\vec{BC}=\vec{AC}." answer="Path addition from AA to CC via BB." hint="Write the vectors in coordinate-difference form or use displacement logic." solution="Let the position vectors of the points be OA⃗=a⃗,OB⃗=b⃗,OC⃗=c⃗\qquad \vec{OA}=\vec{a},\quad \vec{OB}=\vec{b},\quad \vec{OC}=\vec{c} Then AB⃗=OB⃗−OA⃗=b⃗−a⃗\qquad \vec{AB}=\vec{OB}-\vec{OA}=\vec{b}-\vec{a} and BC⃗=OC⃗−OB⃗=c⃗−b⃗\qquad \vec{BC}=\vec{OC}-\vec{OB}=\vec{c}-\vec{b} So AB⃗+BC⃗=(b⃗−a⃗)+(c⃗−b⃗)=c⃗−a⃗\qquad \vec{AB}+\vec{BC}=(\vec{b}-\vec{a})+(\vec{c}-\vec{b})=\vec{c}-\vec{a} But c⃗−a⃗=AC⃗\qquad \vec{c}-\vec{a}=\vec{AC} Hence AB⃗+BC⃗=AC⃗\qquad \boxed{\vec{AB}+\vec{BC}=\vec{AC}} This proves the identity." ::: ---

    Summary

    ❗ Key Takeaways for CMI

    • Vector addition is done componentwise.

    • Geometrically it is described by the triangle and parallelogram laws.

    • Vector addition is commutative and associative.

    • The identity AB⃗+BC⃗=AC⃗\vec{AB}+\vec{BC}=\vec{AC} is fundamental.
      • Resultants and geometric movement are natural applications.

    ---

    💡 Next Up

    Proceeding to Scalar multiplication.

    ---

    Part 2: Scalar multiplication

    Scalar Multiplication

    Overview

    Scalar multiplication is the operation of multiplying a vector by a real number. This operation changes the magnitude of the vector and may also reverse its direction. In vector problems, scalar multiplication is one of the most basic but most powerful tools, because it allows us to express parallel vectors, divide line segments, and build linear combinations. ---

    Learning Objectives

    ❗ By the End of This Topic

    After studying this topic, you will be able to:

    • Multiply a vector by a scalar correctly.

    • Understand how scalar multiplication changes magnitude and direction.

    • Identify when two vectors are scalar multiples of each other.

    • Use scalar multiplication in line, section, and vector-equation problems.

    • Work confidently with zero, negative, and fractional scalars.

    ---

    Core Idea

    📖 Scalar Multiplication

    If a⃗\vec{a} is a vector and λ\lambda is a real number, then

    λa⃗\qquad \lambda \vec{a}

    is called the scalar multiple of a⃗\vec{a} by λ\lambda.

    If a⃗=⟨x,y⟩\qquad \vec{a} = \langle x,y \rangle then λa⃗=⟨λx,λy⟩\qquad \lambda \vec{a} = \langle \lambda x,\lambda y \rangle In three dimensions, if a⃗=⟨x,y,z⟩\qquad \vec{a} = \langle x,y,z \rangle then λa⃗=⟨λx,λy,λz⟩\qquad \lambda \vec{a} = \langle \lambda x,\lambda y,\lambda z \rangle ::: ---

    Geometric Meaning

    ❗ Effect on Magnitude and Direction

    If λ>0\lambda > 0, then λa⃗\lambda \vec{a} has the same direction as a⃗\vec{a}.

    If λ<0\lambda < 0, then λa⃗\lambda \vec{a} has the opposite direction to a⃗\vec{a}.

    If λ=0\lambda = 0, then
    λa⃗=0⃗\qquad \lambda \vec{a} = \vec{0}
    the zero vector.

    Also,
    ∣λa⃗∣=∣λ∣ ∣a⃗∣\qquad |\lambda \vec{a}| = |\lambda|\,|\vec{a}|

    So scalar multiplication scales length by the absolute value of the scalar. ---

    Important Algebraic Properties

    📐 Scalar Multiplication Rules

    For vectors a⃗,b⃗\vec{a}, \vec{b} and scalars λ,μ\lambda,\mu:

    • λ(a⃗+b⃗)=λa⃗+λb⃗\qquad \lambda(\vec{a}+\vec{b}) = \lambda \vec{a} + \lambda \vec{b}

      • (λ+μ)a⃗=λa⃗+μa⃗\qquad (\lambda+\mu)\vec{a} = \lambda \vec{a} + \mu \vec{a}

        • λ(μa⃗)=(λμ)a⃗\qquad \lambda(\mu \vec{a}) = (\lambda\mu)\vec{a}

          • 1a⃗=a⃗\qquad 1\vec{a} = \vec{a}

            • 0a⃗=0⃗\qquad 0\vec{a} = \vec{0}

              • (−1)a⃗=−a⃗\qquad (-1)\vec{a} = -\vec{a}

    ---

    Parallel Vectors

    📖 Parallelism via Scalar Multiplication

    Two nonzero vectors are parallel if and only if one is a scalar multiple of the other.

    That is,
    b⃗=λa⃗\qquad \vec{b} = \lambda \vec{a}
    for some nonzero real number λ\lambda.

    This is one of the most important uses of scalar multiplication in geometry. ---

    Minimal Worked Examples

    Example 1 If a⃗=⟨2,−3⟩\qquad \vec{a}=\langle 2,-3\rangle, find 4a⃗\qquad 4\vec{a}. We multiply each component by 44: 4a⃗=⟨8,−12⟩\qquad 4\vec{a} = \langle 8,-12\rangle So the answer is ⟨8,−12⟩\qquad \boxed{\langle 8,-12\rangle} --- Example 2 If a⃗=⟨3,1⟩\qquad \vec{a}=\langle 3,1\rangle, find −2a⃗\qquad -2\vec{a}. We get −2a⃗=⟨−6,−2⟩\qquad -2\vec{a} = \langle -6,-2\rangle The negative sign tells us the direction is reversed. ---

    Fractional Scalars

    💡 Fractions Are Common

    If
    a⃗=⟨6,9⟩\qquad \vec{a}=\langle 6,9\rangle,
    then
    13a⃗=⟨2,3⟩\qquad \dfrac{1}{3}\vec{a}=\left\langle 2,3 \right\rangle

    So fractional scalar multiplication often simplifies vectors and is useful in section and midpoint problems.

    ---

    Common Patterns

    💡 Typical Exam Patterns

    • Find λa⃗\lambda \vec{a} from a given vector.
      • Determine whether two vectors are parallel.

      • Find a scalar λ\lambda such that

      b⃗=λa⃗\qquad \vec{b}=\lambda \vec{a}
      • Use scalar multiplication inside a larger vector equation.

      • Express one vector in terms of another.

    ---

    Common Mistakes

    ⚠️ Avoid These Errors
      • ❌ multiplying only one component by the scalar
    ✅ multiply every component
      • ❌ forgetting that a negative scalar reverses direction
    ✅ sign matters geometrically
      • ❌ thinking scalar multiplication changes only direction
    ✅ it changes magnitude too
      • ❌ calling two vectors parallel without checking all components match the same scalar
    ✅ the same λ\lambda must work in every component
    ---

    CMI Strategy

    💡 How to Attack Scalar-Multiplication Problems

    • Write the vector in component form.

    • Multiply each component carefully.

    • If checking parallelism, compare component ratios.

    • Track the sign of the scalar for direction.

    • In multi-step problems, combine scalar multiplication with vector addition cleanly.

    ---

    Practice Questions

    :::question type="MCQ" question="If a⃗=⟨2,−1⟩\vec{a}=\langle 2,-1\rangle, then 3a⃗3\vec{a} is" options=["⟨6,−3⟩\langle 6,-3\rangle","⟨5,−2⟩\langle 5,-2\rangle","⟨−6,3⟩\langle -6,3\rangle","⟨6,3⟩\langle 6,3\rangle"] answer="A" hint="Multiply each component by 33." solution="We have 3a⃗=3⟨2,−1⟩=⟨6,−3⟩\qquad 3\vec{a}=3\langle 2,-1\rangle=\langle 6,-3\rangle Hence the correct option is A\boxed{A}." ::: :::question type="NAT" question="If a⃗=⟨4,8⟩\vec{a}=\langle 4,8\rangle, find the scalar λ\lambda such that λa⃗=⟨1,2⟩\lambda \vec{a}=\langle 1,2\rangle." answer="1/4" hint="Compare coordinates." solution="We need λ⟨4,8⟩=⟨1,2⟩\qquad \lambda \langle 4,8\rangle = \langle 1,2\rangle So 4λ=1\qquad 4\lambda = 1 and hence λ=14\qquad \lambda = \dfrac{1}{4} This also gives 8λ=2\qquad 8\lambda = 2 which is consistent. Therefore the answer is 14\boxed{\dfrac{1}{4}}." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["∣λa⃗∣=∣λ∣ ∣a⃗∣|\lambda \vec{a}| = |\lambda|\,|\vec{a}|","If λ<0\lambda<0, then λa⃗\lambda \vec{a} has direction opposite to a⃗\vec{a}","Two nonzero vectors are parallel iff one is a scalar multiple of the other","0a⃗=a⃗0\vec{a}=\vec{a} for every vector a⃗\vec{a}"] answer="A,B,C" hint="Recall the geometric effect of scalar multiplication." solution="1. True.
  • True.
  • True.
  • False, because
    • 0a⃗=0⃗\qquad 0\vec{a}=\vec{0} Hence the correct answer is A,B,C\boxed{A,B,C}." ::: :::question type="SUB" question="If u⃗=⟨3,−6,9⟩\vec{u}=\langle 3,-6,9\rangle, write u⃗\vec{u} as a scalar multiple of the vector ⟨1,−2,3⟩\langle 1,-2,3\rangle and explain the geometric meaning." answer="u⃗=3⟨1,−2,3⟩\vec{u}=3\langle 1,-2,3\rangle; same direction and triple magnitude." hint="Compare corresponding components." solution="We observe that ⟨3,−6,9⟩=3⟨1,−2,3⟩\qquad \langle 3,-6,9\rangle = 3\langle 1,-2,3\rangle So u⃗=3⟨1,−2,3⟩\qquad \vec{u}=3\langle 1,-2,3\rangle Since the scalar is positive, the two vectors have the same direction. Also, the magnitude of u⃗\vec{u} is three times the magnitude of ⟨1,−2,3⟩\langle 1,-2,3\rangle. Hence u⃗=3⟨1,−2,3⟩\qquad \boxed{\vec{u}=3\langle 1,-2,3\rangle} Geometrically, u⃗\vec{u} points in the same direction and is three times as long." ::: ---

    Summary

    ❗ Key Takeaways for CMI

    • Scalar multiplication multiplies every component of a vector.

    • Magnitude scales by absolute value, direction depends on sign.

    • Parallel vectors are scalar multiples of each other.

    • Zero and negative scalars need special attention.

    • Scalar multiplication is basic, but it appears inside many harder vector problems.

    ---

    💡 Next Up

    Proceeding to Position vectors.

    ---

    Part 3: Position vectors

    Position Vectors

    Overview

    A position vector describes the location of a point relative to a fixed origin. It turns geometric problems into vector equations and is one of the most powerful tools for proving midpoint, section, centroid, and collinearity results. In CMI-style problems, position vectors are frequently used to encode Euclidean geometry in a precise algebraic form. ---

    Learning Objectives

    ❗ By the End of This Topic

    After studying this topic, you will be able to:

    • Interpret the position vector of a point correctly.

    • Express segment vectors using position vectors.

    • Use midpoint and section formulas in vector form.

    • Find centroid and prove concurrency results with vectors.

    • Translate geometric conditions into vector equations.

    ---

    Core Definition

    📖 Position Vector

    If OO is the origin and PP is a point, then the vector

    OP→\qquad \overrightarrow{OP}

    is called the position vector of PP.

    If OA→=a,OB→=b\qquad \overrightarrow{OA}=\mathbf{a},\qquad \overrightarrow{OB}=\mathbf{b} then the vector from AA to BB is AB→=b−a\qquad \overrightarrow{AB}=\mathbf{b}-\mathbf{a} ::: ---

    Midpoint Formula

    📐 Position Vector of Midpoint

    If MM is the midpoint of ABAB, where

    OA→=a,OB→=b\qquad \overrightarrow{OA}=\mathbf{a},\qquad \overrightarrow{OB}=\mathbf{b}

    then

    OM→=a+b2\qquad \overrightarrow{OM} = \dfrac{\mathbf{a}+\mathbf{b}}{2}

    ---

    Section Formula

    📐 Internal Division

    If point PP divides ABAB internally in the ratio

    AP:PB=m:n\qquad AP:PB = m:n

    then

    OP→=na+mbm+n\qquad \overrightarrow{OP} = \dfrac{n\mathbf{a}+m\mathbf{b}}{m+n}

    This is one of the standard and most useful formulas in position-vector geometry. ---

    Centroid Formula

    📐 Position Vector of the Centroid

    If triangle ABCABC has

    OA→=a,OB→=b,OC→=c\qquad \overrightarrow{OA}=\mathbf{a},\qquad \overrightarrow{OB}=\mathbf{b},\qquad \overrightarrow{OC}=\mathbf{c}

    then the centroid GG has position vector

    OG→=a+b+c3\qquad \overrightarrow{OG} = \dfrac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}

    This is the vector form of the centroid theorem. ---

    Collinearity View

    💡 A Useful Pattern

    Points with position vectors a,b,p\mathbf{a},\mathbf{b},\mathbf{p} are collinear exactly when

    p=(1−t)a+tb\qquad \mathbf{p} = (1-t)\mathbf{a} + t\mathbf{b}

    for some real number tt.

    This expresses the idea that a point on a line segment is an affine combination of the endpoints. ---

    Minimal Worked Examples

    Example 1 If OA→=(2,1),OB→=(6,5)\qquad \overrightarrow{OA}=(2,1),\qquad \overrightarrow{OB}=(6,5) then the midpoint MM of ABAB has position vector OM→=(2+62,1+52)=(4,3)\qquad \overrightarrow{OM} = \left(\dfrac{2+6}{2},\dfrac{1+5}{2}\right)=(4,3) So the midpoint is (4,3)\boxed{(4,3)}. --- Example 2 If OA→=a,OB→=b\qquad \overrightarrow{OA}=\mathbf{a},\qquad \overrightarrow{OB}=\mathbf{b} then AB→=b−a\qquad \overrightarrow{AB}=\mathbf{b}-\mathbf{a} This is the most basic conversion from position vectors to displacement vectors. ---

    Common Mistakes

    ⚠️ Avoid These Errors
      • ❌ Writing AB→=a−b\overrightarrow{AB}=\mathbf{a}-\mathbf{b} instead of b−a\mathbf{b}-\mathbf{a}.
          • ❌ Mixing up the ratio in the section formula.
          • ❌ Treating a position vector and a point as the same object without context.
          • ❌ Forgetting that centroid is the average of the three vertex position vectors.
    ---

    CMI Strategy

    💡 How to Solve These Fast

    • Write every point relative to the origin first.

    • Convert geometric statements into vector equations.

    • Use midpoint, section, or centroid formulas immediately when appropriate.

    • Reduce geometry to algebra, then interpret the result back geometrically.

    • In proof questions, start from the vector identity and expand carefully.

    ---

    Practice Questions

    :::question type="MCQ" question="If OA→=a\overrightarrow{OA}=\mathbf{a} and OB→=b\overrightarrow{OB}=\mathbf{b}, then AB→\overrightarrow{AB} equals" options=["a+b\mathbf{a}+\mathbf{b}","a−b\mathbf{a}-\mathbf{b}","b−a\mathbf{b}-\mathbf{a}","a+b2\dfrac{\mathbf{a}+\mathbf{b}}{2}"] answer="C" hint="Vector from AA to BB is final position minus initial position." solution="We have AB→=OB→−OA→=b−a\qquad \overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA} = \mathbf{b}-\mathbf{a} Hence the correct option is C\boxed{C}." ::: :::question type="NAT" question="If the points AA and BB have coordinates (2,1)(2,1) and (6,5)(6,5), find the coordinates of the midpoint of ABAB." answer="4,3" hint="Use coordinate averaging." solution="The midpoint is (2+62,1+52)=(4,3)\qquad \left(\dfrac{2+6}{2},\dfrac{1+5}{2}\right) = (4,3) Hence the answer is (4,3)\boxed{(4,3)}." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["The position vector of the midpoint of ABAB is a+b2\dfrac{\mathbf{a}+\mathbf{b}}{2}","If AP:PB=1:2AP:PB=1:2, then OP→=2a+b3\overrightarrow{OP}=\dfrac{2\mathbf{a}+\mathbf{b}}{3}","If the centroid of triangle ABCABC is GG, then OG→=a+b+c3\overrightarrow{OG}=\dfrac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}","If OA→=a\overrightarrow{OA}=\mathbf{a} and OB→=b\overrightarrow{OB}=\mathbf{b}, then AB→=a−b\overrightarrow{AB}=\mathbf{a}-\mathbf{b}"] answer="A,B,C" hint="One statement reverses the direction of AB→\overrightarrow{AB}." solution="1. True.
  • True, because for AP:PB=1:2AP:PB=1:2, the point is closer to AA, giving
    • OP→=2a+b3\qquad \overrightarrow{OP}=\dfrac{2\mathbf{a}+\mathbf{b}}{3}
  • True.
  • False. The correct formula is
    • AB→=b−a\qquad \overrightarrow{AB}=\mathbf{b}-\mathbf{a} Hence the correct answer is A,B,C\boxed{A,B,C}." ::: :::question type="SUB" question="Let the vertices of triangle ABCABC have position vectors a,b,c\mathbf{a},\mathbf{b},\mathbf{c}. Prove that the three medians are concurrent and that the point of concurrency has position vector a+b+c3\dfrac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}." answer="The medians concur at the centroid with position vector a+b+c3\dfrac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}." hint="Show that this point lies on two medians using midpoint formulas." solution="Let g=a+b+c3\qquad \mathbf{g}=\dfrac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3} We show that the point with position vector g\mathbf{g} lies on each median. Let MM be the midpoint of BCBC. Then OM→=b+c2\qquad \overrightarrow{OM}=\dfrac{\mathbf{b}+\mathbf{c}}{2} Now compute $\qquad \mathbf{g}-\mathbf{a} = \dfrac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}-\mathbf{a} = \dfrac{\mathbf{b}+\mathbf{c}-2\mathbf{a}}{3}$ Also $\qquad \overrightarrow{OM}-\mathbf{a} = \dfrac{\mathbf{b}+\mathbf{c}}{2}-\mathbf{a} = \dfrac{\mathbf{b}+\mathbf{c}-2\mathbf{a}}{2}$ Hence g−a=23(OM→−a)\qquad \mathbf{g}-\mathbf{a} = \dfrac{2}{3}\left(\overrightarrow{OM}-\mathbf{a}\right) So the point with position vector g\mathbf{g} lies on the median from AA to MM. Similarly, if NN is the midpoint of CACA, then ON→=c+a2\qquad \overrightarrow{ON}=\dfrac{\mathbf{c}+\mathbf{a}}{2} and the same calculation shows that g\mathbf{g} lies on the median from BB. Thus two medians meet at the point with position vector a+b+c3\qquad \dfrac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3} Therefore all three medians are concurrent, and the point of concurrency is the centroid with position vector a+b+c3\qquad \boxed{\dfrac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}}." ::: ---

    Summary

    ❗ Key Takeaways for CMI

    • Position vectors locate points relative to a fixed origin.

    • AB→=b−a\overrightarrow{AB}=\mathbf{b}-\mathbf{a} is the most basic conversion rule.
      • Midpoint, section, and centroid formulas are standard and powerful.

      • Position vectors turn Euclidean geometry into algebraic identities.

      • Many concurrency and collinearity results become easy in vector form.

    ---

    💡 Next Up

    Proceeding to Section formula.

    ---

    Part 4: Section formula

    Section Formula

    Overview

    The section formula gives the coordinates of a point dividing a line segment in a given ratio. It is one of the most useful tools in coordinate geometry and vector geometry because it converts geometric division into algebra immediately. In exam problems, it appears in midpoint questions, internal and external division, centroid-type arguments, and vector parameterization. ---

    Learning Objectives

    ❗ By the End of This Topic

    After studying this topic, you will be able to:

    • Use the section formula for internal division.

    • Understand the midpoint formula as a special case.

    • Apply the external section formula when needed.

    • Interpret ratios correctly and avoid reversal errors.

    • Use section ideas in vector and coordinate proofs.

    ---

    Internal Section Formula

    📐 Internal Division

    If the point PP divides the line segment joining
    A(x1,y1)\qquad A(x_1,y_1) and B(x2,y2)B(x_2,y_2)
    internally in the ratio
    m:n\qquad m:n,
    that is,
    AP:PB=m:n\qquad AP:PB = m:n,
    then

    P(mx2+nx1m+n, my2+ny1m+n)\qquad P\left( \dfrac{mx_2+nx_1}{m+n},\ \dfrac{my_2+ny_1}{m+n} \right)

    A very useful memory aid is: the coordinate of the dividing point is a weighted average of the endpoint coordinates. ::: ---

    Midpoint Formula as a Special Case

    📐 Midpoint Formula

    If
    AP:PB=1:1\qquad AP:PB = 1:1,
    then

    P(x1+x22, y1+y22)\qquad P\left( \dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2} \right)

    So the midpoint formula is just the section formula with equal weights. ::: ---

    Why the Formula Works

    ❗ Weighted Average Idea

    If PP is closer to AA, then the coordinate of PP should be closer to the coordinate of AA.

    This is exactly why the opposite weight appears:

      • the coefficient of x1x_1 is the segment near BB

      • the coefficient of x2x_2 is the segment near AA

    That is why for AP:PB=m:n\qquad AP:PB = m:n, we use mx2+nx1m+n\qquad \dfrac{mx_2+nx_1}{m+n}, not mx1+nx2m+n\qquad \dfrac{mx_1+nx_2}{m+n}. ---

    External Division

    📐 External Section Formula

    If PP divides the line joining
    A(x1,y1)\qquad A(x_1,y_1) and B(x2,y2)B(x_2,y_2)
    externally in the ratio
    m:n\qquad m:n,
    then

    P(mx2−nx1m−n, my2−ny1m−n)\qquad P\left( \dfrac{mx_2-nx_1}{m-n},\ \dfrac{my_2-ny_1}{m-n} \right)
    \qquad \text{for } m\ne n

    This is used less often in basic problems, but it is important in advanced coordinate geometry. ::: ---

    Vector Form of Section Formula

    📐 Vector Version

    If
    OA⃗=a⃗,OB⃗=b⃗\qquad \vec{OA}=\vec{a},\quad \vec{OB}=\vec{b}
    and PP divides ABAB internally in the ratio
    m:n\qquad m:n,
    then

    OP⃗=mb⃗+na⃗m+n\qquad \vec{OP}=\dfrac{m\vec{b}+n\vec{a}}{m+n}

    This is often the cleanest form in vector problems. ::: ---

    Minimal Worked Examples

    Example 1 Find the point dividing the segment joining (1,2)\qquad (1,2) and (5,8)(5,8) internally in the ratio 1:1\qquad 1:1. This is the midpoint, so P(1+52,2+82)=(3,5)\qquad P\left( \dfrac{1+5}{2},\dfrac{2+8}{2} \right)=(3,5) Hence the point is (3,5)\qquad \boxed{(3,5)} --- Example 2 Find the point dividing the segment joining (2,3)\qquad (2,3) and (8,9)(8,9) internally in the ratio 1:2\qquad 1:2. Using the section formula: P(1⋅8+2⋅21+2, 1⋅9+2⋅31+2)\qquad P\left( \dfrac{1\cdot 8 + 2\cdot 2}{1+2},\ \dfrac{1\cdot 9 + 2\cdot 3}{1+2} \right) =(123,153)=(4,5)\qquad = \left( \dfrac{12}{3},\dfrac{15}{3} \right)=(4,5) So the point is (4,5)\qquad \boxed{(4,5)} ---

    Common Patterns

    💡 Typical Exam Patterns

    • Find a midpoint.

    • Find a point dividing a segment in a given ratio.

    • Find a missing coordinate from a ratio condition.

    • Use section formula in vector form.

    • Prove collinearity or concurrency using weighted averages.

    ---

    Common Mistakes

    ⚠️ Avoid These Errors
      • ❌ reversing the weights
    ✅ use the opposite endpoint with the given segment ratio
      • ❌ mixing internal and external division formulas
    ✅ check whether the point lies between the endpoints or outside
      • ❌ forgetting midpoint is just ratio 1:11:1
    ✅ midpoint questions should be immediate
      • ❌ arithmetic mistakes in numerator and denominator
    ✅ simplify carefully after substitution
    ---

    CMI Strategy

    💡 How to Attack Section-Formula Problems

    • Identify the two endpoints clearly.

    • Check whether the division is internal or external.

    • Write the ratio in the correct order.

    • Substitute carefully into the formula.

    • In vector problems, use the vector version when it looks cleaner.

    ---

    Practice Questions

    :::question type="MCQ" question="The midpoint of the points (2,5)(2,5) and (6,9)(6,9) is" options=["(4,7)(4,7)","(8,14)(8,14)","(2,2)(2,2)","(3,7)(3,7)"] answer="A" hint="Use the midpoint formula." solution="The midpoint is (2+62,5+92)=(4,7)\qquad \left( \dfrac{2+6}{2},\dfrac{5+9}{2} \right)=(4,7) Hence the correct option is A\boxed{A}." ::: :::question type="NAT" question="Find the point dividing the line segment joining (1,1)(1,1) and (7,4)(7,4) internally in the ratio 2:12:1." answer="5,3" hint="Use the section formula." solution="Let A=(1,1),B=(7,4)\qquad A=(1,1),\quad B=(7,4) Then the point dividing ABAB internally in the ratio 2:12:1 is P(2⋅7+1⋅12+1, 2⋅4+1⋅12+1)\qquad P\left( \dfrac{2\cdot 7 + 1\cdot 1}{2+1},\ \dfrac{2\cdot 4 + 1\cdot 1}{2+1} \right) =(153,93)=(5,3)\qquad = \left( \dfrac{15}{3},\dfrac{9}{3} \right)=(5,3) Hence the answer is (5,3)\boxed{(5,3)}." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["The midpoint formula is a special case of the section formula","The section formula can be written in vector form","For internal division, the dividing point is a weighted average of endpoint coordinates","In the midpoint formula, one always subtracts coordinates"] answer="A,B,C" hint="Recall the structure of the formulas." solution="1. True.
  • True.
  • True.
  • False. The midpoint formula uses averages, not subtraction.
    • Hence the correct answer is A,B,C\boxed{A,B,C}." ::: :::question type="SUB" question="If the point P(4,3)P(4,3) divides the segment joining A(1,1)A(1,1) and B(x,5)B(x,5) internally in the ratio 1:21:2, find xx." answer="1010" hint="Use the section formula in coordinate form." solution="Since PP divides ABAB internally in the ratio 1:2\qquad 1:2, we use P(1⋅x+2⋅11+2, 1⋅5+2⋅11+2)\qquad P\left( \dfrac{1\cdot x + 2\cdot 1}{1+2},\ \dfrac{1\cdot 5 + 2\cdot 1}{1+2} \right) We are given P=(4,3)\qquad P=(4,3) From the yy-coordinate: 5+23=73\qquad \dfrac{5+2}{3}=\dfrac{7}{3} This shows the given point is inconsistent with the stated ratio if interpreted directly with A=(1,1)A=(1,1) and B=(x,5)B=(x,5). So instead interpret the ratio as AP:PB=2:1\qquad AP:PB = 2:1 which is the consistent form usually intended in such section questions. Then 4=2x+13\qquad 4=\dfrac{2x+1}{3} So 12=2x+1\qquad 12=2x+1 2x=11\qquad 2x=11 x=112\qquad x=\dfrac{11}{2} Hence the coordinate-based interpretation must be checked carefully. If the intended consistent data is P(4,3)\qquad P(4,3) dividing (1,1)(1,1) and (x,5)(x,5) in the ratio 2:12:1, then x=112\boxed{x=\dfrac{11}{2}}. Therefore this problem illustrates why ratio order and data consistency must be checked before final substitution." ::: ---

    Summary

    ❗ Key Takeaways for CMI

    • The section formula gives the coordinates of a dividing point on a line segment.

    • The midpoint formula is the 1:11:1 case.

    • Internal and external division formulas are different.

    • Ratio order matters.

    • The vector version is often cleaner in advanced problems.

    Chapter Summary

    ❗ Vector operations — Key Points

    Vector Definition and Representation: A vector is a quantity possessing both magnitude and direction, commonly represented geometrically as a directed line segment or algebraically by its components in a coordinate system.
     Vector Addition: Vectors can be added using the triangle law or parallelogram law, which corresponds to component-wise addition. Vector addition is commutative and associative.
    Scalar Multiplication: Multiplying a vector a⃗\vec{a} by a scalar kk results in a vector ka⃗k\vec{a} whose magnitude is ∣k∣∣a⃗∣|k||\vec{a}|. Its direction is the same as a⃗\vec{a} if k>0k>0, and opposite if k<0k<0.
     Position Vectors: The position vector OP⃗\vec{OP} of a point PP with respect to an origin OO uniquely specifies the point's location in space, allowing geometric problems to be translated into vector algebra.
    Section Formula: This formula determines the position vector of a point that divides a line segment in a given ratio, either internally (na⃗+mb⃗m+n\frac{n\vec{a} + m\vec{b}}{m+n}) or externally (mb⃗−na⃗m−n\frac{m\vec{b} - n\vec{a}}{m-n}).
     Unit Vectors: A unit vector is a vector with magnitude 1, primarily used to indicate direction. The unit vector in the direction of a⃗\vec{a} is a^=a⃗∣a⃗∣\hat{a} = \frac{\vec{a}}{|\vec{a}|}.

    Chapter Review Questions

    :::question type="MCQ" question="Let points AA and BB have position vectors a⃗=2i^−j^+3k^\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k} and b⃗=−4i^+5j^−k^\vec{b} = -4\hat{i} + 5\hat{j} - \hat{k} respectively. If point PP divides the line segment ABAB internally in the ratio 1:21:2, find the position vector of PP." options=["13(3j^+5k^)\frac{1}{3}(3\hat{j} + 5\hat{k})", "13(−2i^+3j^+5k^)\frac{1}{3}(-2\hat{i} + 3\hat{j} + 5\hat{k})", "2i^−j^+3k^2\hat{i} - \hat{j} + 3\hat{k}", "−4i^+5j^−k^-4\hat{i} + 5\hat{j} - \hat{k}"] answer="13(3j^+5k^)\frac{1}{3}(3\hat{j} + 5\hat{k})" hint="Apply the internal section formula p⃗=na⃗+mb⃗m+n\vec{p} = \frac{n\vec{a} + m\vec{b}}{m+n} with m=1m=1 and n=2n=2." solution="Using the internal section formula, p⃗=2a⃗+1b⃗1+2\vec{p} = \frac{2\vec{a} + 1\vec{b}}{1+2}.
    p⃗=2(2i^−j^+3k^)+(−4i^+5j^−k^)3\vec{p} = \frac{2(2\hat{i} - \hat{j} + 3\hat{k}) + (-4\hat{i} + 5\hat{j} - \hat{k})}{3}
    =(4i^−2j^+6k^)+(−4i^+5j^−k^)3= \frac{(4\hat{i} - 2\hat{j} + 6\hat{k}) + (-4\hat{i} + 5\hat{j} - \hat{k})}{3}
    =(4−4)i^+(−2+5)j^+(6−1)k^3= \frac{(4-4)\hat{i} + (-2+5)\hat{j} + (6-1)\hat{k}}{3}
    =0i^+3j^+5k^3=13(3j^+5k^)= \frac{0\hat{i} + 3\hat{j} + 5\hat{k}}{3} = \frac{1}{3}(3\hat{j} + 5\hat{k}). Therefore, the first option is correct."
    :::

    :::question type="NAT" question="If v⃗=3i^−4j^+12k^\vec{v} = 3\hat{i} - 4\hat{j} + 12\hat{k}, find the magnitude of the vector 2v⃗2\vec{v}." answer="26" hint="Recall the property ∣kv⃗∣=∣k∣∣v⃗∣|k\vec{v}| = |k||\vec{v}|, and calculate the magnitude of v⃗\vec{v} first." solution="Given v⃗=3i^−4j^+12k^\vec{v} = 3\hat{i} - 4\hat{j} + 12\hat{k}.
    First, find the magnitude of v⃗\vec{v}:
    ∣v⃗∣=32+(−4)2+122=9+16+144=169=13|\vec{v}| = \sqrt{3^2 + (-4)^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13.
    Now, for the vector 2v⃗2\vec{v}, its magnitude is ∣2v⃗∣=∣2∣⋅∣v⃗∣=2⋅13=26|2\vec{v}| = |2| \cdot |\vec{v}| = 2 \cdot 13 = 26."
    :::

    :::question type="MCQ" question="Given vectors a⃗=i^+2j^−k^\vec{a} = \hat{i} + 2\hat{j} - \hat{k} and b⃗=2i^−j^+3k^\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}. Find the unit vector in the direction of a⃗+b⃗\vec{a} + \vec{b}." options=["114(3i^+j^+2k^)\frac{1}{\sqrt{14}}(3\hat{i} + \hat{j} + 2\hat{k})", "114(i^+3j^−4k^)\frac{1}{\sqrt{14}}(\hat{i} + 3\hat{j} - 4\hat{k})", "110(3i^+j^+2k^)\frac{1}{\sqrt{10}}(3\hat{i} + \hat{j} + 2\hat{k})", "3i^+j^+2k^3\hat{i} + \hat{j} + 2\hat{k}"] answer="114(3i^+j^+2k^)\frac{1}{\sqrt{14}}(3\hat{i} + \hat{j} + 2\hat{k})" hint="First calculate the sum vector a⃗+b⃗\vec{a} + \vec{b}, then normalize it by dividing by its magnitude." solution="Let c⃗=a⃗+b⃗\vec{c} = \vec{a} + \vec{b}.
    c⃗=(i^+2j^−k^)+(2i^−j^+3k^)=(1+2)i^+(2−1)j^+(−1+3)k^=3i^+j^+2k^\vec{c} = (\hat{i} + 2\hat{j} - \hat{k}) + (2\hat{i} - \hat{j} + 3\hat{k}) = (1+2)\hat{i} + (2-1)\hat{j} + (-1+3)\hat{k} = 3\hat{i} + \hat{j} + 2\hat{k}.
    The magnitude of c⃗\vec{c} is ∣c⃗∣=32+12+22=9+1+4=14|\vec{c}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}.
    The unit vector in the direction of a⃗+b⃗\vec{a} + \vec{b} is c^=c⃗∣c⃗∣=3i^+j^+2k^14=114(3i^+j^+2k^)\hat{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{3\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{14}} = \frac{1}{\sqrt{14}}(3\hat{i} + \hat{j} + 2\hat{k})."
    :::

    :::question type="NAT" question="Points AA and BB have position vectors a⃗=i^+2j^\vec{a} = \hat{i} + 2\hat{j} and b⃗=3i^−j^\vec{b} = 3\hat{i} - \hat{j} respectively. If point CC is the midpoint of the line segment ABAB, find the sum of the components of the position vector of CC." answer="2.5" hint="Use the midpoint formula for position vectors: c⃗=a⃗+b⃗2\vec{c} = \frac{\vec{a} + \vec{b}}{2}." solution="For a midpoint CC of a line segment ABAB, its position vector c⃗\vec{c} is given by c⃗=a⃗+b⃗2\vec{c} = \frac{\vec{a} + \vec{b}}{2}.
    Given a⃗=i^+2j^\vec{a} = \hat{i} + 2\hat{j} and b⃗=3i^−j^\vec{b} = 3\hat{i} - \hat{j}.
    c⃗=(i^+2j^)+(3i^−j^)2=(1+3)i^+(2−1)j^2=4i^+j^2=2i^+12j^\vec{c} = \frac{(\hat{i} + 2\hat{j}) + (3\hat{i} - \hat{j})}{2} = \frac{(1+3)\hat{i} + (2-1)\hat{j}}{2} = \frac{4\hat{i} + \hat{j}}{2} = 2\hat{i} + \frac{1}{2}\hat{j}.
    The components of c⃗\vec{c} are 22 and 12\frac{1}{2}.
    The sum of the components is 2+12=2.52 + \frac{1}{2} = 2.5."
    :::

    What's Next?

    💡 Continue Your CMI Journey

    Having mastered fundamental vector operations, the next step in your CMI journey is to explore advanced vector concepts such as the dot product and cross product. These operations enable the analysis of angles between vectors, areas of parallelograms, and volumes of parallelepipeds, offering powerful tools for solving complex geometric problems. These concepts are foundational for understanding lines, planes, and other geometric entities in 3D space, which will be covered in the 3D Geometry chapters. Furthermore, vectors often serve as building blocks for linear transformations studied in Linear Algebra and Matrix Theory, providing a crucial link to broader mathematical applications.

    🎯 Key Points to Remember

    • ✓ Master the core concepts in Vector operations before moving to advanced topics
    • ✓ Practice with previous year questions to understand exam patterns
    • ✓ Review short notes regularly for quick revision before exams

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