The Hook: Building a Room - Volume from Vectors
Imagine you’re designing a room in a video game like Minecraft or Fortnite. You’re given three edges:
- Edge 1: 3 blocks East
- Edge 2: 4 blocks North
- Edge 3: 5 blocks Up
Question: What’s the volume of this room?
You could multiply 3 × 4 × 5 = 60 cubic blocks if edges are perpendicular. But what if the room is slanted (a parallelepiped)?
Answer: Use Scalar Triple Product - it gives volume for ANY three edges!
$$\text{Volume} = |\vec{a} \cdot (\vec{b} \times \vec{c})|$$This combines both dot and cross products - the ultimate vector operation!
JEE Weightage: Triple products appear in 2-3 questions in JEE Main, 3-4 in JEE Advanced. High-difficulty, high-reward topic!
The Core Concept
What are Triple Products?
When you combine three vectors, you get:
- Scalar Triple Product: $\vec{a} \cdot (\vec{b} \times \vec{c})$ → gives a scalar (volume)
- Vector Triple Product: $\vec{a} \times (\vec{b} \times \vec{c})$ → gives a vector
Key Difference:
- Scalar triple = dot of (cross) → scalar
- Vector triple = cross of (cross) → vector
Part 1: Scalar Triple Product
Definition
Also written as: $[\vec{a} \, \vec{b} \, \vec{c}]$ or $(\vec{a}, \vec{b}, \vec{c})$
Result: A scalar (number)
Geometric Meaning: Volume of parallelepiped with edges $\vec{a}, \vec{b}, \vec{c}$
Interactive Demo: Triple Product Visualization
Visualize how three vectors form a parallelepiped and compute volume.
Formula Using Determinant
If $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$:
$$\boxed{[\vec{a} \, \vec{b} \, \vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}}$$This is the fastest method for JEE!
Memory Trick: “Row-wise write components” → First vector in first row, etc.
Find $[\vec{a} \, \vec{b} \, \vec{c}]$ where $\vec{a} = \hat{i}+\hat{j}+\hat{k}$, $\vec{b} = 2\hat{i}+3\hat{j}+\hat{k}$, $\vec{c} = \hat{i}+\hat{k}$.
Solution:
$$[\vec{a} \, \vec{b} \, \vec{c}] = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 1 \\ 1 & 0 & 1 \end{vmatrix}$$Expanding along first row:
$$= 1(3-0) - 1(2-1) + 1(0-3)$$ $$= 3 - 1 - 3 = -1$$Geometric Interpretation: Volume
Take absolute value because volume is positive!
Parallelepiped: 3D figure with 6 parallelogram faces (like a slanted box)
Volume of Tetrahedron:
$$\boxed{\text{Volume} = \frac{1}{6}|[\vec{a} \, \vec{b} \, \vec{c}]|}$$Tetrahedron is 1/6 of parallelepiped (like pyramid vs rectangular prism).
Find volume of parallelepiped with edges $\vec{a} = 2\hat{i}+3\hat{j}$, $\vec{b} = \hat{i}+\hat{j}+\hat{k}$, $\vec{c} = 3\hat{i}+\hat{j}+2\hat{k}$.
Solution:
$$V = \left|\begin{vmatrix} 2 & 3 & 0 \\ 1 & 1 & 1 \\ 3 & 1 & 2 \end{vmatrix}\right|$$Expanding:
$$= |2(2-1) - 3(2-3) + 0|$$ $$= |2(1) - 3(-1)|$$ $$= |2 + 3| = 5 \text{ cubic units}$$Properties of Scalar Triple Product
1. Cyclic Property
Cyclic rotation doesn’t change value!
Memory Trick: “Cycle Keeps Value” → CKV
Why? Volume of room doesn’t depend on which edge you call “first”!
2. Anti-Cyclic Property
$$\boxed{[\vec{a} \, \vec{b} \, \vec{c}] = -[\vec{b} \, \vec{a} \, \vec{c}]}$$Swapping any two vectors changes sign (like determinants).
3. Dot and Cross are Interchangeable
$$\boxed{\vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c}}$$Amazing property: Can move the cross product around!
4. Zero Scalar Triple Product
$$\boxed{[\vec{a} \, \vec{b} \, \vec{c}] = 0 \iff \text{vectors are coplanar}}$$Coplanar = all three vectors lie in same plane.
Why? If coplanar, they can’t form a 3D box → volume = 0!
To check if three vectors are coplanar: Calculate scalar triple product. If = 0, they’re coplanar!
Appears in: 40% of coplanarity problems
5. Scalar Multiplication
$$[\lambda\vec{a} \, \vec{b} \, \vec{c}] = \lambda[\vec{a} \, \vec{b} \, \vec{c}]$$Can take scalar out (like determinants).
Coplanarity of Four Points
Points A, B, C, D are coplanar if:
$$\boxed{[\overrightarrow{AB} \, \overrightarrow{AC} \, \overrightarrow{AD}] = 0}$$Use: Position vectors to form edge vectors, then check scalar triple product!
Check if points A(1,2,3), B(2,3,1), C(3,1,2), D(0,4,6) are coplanar.
Solution:
$$\overrightarrow{AB} = \hat{i} + \hat{j} - 2\hat{k}$$ $$\overrightarrow{AC} = 2\hat{i} - \hat{j} - \hat{k}$$ $$\overrightarrow{AD} = -\hat{i} + 2\hat{j} + 3\hat{k}$$ $$[\overrightarrow{AB} \, \overrightarrow{AC} \, \overrightarrow{AD}] = \begin{vmatrix} 1 & 1 & -2 \\ 2 & -1 & -1 \\ -1 & 2 & 3 \end{vmatrix}$$ $$= 1(-3+2) - 1(6-1) + (-2)(4-1)$$ $$= -1 - 5 - 6 = -12 \neq 0$$Not coplanar! (volume would be 12 cubic units)
Part 2: Vector Triple Product
Definition
Result: A vector (not a scalar!)
Warning: $(\vec{a} \times \vec{b}) \times \vec{c} \neq \vec{a} \times (\vec{b} \times \vec{c})$ → NOT associative!
BAC-CAB Rule (Most Important!)
Memory Trick: “BAC minus CAB”
- B comes first, multiply by A·C
- C comes second, multiply by A·B
This is THE most important vector identity for JEE Advanced!
In words: The result is a linear combination of $\vec{b}$ and $\vec{c}$ (the two vectors being crossed).
Why “BAC-CAB”?
$$\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$$Reading the vectors: B-A·C minus C-A·B → BAC-CAB
Simplify $\hat{i} \times (\hat{j} \times \hat{k})$ using BAC-CAB rule.
Solution:
$$\hat{i} \times (\hat{j} \times \hat{k}) = \hat{j}(\hat{i} \cdot \hat{k}) - \hat{k}(\hat{i} \cdot \hat{j})$$ $$= \hat{j}(0) - \hat{k}(0) = \vec{0}$$Alternatively: $\hat{j} \times \hat{k} = \hat{i}$, so $\hat{i} \times \hat{i} = \vec{0}$ ✓
Other Form of Vector Triple Product
$$\boxed{(\vec{a} \times \vec{b}) \times \vec{c} = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{a}(\vec{b} \cdot \vec{c})}$$Different from $\vec{a} \times (\vec{b} \times \vec{c})$!
Memory trick: The result is a combination of the first two vectors ($\vec{a}$ and $\vec{b}$).
Properties of Vector Triple Product
1. NOT Associative
$$\vec{a} \times (\vec{b} \times \vec{c}) \neq (\vec{a} \times \vec{b}) \times \vec{c}$$Parentheses matter! Always use BAC-CAB for the correct form.
2. Result is Coplanar
is coplanar with $\vec{b}$ and $\vec{c}$.
Why? It’s a linear combination of $\vec{b}$ and $\vec{c}$!
3. Perpendicular to First Vector
$$\vec{a} \times (\vec{b} \times \vec{c})$$is perpendicular to $\vec{a}$.
Why? It’s a cross product with $\vec{a}$!
Important Identities
Lagrange’s Identity
$$\boxed{(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c})}$$High-level identity - appears in JEE Advanced only.
Jacobi Identity
$$\vec{a} \times (\vec{b} \times \vec{c}) + \vec{b} \times (\vec{c} \times \vec{a}) + \vec{c} \times (\vec{a} \times \vec{b}) = \vec{0}$$Cyclic sum equals zero - rarely asked but elegant!
Memory Tricks & Patterns
Mnemonic for Scalar Triple Product
Memory Trick: “Scalar Triple = Volume” → STV
Memory Trick for Zero: “Coplanar → Zero” → If all vectors in same plane, volume = 0
Pattern Recognition
If problem asks “Are vectors/points coplanar?”
- Form scalar triple product
- If = 0, coplanar ✓
- If ≠ 0, not coplanar ✗
Appears in: 60% of coplanarity problems
See $\vec{a} \times (\vec{b} \times \vec{c})$ in problem?
- Immediately apply BAC-CAB: $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$
- Calculate dot products
- Simplify
Don’t try to calculate cross products directly! BAC-CAB is always faster.
Keywords indicating scalar triple product:
- “Volume of parallelepiped”
- “Volume of tetrahedron” (use 1/6 factor)
- “Are points coplanar?”
Direct application - write determinant, evaluate!
When to Use Triple Products
Use Scalar Triple Product when:
- Finding volume of parallelepiped/tetrahedron
- Checking if vectors/points are coplanar
- Need to evaluate $\vec{a} \cdot (\vec{b} \times \vec{c})$
Use Vector Triple Product when:
- Simplifying $\vec{a} \times (\vec{b} \times \vec{c})$
- Need vector coplanar with two given vectors
- Solving vector equations with nested cross products
Use BAC-CAB rule always for vector triple product!
Common Mistakes to Avoid
Wrong: Volume = $[\vec{a} \, \vec{b} \, \vec{c}]$ (could be negative!)
Right: Volume = $|[\vec{a} \, \vec{b} \, \vec{c}]|$ (absolute value!)
Scalar triple product can be negative, but volume is always positive!
Wrong: $\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) + \vec{c}(\vec{a} \cdot \vec{b})$ (plus!)
Right: $\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})$ (minus!)
It’s BAC minus CAB!
Wrong: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \times \vec{c}$
Right: These are DIFFERENT vectors! Parentheses matter.
Always expand using BAC-CAB for the correct form!
Wrong: Writing vectors in determinant randomly
Right: First vector → first row, second → second row, third → third row
Order matters! Swapping rows changes sign.
Wrong: Volume of tetrahedron = $|[\vec{a} \, \vec{b} \, \vec{c}]|$
Right: Volume of tetrahedron = $\frac{1}{6}|[\vec{a} \, \vec{b} \, \vec{c}]|$
Don’t forget the 1/6 factor!
Practice Problems
Level 1: Foundation (NCERT Style)
Find $[\vec{a} \, \vec{b} \, \vec{c}]$ where $\vec{a} = \hat{i}$, $\vec{b} = \hat{j}$, $\vec{c} = \hat{k}$.
Solution:
$$[\vec{a} \, \vec{b} \, \vec{c}] = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 1$$(This is volume of unit cube!)
If $[\vec{a} \, \vec{b} \, \vec{c}] = 5$, find $[\vec{b} \, \vec{c} \, \vec{a}]$.
Solution: By cyclic property: $[\vec{b} \, \vec{c} \, \vec{a}] = [\vec{a} \, \vec{b} \, \vec{c}] = 5$
Simplify $\hat{i} \times (\hat{i} \times \hat{j})$.
Solution: Using BAC-CAB:
$$\hat{i} \times (\hat{i} \times \hat{j}) = \hat{i}(\hat{i} \cdot \hat{j}) - \hat{j}(\hat{i} \cdot \hat{i})$$ $$= \hat{i}(0) - \hat{j}(1) = -\hat{j}$$Verify: $\hat{i} \times \hat{j} = \hat{k}$, so $\hat{i} \times \hat{k} = -\hat{j}$ ✓
Level 2: JEE Main Type
Find volume of parallelepiped with edges:
- $\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$
- $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$
- $\vec{c} = 3\hat{i} + \hat{j} + 2\hat{k}$
Solution:
$$V = \left|\begin{vmatrix} 2 & 3 & 4 \\ 1 & 2 & 3 \\ 3 & 1 & 2 \end{vmatrix}\right|$$Expanding along first row:
$$= |2(4-3) - 3(2-9) + 4(1-6)|$$ $$= |2(1) - 3(-7) + 4(-5)|$$ $$= |2 + 21 - 20| = |3| = 3 \text{ cubic units}$$Check if $\vec{a} = \hat{i}+2\hat{j}+3\hat{k}$, $\vec{b} = 2\hat{i}+3\hat{j}+4\hat{k}$, $\vec{c} = 3\hat{i}+4\hat{j}+5\hat{k}$ are coplanar.
Solution:
$$[\vec{a} \, \vec{b} \, \vec{c}] = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}$$Notice: Each row is previous row + $(\hat{i}+\hat{j}+\hat{k})$ This means vectors are in arithmetic progression → coplanar!
Expanding: $= 1(15-16) - 2(10-12) + 3(8-9)$
$$= -1 + 4 - 3 = 0$$Coplanar! ✓
Find volume of tetrahedron with vertices at O(0,0,0), A(1,2,3), B(2,1,3), C(3,2,1).
Solution:
$$\overrightarrow{OA} = \hat{i}+2\hat{j}+3\hat{k}$$ $$\overrightarrow{OB} = 2\hat{i}+\hat{j}+3\hat{k}$$ $$\overrightarrow{OC} = 3\hat{i}+2\hat{j}+\hat{k}$$ $$[\overrightarrow{OA} \, \overrightarrow{OB} \, \overrightarrow{OC}] = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 2 & 1 \end{vmatrix}$$ $$= 1(1-6) - 2(2-9) + 3(4-3)$$ $$= -5 + 14 + 3 = 12$$Volume = $\frac{1}{6}|12| = 2$ cubic units
If $\vec{a} = \hat{i}+\hat{j}$, $\vec{b} = \hat{j}+\hat{k}$, $\vec{c} = \hat{k}+\hat{i}$, find $\vec{a} \times (\vec{b} \times \vec{c})$.
Solution: Using BAC-CAB:
$$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})$$ $$\vec{a} \cdot \vec{c} = (1)(1) + (1)(0) + (0)(1) = 1$$ $$\vec{a} \cdot \vec{b} = (1)(0) + (1)(1) + (0)(1) = 1$$ $$= \vec{b}(1) - \vec{c}(1) = (\hat{j}+\hat{k}) - (\hat{k}+\hat{i})$$ $$= -\hat{i} + \hat{j}$$Level 3: JEE Advanced Type
Show that A(1,1,1), B(2,3,5), C(4,0,-1), D(0,5,3) are coplanar.
Solution:
$$\overrightarrow{AB} = \hat{i}+2\hat{j}+4\hat{k}$$ $$\overrightarrow{AC} = 3\hat{i}-\hat{j}-2\hat{k}$$ $$\overrightarrow{AD} = -\hat{i}+4\hat{j}+2\hat{k}$$ $$[\overrightarrow{AB} \, \overrightarrow{AC} \, \overrightarrow{AD}] = \begin{vmatrix} 1 & 2 & 4 \\ 3 & -1 & -2 \\ -1 & 4 & 2 \end{vmatrix}$$ $$= 1(-2+8) - 2(6-2) + 4(12-1)$$ $$= 6 - 8 + 44 = 42 \neq 0$$Not coplanar!
(There must be calculation error in problem statement, or points given are incorrect)
If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors satisfying $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, find $\vec{a} \times (\vec{b} \times \vec{c})$.
Solution: From $\vec{a} + \vec{b} + \vec{c} = \vec{0}$:
$$\vec{b} + \vec{c} = -\vec{a}$$Using BAC-CAB:
$$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})$$From $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, taking dot with $\vec{a}$:
$$\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0$$ $$1 + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0$$Similarly, $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = -\frac{1}{2}$
$$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(-\frac{1}{2}) - \vec{c}(-\frac{1}{2})$$ $$= -\frac{1}{2}\vec{b} + \frac{1}{2}\vec{c} = \frac{1}{2}(\vec{c} - \vec{b})$$If $\vec{a} = \hat{i}+\hat{j}$, $\vec{b} = \hat{j}+\hat{k}$, $\vec{c} = \hat{k}+\hat{i}$, find $\vec{a}'$ such that $\vec{a}' \cdot \vec{a} = 1$, $\vec{a}' \cdot \vec{b} = 0$, $\vec{a}' \cdot \vec{c} = 0$.
Solution: $\vec{a}'$ is perpendicular to both $\vec{b}$ and $\vec{c}$, so:
$$\vec{a}' = \lambda(\vec{b} \times \vec{c})$$ $$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix} = \hat{i}(1-0) - \hat{j}(0-1) + \hat{k}(0-1)$$ $$= \hat{i} + \hat{j} - \hat{k}$$From $\vec{a}' \cdot \vec{a} = 1$:
$$\lambda(\hat{i}+\hat{j}-\hat{k}) \cdot (\hat{i}+\hat{j}) = 1$$ $$\lambda(1+1) = 1 \Rightarrow \lambda = \frac{1}{2}$$ $$\vec{a}' = \frac{1}{2}(\hat{i}+\hat{j}-\hat{k})$$This is called reciprocal vector system!
Connection to Physics & 3D Geometry
Applications:
Volume Calculations: Parallelepiped, tetrahedron
Coplanarity Tests: Points/vectors in same plane
- See: Plane Equations
Torque in Physics: Sometimes involves vector triple products
- See: Rotational Dynamics
Electromagnetic Theory: Advanced E&M uses vector identities
- See: Maxwell’s Equations
Equation of Plane: Normal vector using cross product
- See: 3D Geometry
Master triple products → Excel in 3D geometry & advanced physics!
Quick Revision Box
| Concept | Formula |
|---|---|
| Scalar triple product | $[\vec{a} \, \vec{b} \, \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c})$ |
| Determinant form | $\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$ |
| Volume of parallelepiped | $\|[\vec{a} \, \vec{b} \, \vec{c}]\|$ |
| Volume of tetrahedron | $\frac{1}{6}\|[\vec{a} \, \vec{b} \, \vec{c}]\|$ |
| Coplanar condition | $[\vec{a} \, \vec{b} \, \vec{c}] = 0$ |
| Cyclic property | $[\vec{a} \, \vec{b} \, \vec{c}] = [\vec{b} \, \vec{c} \, \vec{a}] = [\vec{c} \, \vec{a} \, \vec{b}]$ |
| Vector triple product | $\vec{a} \times (\vec{b} \times \vec{c})$ |
| BAC-CAB rule | $\vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})$ |
What’s Next?
You’ve completed the Vector Algebra chapter! Apply these concepts:
Mathematics:
- 3D Coordinate Geometry - Equations of lines, planes using vectors
- Matrices & Determinants - Determinant concepts used in scalar triple product
Physics:
- Rotational Dynamics - Torque, angular momentum (cross products)
- Magnetism - Magnetic force F = qv × B
- Electrostatics - Electric field, potential (dot products)
Practice:
- Solve previous year JEE questions on vectors
- Focus on volume and coplanarity problems
Teacher’s Summary
- Scalar triple product = Volume - Use determinant form for speed
- Coplanar ⟺ scalar triple product = 0 - Instant coplanarity test
- BAC-CAB rule is mandatory - Don’t calculate vector triple product directly
- Absolute value for volume - Scalar triple product can be negative
- Cyclic property saves time - $[\vec{a} \, \vec{b} \, \vec{c}] = [\vec{b} \, \vec{c} \, \vec{a}]$
- Tetrahedron = (1/6) parallelepiped - Don’t forget factor!
“In JEE, see ‘volume’? Calculate scalar triple product determinant. See ‘coplanar’? Check if scalar triple = 0. See vector triple? Apply BAC-CAB immediately. Master these three rules, ace all triple product questions!”
Time-saving tip: For coplanarity, just check if determinant = 0. No need to calculate full value!
Pro Tip: BAC-CAB appears in JEE Advanced every 2-3 years. When you see $\vec{a} \times (\vec{b} \times \vec{c})$, your reflex should be: “Write $\vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})$”. Practice until it’s automatic!
Congratulations! You’ve mastered all of Vector Algebra - from basics to triple products. Vectors are now your superpower for mechanics and 3D geometry!