The Hook: Opening a Door - Why Position Matters
Try this experiment: Push a door open from two different positions:
- Push at the edge (far from hinges) - Door opens easily
- Push near the hinges - Much harder to open!
Same force, but different torque. Why?
Torque = Force × Distance × sin(angle)
This “multiplication giving perpendicular vector” is the vector product or cross product.
Real example: When you use a wrench, you instinctively push perpendicular to the handle and far from the bolt. That’s maximizing cross product!
JEE Weightage: Vector product appears in 4-5 questions in JEE Main, 5-6 in JEE Advanced. Essential for rotational motion, magnetism!
The Core Concept
What is Vector Product?
The vector product (or cross product) of two vectors $\vec{a}$ and $\vec{b}$ is defined as:
$$\boxed{\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta \, \hat{n}}$$where:
- θ is the angle between vectors (0° ≤ θ ≤ 180°)
- $\hat{n}$ is unit vector perpendicular to both $\vec{a}$ and $\vec{b}$
Result: A vector perpendicular to both original vectors!
In simple terms: Cross product gives a vector pointing “out” of the plane containing the two vectors.
Why “Cross” Product?
Written with a cross: $\vec{a} \times \vec{b}$ (read as “a cross b”)
Magnitude
$$\boxed{|\vec{a} \times \vec{b}| = ab\sin\theta}$$where $a = |\vec{a}|$ and $b = |\vec{b}|$
Direction: The Right-Hand Rule
To find direction of $\vec{a} \times \vec{b}$:
Steps:
- Point fingers of right hand in direction of $\vec{a}$
- Curl fingers toward $\vec{b}$ (through smaller angle)
- Thumb points in direction of $\vec{a} \times \vec{b}$
Memory Trick: “First vector Curl to second Thumb out” → FCT
Important: Direction follows right-hand rule - this is a convention!
Component Form (Most Useful for JEE!)
If $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$:
$$\boxed{\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}}$$Expanding:
$$\vec{a} \times \vec{b} = (a_2b_3 - a_3b_2)\hat{i} - (a_1b_3 - a_3b_1)\hat{j} + (a_1b_2 - a_2b_1)\hat{k}$$Memory Trick: “Cover i, Cross rest” → Cover column, cross multiply!
Determinant Expansion Shortcut
For $\hat{i}$ component: Cover first column, cross multiply:
$$\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix} = a_2b_3 - a_3b_2$$For $\hat{j}$ component: Cover second column, cross multiply, add minus sign:
$$-\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix} = -(a_1b_3 - a_3b_1)$$For $\hat{k}$ component: Cover third column, cross multiply:
$$\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} = a_1b_2 - a_2b_1$$Find $\vec{a} \times \vec{b}$ where $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$.
Solution:
$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & -1 & 2 \end{vmatrix}$$ $$= \hat{i}(3 \cdot 2 - 1 \cdot (-1)) - \hat{j}(2 \cdot 2 - 1 \cdot 1) + \hat{k}(2 \cdot (-1) - 3 \cdot 1)$$ $$= \hat{i}(6 + 1) - \hat{j}(4 - 1) + \hat{k}(-2 - 3)$$ $$= 7\hat{i} - 3\hat{j} - 5\hat{k}$$Properties of Vector Product
1. Anti-Commutative Law
$$\boxed{\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})}$$Meaning: Order matters! Reversing gives opposite direction.
Why? Right-hand rule reverses when you swap vectors.
2. Distributive Law
$$\boxed{\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}}$$Meaning: Can distribute cross product over addition.
3. Scalar Multiplication
$$\boxed{(m\vec{a}) \times \vec{b} = m(\vec{a} \times \vec{b}) = \vec{a} \times (m\vec{b})}$$where m is a scalar.
4. Not Associative
$$\boxed{(\vec{a} \times \vec{b}) \times \vec{c} \neq \vec{a} \times (\vec{b} \times \vec{c})}$$Important: Grouping matters in cross product!
5. Cross Product with Itself
$$\boxed{\vec{a} \times \vec{a} = \vec{0}}$$Why? θ = 0° → sin(0°) = 0
Cross Product of Unit Vectors
Following cyclic order (i → j → k → i):
$$\hat{i} \times \hat{j} = \hat{k}$$ $$\hat{j} \times \hat{k} = \hat{i}$$ $$\hat{k} \times \hat{i} = \hat{j}$$Reverse cyclic order (add minus):
$$\hat{j} \times \hat{i} = -\hat{k}$$ $$\hat{k} \times \hat{j} = -\hat{i}$$ $$\hat{i} \times \hat{k} = -\hat{j}$$Cross with itself:
$$\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = \vec{0}$$Memory Trick: Draw circle: i → j → k → i
- Clockwise: Positive
- Counter-clockwise: Negative
j
↑
|
i ← + → k
Following arrows (i to j to k): Positive Against arrows: Negative
Geometric Interpretations
1. Area of Parallelogram
Area of parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$:
$$\boxed{\text{Area} = |\vec{a} \times \vec{b}| = ab\sin\theta}$$This is the magnitude of cross product!
Why? Area of parallelogram = base × height = $b \times (a\sin\theta)$
2. Area of Triangle
$$\boxed{\text{Area of triangle} = \frac{1}{2}|\vec{a} \times \vec{b}|}$$Triangle is half of parallelogram!
Find area of triangle with vertices A(1, 2, 3), B(2, 3, 1), C(3, 1, 2).
Solution:
$$\overrightarrow{AB} = (2-1)\hat{i} + (3-2)\hat{j} + (1-3)\hat{k} = \hat{i} + \hat{j} - 2\hat{k}$$ $$\overrightarrow{AC} = (3-1)\hat{i} + (1-2)\hat{j} + (2-3)\hat{k} = 2\hat{i} - \hat{j} - \hat{k}$$ $$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \\ 2 & -1 & -1 \end{vmatrix}$$ $$= \hat{i}(-1-2) - \hat{j}(-1+4) + \hat{k}(-1-2)$$ $$= -3\hat{i} - 3\hat{j} - 3\hat{k}$$ $$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{9+9+9} = 3\sqrt{3}$$Area = $\frac{1}{2} \times 3\sqrt{3} = \frac{3\sqrt{3}}{2}$ square units
Perpendicular and Parallel Vectors
Parallel Vectors (Collinear)
$$\boxed{\vec{a} \parallel \vec{b} \iff \vec{a} \times \vec{b} = \vec{0}}$$Why? θ = 0° or 180° → sin θ = 0
To check if vectors parallel: Calculate cross product!
- If $\vec{a} \times \vec{b} = \vec{0}$ → Parallel ✓
- If $\vec{a} \times \vec{b} \neq \vec{0}$ → Not parallel ✗
Finding Perpendicular Vector
Need vector perpendicular to both $\vec{a}$ and $\vec{b}$?
$$\boxed{\vec{n} = \vec{a} \times \vec{b}}$$This is perpendicular to BOTH!
Find vector perpendicular to both $\vec{a} = \hat{i} + 2\hat{j}$ and $\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$.
Solution:
$$\vec{n} = \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 0 \\ 2 & -1 & 3 \end{vmatrix}$$ $$= \hat{i}(6-0) - \hat{j}(3-0) + \hat{k}(-1-4)$$ $$= 6\hat{i} - 3\hat{j} - 5\hat{k}$$Verify: Check $\vec{n} \cdot \vec{a} = 0$ and $\vec{n} \cdot \vec{b} = 0$ ✓
Physics Applications
1. Torque (Rotational Force)
where:
- $\vec{r}$ = position vector from pivot to point of application
- $\vec{F}$ = applied force
- $\vec{\tau}$ = torque (turning effect)
Magnitude: $\tau = rF\sin\theta$
See: Rotational Dynamics
Maximum torque: When force is perpendicular to position vector (θ = 90°)
Interactive Demo: Vector Operations Visualizer
Visualize cross product, parallelogram area, and the perpendicular result vector. Switch to “A x B” tab and try 3D mode for full visualization.
2. Angular Momentum
$$\boxed{\vec{L} = \vec{r} \times \vec{p}}$$where $\vec{p} = m\vec{v}$ is linear momentum.
3. Magnetic Force on Moving Charge
where:
- q = charge
- $\vec{v}$ = velocity
- $\vec{B}$ = magnetic field
Force is perpendicular to both velocity and magnetic field!
See: Magnetic Force
4. Moment of Force
For multiple forces creating rotation, net torque:
$$\vec{\tau}_{net} = \sum (\vec{r}_i \times \vec{F}_i)$$Memory Tricks & Patterns
Mnemonic for Cross Product
Memory Trick: “Cross gives Cross-wise vector” → Result is perpendicular
Memory Trick for Sign: “i → j → k → i” (cycle forward = positive)
Pattern Recognition
If problem says “parallel” or “collinear”:
- Immediately write: $\vec{a} \times \vec{b} = \vec{0}$
- Expand determinant, set components = 0
Appears in: 40% of vector parallel/collinear problems
See $\hat{i}, \hat{j}, \hat{k}$ in calculation? Use cyclic order directly!
Example: $(\hat{i}+2\hat{j}) \times (3\hat{i}-\hat{k})$
$$= 3(\hat{i} \times \hat{i}) - (\hat{i} \times \hat{k}) + 6(\hat{j} \times \hat{i}) - 2(\hat{j} \times \hat{k})$$ $$= 0 + \hat{j} - 6\hat{k} - 2\hat{i}$$ $$= -2\hat{i} + \hat{j} - 6\hat{k}$$Need area of parallelogram/triangle with sides as vectors?
- Directly use $|\vec{a} \times \vec{b}|$ (parallelogram)
- Or $\frac{1}{2}|\vec{a} \times \vec{b}|$ (triangle)
No need for base-height formula!
When to Use Cross Product
Use cross product when:
- Finding area of parallelogram/triangle
- Need vector perpendicular to two given vectors
- Calculating torque (τ = r × F)
- Finding angular momentum (L = r × p)
- Solving magnetic force problems (F = qv × B)
- Checking if vectors are parallel (cross product = 0)
Use dot product when:
- Finding angle between vectors → See Scalar Product
- Checking perpendicularity (dot product = 0)
- Calculating work done
Common Mistakes to Avoid
Wrong: $\vec{a} \times \vec{b} = \vec{b} \times \vec{a}$
Right: $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$ (opposite directions!)
JEE loves to test this in MCQs!
Wrong: $\vec{a} \times \vec{b} = (a_2b_3 - a_3b_2)\hat{i} + (a_1b_3 - a_3b_1)\hat{j} + ...$
Right: $\vec{a} \times \vec{b} = (a_2b_3 - a_3b_2)\hat{i} - (a_1b_3 - a_3b_1)\hat{j} + ...$
The $\hat{j}$ term has MINUS sign!
Wrong: $(\vec{a} \times \vec{b}) \times \vec{c} = \vec{a} \times (\vec{b} \times \vec{c})$
Right: These are generally different! Parentheses matter.
Use: Triple product formulas if needed → See Triple Products
Question: “Find cross product of $\vec{a}$ and $\vec{b}$”
Wrong: Writing just the magnitude $ab\sin\theta$
Right: Must include direction! Either:
- Full vector form: $(a_2b_3-a_3b_2)\hat{i} - ...$
- Or magnitude + direction: $ab\sin\theta \, \hat{n}$
Wrong: $\hat{i} \times \hat{k} = \hat{j}$ (wrong direction!)
Right: $\hat{i} \times \hat{k} = -\hat{j}$ (against cycle)
Remember: i → j → k → i (forward = positive)
Practice Problems
Level 1: Foundation (NCERT Style)
Find $\vec{a} \times \vec{b}$ where $\vec{a} = 2\hat{i} + \hat{j}$ and $\vec{b} = \hat{i} + 3\hat{k}$.
Solution:
$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 0 \\ 1 & 0 & 3 \end{vmatrix}$$ $$= \hat{i}(3-0) - \hat{j}(6-0) + \hat{k}(0-1)$$ $$= 3\hat{i} - 6\hat{j} - \hat{k}$$Calculate $(\hat{i} + \hat{j}) \times (\hat{j} + \hat{k})$.
Solution:
$$= \hat{i} \times \hat{j} + \hat{i} \times \hat{k} + \hat{j} \times \hat{j} + \hat{j} \times \hat{k}$$ $$= \hat{k} - \hat{j} + \vec{0} + \hat{i}$$ $$= \hat{i} - \hat{j} + \hat{k}$$Find area of parallelogram with adjacent sides $\vec{a} = \hat{i} + 2\hat{j}$ and $\vec{b} = 2\hat{i} + \hat{j}$.
Solution:
$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 0 \\ 2 & 1 & 0 \end{vmatrix} = \hat{k}(1-4) = -3\hat{k}$$Area = $|\vec{a} \times \vec{b}| = |-3| = 3$ square units
Level 2: JEE Main Type
Find unit vector perpendicular to both $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$.
Solution:
$$\vec{n} = \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{vmatrix}$$ $$= \hat{i}(2+1) - \hat{j}(4-1) + \hat{k}(-2-1)$$ $$= 3\hat{i} - 3\hat{j} - 3\hat{k}$$Magnitude: $|\vec{n}| = \sqrt{9+9+9} = 3\sqrt{3}$
Unit vector: $\hat{n} = \frac{3\hat{i}-3\hat{j}-3\hat{k}}{3\sqrt{3}} = \frac{\hat{i}-\hat{j}-\hat{k}}{\sqrt{3}}$
Find area of triangle with vertices A(1, 0, 0), B(0, 1, 0), C(0, 0, 1).
Solution:
$$\overrightarrow{AB} = -\hat{i} + \hat{j}$$ $$\overrightarrow{AC} = -\hat{i} + \hat{k}$$ $$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix}$$ $$= \hat{i}(1-0) - \hat{j}(-1-0) + \hat{k}(0+1)$$ $$= \hat{i} + \hat{j} + \hat{k}$$Magnitude: $\sqrt{1+1+1} = \sqrt{3}$
Area = $\frac{1}{2}\sqrt{3} = \frac{\sqrt{3}}{2}$ square units
Check if vectors $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{b} = 4\hat{i} + 6\hat{j} + 2\hat{k}$ are parallel.
Solution:
$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 4 & 6 & 2 \end{vmatrix}$$ $$= \hat{i}(6-6) - \hat{j}(4-4) + \hat{k}(12-12)$$ $$= \vec{0}$$Since $\vec{a} \times \vec{b} = \vec{0}$, vectors are parallel ✓
(Also observe: $\vec{b} = 2\vec{a}$)
Force $\vec{F} = 3\hat{i} + 4\hat{j}$ N acts at position $\vec{r} = 2\hat{i} - \hat{j}$ m from pivot. Find torque.
Solution:
$$\vec{\tau} = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 0 \\ 3 & 4 & 0 \end{vmatrix}$$ $$= \hat{k}(8 + 3) = 11\hat{k} \text{ N·m}$$Magnitude: 11 N·m (perpendicular to plane, outward)
Level 3: JEE Advanced Type
If $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$, find $\vec{c}$ such that $\vec{a} \times \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = 3$.
Solution: Let $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$
From $\vec{a} \times \vec{c} = \vec{b}$:
$$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ x & y & z \end{vmatrix} = \hat{i} + 2\hat{j} - \hat{k}$$Comparing components:
- $(z-y)\hat{i} = \hat{i}$ → $z - y = 1$ … (1)
- $(x-2z)\hat{j} = 2\hat{j}$ → $x - 2z = 2$ … (2)
- $(2y-x)\hat{k} = -\hat{k}$ → $2y - x = -1$ … (3)
From $\vec{a} \cdot \vec{c} = 3$:
$$2x + y + z = 3$$… (4)
Solving: From (2) and (3): $x = 2z + 2$ and $2y = x - 1 = 2z + 1$ So $y = z + \frac{1}{2}$
From (1): $z - (z + \frac{1}{2}) = 1$ → $-\frac{1}{2} = 1$ (contradiction!)
This system has no solution. Such $\vec{c}$ doesn’t exist!
Find sin θ where θ is angle between $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}$.
Solution:
$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 1 & -2 \end{vmatrix}$$ $$= \hat{i}(-4-2) - \hat{j}(-2-4) + \hat{k}(1-4)$$ $$= -6\hat{i} + 6\hat{j} - 3\hat{k}$$ $$|\vec{a} \times \vec{b}| = \sqrt{36+36+9} = \sqrt{81} = 9$$ $$|\vec{a}| = \sqrt{1+4+4} = 3, \quad |\vec{b}| = \sqrt{4+1+4} = 3$$ $$\sin\theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|} = \frac{9}{3 \times 3} = 1$$ $$\theta = 90°$$(vectors are perpendicular!)
Find volume of parallelepiped with adjacent edges $\vec{a} = \hat{i}+\hat{j}$, $\vec{b} = \hat{j}+\hat{k}$, $\vec{c} = \hat{k}+\hat{i}$.
Solution: Volume = $|\vec{a} \cdot (\vec{b} \times \vec{c})|$ (scalar triple product)
$$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix} = \hat{i}(1-0) - \hat{j}(0-1) + \hat{k}(0-1)$$ $$= \hat{i} + \hat{j} - \hat{k}$$ $$\vec{a} \cdot (\vec{b} \times \vec{c}) = (1)(1) + (1)(1) + (0)(-1) = 2$$Volume = $|2| = 2$ cubic units
(See Triple Products for more on this!)
Connection to Physics
Critical Applications:
Torque: $\vec{\tau} = \vec{r} \times \vec{F}$
- See: Rotational Dynamics
Angular Momentum: $\vec{L} = \vec{r} \times \vec{p}$
- See: Angular Momentum
Magnetic Force on Charge: $\vec{F} = q\vec{v} \times \vec{B}$
- See: Magnetic Force
Magnetic Force on Current: $\vec{F} = I\vec{L} \times \vec{B}$
Angular Velocity: $\vec{v} = \vec{\omega} \times \vec{r}$
- See: Circular Motion
Master cross product → Solve 90% of rotational mechanics & all magnetism problems!
Quick Revision Box
| Concept | Formula |
|---|---|
| Definition | $\vec{a} \times \vec{b} = ab\sin\theta \, \hat{n}$ |
| Magnitude | $ |
| Component form | $(a_2b_3-a_3b_2)\hat{i} - (a_1b_3-a_3b_1)\hat{j} + (a_1b_2-a_2b_1)\hat{k}$ |
| Order matters | $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$ |
| Parallel condition | $\vec{a} \times \vec{b} = \vec{0}$ |
| Unit vectors | $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$ |
| Area of parallelogram | $ |
| Area of triangle | $\frac{1}{2} |
| Torque | $\vec{\tau} = \vec{r} \times \vec{F}$ |
| Perpendicular vector | $\vec{n} = \vec{a} \times \vec{b}$ |
What’s Next?
Now combine dot and cross products:
Triple Products - Scalar triple product (volume), vector triple product (important identities)
Also explore:
- Rotational Dynamics - Uses torque = r × F extensively
- Magnetic Effects - F = qv × B is fundamental
- 3D Geometry - Equation of plane using cross product
Teacher’s Summary
- Cross product gives perpendicular vector - that’s its superpower!
- Order matters: $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$ - don’t forget minus!
- Determinant method is fastest - master the 3×3 expansion
- Middle term has minus sign - most common calculation error!
- Area = |a × b| - instant area of parallelogram/triangle
- Parallel ⟺ cross product = 0 - instant collinearity check
“In JEE, see ‘area’? Think cross product magnitude. See ’torque’? Think r × F. See ‘perpendicular vector’? Calculate cross product. Master these, clear rotational mechanics & magnetism!”
Time-saving tip: Use cyclic order for unit vectors - don’t expand determinant!
Pro Tip: In physics problems, if answer asks for “magnitude of torque/angular momentum/magnetic force”, you only need $|\vec{a} \times \vec{b}| = ab\sin\theta$ - no need to find full vector! Saves 1 minute!