Vector Basics: Definition, Types & Position Vectors

Master vector fundamentals with movie examples - understand scalars vs vectors, position vectors, and types with Iron Man flight trajectories

9 min read

The Hook: Iron Man’s Flight Path

Connect: Real Life → Vectors

When Tony Stark (Iron Man) flies from Stark Tower to the Avengers compound, saying “I flew 500 km” isn’t enough information. JARVIS needs:

  • Distance: 500 km (how far)
  • Direction: Northeast bearing 45°

“500 km” alone is a scalar. “500 km Northeast” is a vector.

The Question: Why does physics (and JEE) care so much about direction?

JEE Weightage: Vector Basics appears in 2-3 questions in JEE Main, 3-4 in JEE Advanced. Foundation topic for mechanics!

The Core Concept

What is a Vector?

Definition

A vector is a quantity that has both:

  1. Magnitude (size/length)
  2. Direction (orientation in space)

A scalar has only magnitude, no direction.

In simple terms: If you can answer “which way?” about it, it’s a vector. If you can’t, it’s a scalar.

Scalars vs Vectors

TypeExamplesJEE Appearance
ScalarsDistance, Speed, Mass, Temperature, EnergyWork, Power, Kinetic Energy
VectorsDisplacement, Velocity, Force, AccelerationAll mechanics problems

Vector Notation

$$\boxed{\vec{a} \text{ or } \mathbf{a} \text{ or } \overrightarrow{AB}}$$
  • $\vec{a}$ - General vector notation
  • $|\vec{a}|$ or $a$ - Magnitude of vector
  • $\hat{a}$ - Unit vector (magnitude = 1)
  • $\overrightarrow{AB}$ - Vector from point A to B

Types of Vectors

1. Zero Vector (Null Vector)

$$\boxed{\vec{0} = 0\hat{i} + 0\hat{j} + 0\hat{k}}$$

Properties:

  • Magnitude = 0
  • Direction is undefined (no specific direction)
  • Adding zero vector to any vector leaves it unchanged: $\vec{a} + \vec{0} = \vec{a}$

JEE Trap: Zero vector is the only vector whose direction cannot be determined!

2. Unit Vector

$$\boxed{\hat{a} = \frac{\vec{a}}{|\vec{a}|}}$$

Definition: A vector with magnitude 1 in the direction of $\vec{a}$.

Standard Unit Vectors:

  • $\hat{i}$ - along positive x-axis
  • $\hat{j}$ - along positive y-axis
  • $\hat{k}$ - along positive z-axis

Properties: $|\hat{i}| = |\hat{j}| = |\hat{k}| = 1$

Quick Example

If $\vec{a} = 3\hat{i} + 4\hat{j}$, find the unit vector.

Solution:

  • Magnitude: $|\vec{a}| = \sqrt{3^2 + 4^2} = \sqrt{9+16} = 5$
  • Unit vector: $\hat{a} = \frac{3\hat{i} + 4\hat{j}}{5} = \frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}$

3. Equal Vectors

$$\boxed{\vec{a} = \vec{b} \iff |\vec{a}| = |\vec{b}| \text{ and same direction}}$$

Important: Vectors can be equal even if they start at different points - only magnitude and direction matter!

4. Negative Vector

$$\boxed{-\vec{a} \text{ has same magnitude but opposite direction to } \vec{a}}$$

Example: If $\vec{a} = 3\hat{i} + 4\hat{j}$, then $-\vec{a} = -3\hat{i} - 4\hat{j}$

5. Parallel Vectors (Collinear)

$$\boxed{\vec{a} \parallel \vec{b} \iff \vec{a} = \lambda\vec{b} \text{ for some scalar } \lambda}$$

Cases:

  • If $\lambda > 0$: Same direction
  • If $\lambda < 0$: Opposite direction

6. Position Vector

Most Important for JEE!

The position vector of a point P(x, y, z) with respect to origin O is:

$$\boxed{\overrightarrow{OP} = x\hat{i} + y\hat{j} + z\hat{k}}$$

Simply: It’s the vector from origin to the point.

Position Vectors: The Foundation

Position Vector of a Point

For point P(x, y, z):

$$\vec{r} = \overrightarrow{OP} = x\hat{i} + y\hat{j} + z\hat{k}$$

Magnitude: $|\vec{r}| = \sqrt{x^2 + y^2 + z^2}$ (distance from origin)

Vector Joining Two Points

Super Important Formula

Vector from point A$(x_1, y_1, z_1)$ to B$(x_2, y_2, z_2)$:

$$\boxed{\overrightarrow{AB} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}}$$

Memory Trick: “Tail to Head” → Subtract tail coordinates from head coordinates

Length of AB:

$$|\overrightarrow{AB}| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$

This is just the distance formula!

Interactive Demo: Visualize Vector Addition

Explore how vectors add geometrically using the interactive visualization below.

Example: Find Vector AB

A(1, 2, 3) and B(4, 6, 9). Find $\overrightarrow{AB}$ and its magnitude.

Solution:

$$\overrightarrow{AB} = (4-1)\hat{i} + (6-2)\hat{j} + (9-3)\hat{k} = 3\hat{i} + 4\hat{j} + 6\hat{k}$$ $$|\overrightarrow{AB}| = \sqrt{3^2 + 4^2 + 6^2} = \sqrt{9+16+36} = \sqrt{61}$$

Component Form of Vectors

2D Vector

$$\vec{a} = a_x\hat{i} + a_y\hat{j}$$

Magnitude: $|\vec{a}| = \sqrt{a_x^2 + a_y^2}$

Direction: $\tan\theta = \frac{a_y}{a_x}$ (angle with x-axis)

3D Vector

$$\vec{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}$$

Magnitude: $|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}$

Direction Cosines:

$$\boxed{l = \frac{a_x}{|\vec{a}|}, \quad m = \frac{a_y}{|\vec{a}|}, \quad n = \frac{a_z}{|\vec{a}|}}$$

Important Property: $l^2 + m^2 + n^2 = 1$

Memory Tricks & Patterns

Mnemonic for Types of Vectors

Memory Trick: “Zero Unit Equal Negative Parallel Position” → ZUENPP

  1. Zero vector - magnitude 0
  2. Unit vector - magnitude 1
  3. Equal vectors - same magnitude & direction
  4. Negative vector - opposite direction
  5. Parallel vectors - $\vec{a} = \lambda\vec{b}$
  6. Position vector - from origin

Pattern Recognition

JEE Pattern #1: 3-4-5 Triangle

Vector $3\hat{i} + 4\hat{j}$ appears frequently because:

  • Magnitude = 5 (Pythagorean triplet)
  • Easy to work with in calculations
  • Unit vector is $\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}$

Other common triplets: (5, 12, 13), (8, 15, 17), (7, 24, 25)

JEE Pattern #2: Special Angles

Vectors at 30°, 45°, 60° with x-axis appear often:

  • 45°: Components are equal, $\vec{a} = a\hat{i} + a\hat{j}$
  • 30°: $\vec{a} = a\sqrt{3}\hat{i} + a\hat{j}$ (magnitude 2a)
  • 60°: $\vec{a} = a\hat{i} + a\sqrt{3}\hat{j}$ (magnitude 2a)

When to Use Position Vectors

Decision Tree

Use position vectors when:

  • Problem gives coordinates of points
  • Need to find distance between points
  • Working with centroids, midpoints
  • Proving collinearity of points

Use general vectors when:

  • Forces, velocities given as magnitudes
  • Direction is given as angles
  • Abstract vector algebra problems

Common Mistakes to Avoid

Trap #1: Vector vs Magnitude Confusion

Wrong: $\vec{a} + \vec{b} = |\vec{a}| + |\vec{b}|$

Right: Vectors add by components, not magnitudes!

If $\vec{a} = 3\hat{i}$ and $\vec{b} = 4\hat{j}$:

  • $\vec{a} + \vec{b} = 3\hat{i} + 4\hat{j}$ (magnitude = 5, NOT 7!)
Trap #2: Direction of AB vs BA

Common Error: Thinking $\overrightarrow{AB} = \overrightarrow{BA}$

Truth: $\overrightarrow{AB} = -\overrightarrow{BA}$ (opposite directions)

Always: Final point - Initial point

Trap #3: Zero Vector Direction

Question Type: “What is the direction of zero vector?”

Wrong Answer: “All directions” or “No direction”

Correct: Direction is undefined (not determinable)

Trap #4: Unit Vector Calculation

Wrong: Unit vector of $\vec{a} = 3\hat{i} + 4\hat{j}$ is $\hat{i} + \hat{j}$

Right: First find magnitude (5), then divide: $\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}$

Practice Problems

Level 1: Foundation (NCERT Style)

Problem 1: Basic Position Vector

Find the position vector of point P(2, -3, 5).

Solution:

$$\vec{r} = 2\hat{i} - 3\hat{j} + 5\hat{k}$$

Magnitude: $|\vec{r}| = \sqrt{4 + 9 + 25} = \sqrt{38}$

Problem 2: Unit Vector

Find unit vector in the direction of $\vec{a} = 2\hat{i} - 2\hat{j} + \hat{k}$.

Solution:

  • $|\vec{a}| = \sqrt{4 + 4 + 1} = 3$
  • $\hat{a} = \frac{2\hat{i} - 2\hat{j} + \hat{k}}{3} = \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} + \frac{1}{3}\hat{k}$
Problem 3: Vector Between Points

Find $\overrightarrow{AB}$ where A(1, 2, 3) and B(3, 4, 5).

Solution:

$$\overrightarrow{AB} = (3-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = 2\hat{i} + 2\hat{j} + 2\hat{k}$$

Length: $|\overrightarrow{AB}| = \sqrt{4+4+4} = 2\sqrt{3}$

Level 2: JEE Main Type

Problem 4: Collinear Vectors

If $\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b} = 4\hat{i} + 6\hat{j} + 8\hat{k}$, show that they are parallel.

Solution:

$$\vec{b} = 4\hat{i} + 6\hat{j} + 8\hat{k} = 2(2\hat{i} + 3\hat{j} + 4\hat{k}) = 2\vec{a}$$

Since $\vec{b} = 2\vec{a}$ (where $\lambda = 2$), vectors are parallel. ✓

Problem 5: Direction Cosines

Find direction cosines of $\vec{a} = 6\hat{i} - 3\hat{j} + 2\hat{k}$.

Solution:

  • Magnitude: $|\vec{a}| = \sqrt{36 + 9 + 4} = 7$
  • Direction cosines:
    • $l = \frac{6}{7}$
    • $m = \frac{-3}{7}$
    • $n = \frac{2}{7}$

Verify: $l^2 + m^2 + n^2 = \frac{36 + 9 + 4}{49} = 1$ ✓

Problem 6: Finding Point from Position Vector

A point P has position vector $\vec{r} = 3\hat{i} - 2\hat{j} + 5\hat{k}$. Find coordinates of P and distance from origin.

Solution:

  • Coordinates: P(3, -2, 5)
  • Distance: $|\vec{r}| = \sqrt{9 + 4 + 25} = \sqrt{38}$

Level 3: JEE Advanced Type

Problem 7: Vector Equality Condition

If $\vec{a} = (x+3)\hat{i} + 2\hat{j} + z\hat{k}$ and $\vec{b} = \hat{i} + y\hat{j} - 2\hat{k}$ are equal vectors, find x, y, z.

Solution: For equal vectors, all components must be equal:

  • $x + 3 = 1 \Rightarrow x = -2$
  • $2 = y \Rightarrow y = 2$
  • $z = -2$

Answer: $x = -2, y = 2, z = -2$

Problem 8: Midpoint Using Position Vectors

Points A and B have position vectors $\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b} = 4\hat{i} + \hat{j} + 2\hat{k}$. Find position vector of midpoint M.

Solution:

$$\vec{r}_M = \frac{\vec{a} + \vec{b}}{2} = \frac{(2\hat{i}+3\hat{j}+4\hat{k}) + (4\hat{i}+\hat{j}+2\hat{k})}{2}$$ $$= \frac{6\hat{i} + 4\hat{j} + 6\hat{k}}{2} = 3\hat{i} + 2\hat{j} + 3\hat{k}$$

Midpoint: M(3, 2, 3)

Problem 9: Vector Magnitude Condition

Find values of k such that $|\vec{a}| = 5$ where $\vec{a} = k\hat{i} + 3\hat{j} + 4\hat{k}$.

Solution:

$$|\vec{a}| = \sqrt{k^2 + 9 + 16} = 5$$ $$k^2 + 25 = 25$$ $$k^2 = 0 \Rightarrow k = 0$$

Answer: k = 0

Connection to Physics

Why Vectors Matter in JEE Physics

Every mechanics problem uses vectors:

  1. Kinematics: Displacement, velocity, acceleration are ALL vectors

  2. Forces: Newton’s laws work with vector addition

  3. Work & Energy: Work = Force · Displacement (dot product!)

  4. Rotational Motion: Torque, angular momentum are cross products

Master vectors → Excel in Physics!

Quick Revision Box

ConceptFormula/Key Point
Vector notation$\vec{a}$ or $\overrightarrow{AB}$
Magnitude$
Unit vector$\hat{a} = \frac{\vec{a}}{
Position vectorP(x,y,z) → $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$
Vector AB$\overrightarrow{AB} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$
Parallel vectors$\vec{a} = \lambda\vec{b}$
Direction cosines$l = \frac{a_x}{
Property$l^2 + m^2 + n^2 = 1$
Zero vectorMagnitude = 0, direction undefined

What’s Next?

Now that you understand vector basics, learn how to add and subtract vectors:

Next Topic

Vector Addition & Subtraction - Triangle law, parallelogram law, component method

Also explore:

Teacher’s Summary

Key Takeaways
  1. Vectors have magnitude AND direction - that’s what makes them powerful
  2. Position vectors are the bridge between coordinate geometry and vectors
  3. Unit vectors are your friends - they simplify direction calculations
  4. The formula $\overrightarrow{AB} = B - A$ is used in 80% of vector problems
  5. Direction undefined vs no direction - zero vector direction is undefined!

“In JEE, if you see coordinates, think position vectors. If you see forces/velocities, think component vectors. Master both, ace mechanics!”

Time-saving tip: For 3-4-5, 5-12-13 vectors, write magnitude directly. Saves 30 seconds!

Pro Tip: When solving vector problems, always draw a diagram first. 60% of mistakes happen because students skip visualization!