Quick answer: The scalar triple product [ a ⃗ b ⃗ c ⃗ ] = a ⃗ ⋅ ( b ⃗ × c ⃗ ) [\vec{a}\ \vec{b}\ \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) [ a b c ] = a ⋅ ( b × c ) a ⃗ , b ⃗ , c ⃗ \vec{a}, \vec{b}, \vec{c} a , b , c coplanar . The vector triple product expands as a ⃗ × ( b ⃗ × c ⃗ ) = ( a ⃗ ⋅ c ⃗ ) b ⃗ − ( a ⃗ ⋅ b ⃗ ) c ⃗ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} a × ( b × c ) = ( a ⋅ c ) b − ( a ⋅ b ) c
Syllabus note: Several portals report that Vectors was removed in the revised NIMCET syllabus. Confirm with the official information brochure, and see our revised syllabus update .
This uses the dot and cross products .
Scalar triple product
[ a ⃗ b ⃗ c ⃗ ] = ∣ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ∣ [\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} [ a b c ] = a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3
Property Result Cyclic order [ a ⃗ b ⃗ c ⃗ ] = [ b ⃗ c ⃗ a ⃗ ] = [ c ⃗ a ⃗ b ⃗ ] [\vec{a}\ \vec{b}\ \vec{c}] = [\vec{b}\ \vec{c}\ \vec{a}] = [\vec{c}\ \vec{a}\ \vec{b}] [ a b c ] = [ b c a ] = [ c a b ] Swapping two vectors Changes the sign Dot and cross interchange a ⃗ ⋅ ( b ⃗ × c ⃗ ) = ( a ⃗ × b ⃗ ) ⋅ c ⃗ \vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c} a ⋅ ( b × c ) = ( a × b ) ⋅ c Repeated vector [ a ⃗ a ⃗ b ⃗ ] = 0 [\vec{a}\ \vec{a}\ \vec{b}] = 0 [ a a b ] = 0 Scalar multiple [ λ a ⃗ b ⃗ c ⃗ ] = λ [ a ⃗ b ⃗ c ⃗ ] [\lambda\vec{a}\ \vec{b}\ \vec{c}] = \lambda[\vec{a}\ \vec{b}\ \vec{c}] [ λ a b c ] = λ [ a b c ] Unit vectors [ i ^ j ^ k ^ ] = 1 [\hat{i}\ \hat{j}\ \hat{k}] = 1 [ i ^ j ^ k ^ ] = 1
Geometric meaning
Quantity Formula Volume of parallelepiped ∣ [ a ⃗ b ⃗ c ⃗ ] ∣ \lvert[\vec{a}\ \vec{b}\ \vec{c}]\rvert ∣[ a b c ]∣ Volume of tetrahedron 1 6 ∣ [ a ⃗ b ⃗ c ⃗ ] ∣ \frac{1}{6}\lvert[\vec{a}\ \vec{b}\ \vec{c}]\rvert 6 1 ∣[ a b c ]∣ Volume of triangular prism 1 2 ∣ [ a ⃗ b ⃗ c ⃗ ] ∣ \frac{1}{2}\lvert[\vec{a}\ \vec{b}\ \vec{c}]\rvert 2 1 ∣[ a b c ]∣
Coplanarity: three vectors are coplanar if and only if [ a ⃗ b ⃗ c ⃗ ] = 0 [\vec{a}\ \vec{b}\ \vec{c}] = 0 [ a b c ] = 0 [ A B → A C → A D → ] = 0 [\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}] = 0 [ A B A C A D ] = 0
Vector triple product
a ⃗ × ( b ⃗ × c ⃗ ) = ( a ⃗ ⋅ c ⃗ ) b ⃗ − ( a ⃗ ⋅ b ⃗ ) c ⃗ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} a × ( b × c ) = ( a ⋅ c ) b − ( a ⋅ b ) c
( a ⃗ × b ⃗ ) × c ⃗ = ( a ⃗ ⋅ c ⃗ ) b ⃗ − ( b ⃗ ⋅ c ⃗ ) a ⃗ (\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a} ( a × b ) × c = ( a ⋅ c ) b − ( b ⋅ c ) a
So the cross product is not associative . The result of a ⃗ × ( b ⃗ × c ⃗ ) \vec{a} \times (\vec{b} \times \vec{c}) a × ( b × c ) b ⃗ \vec{b} b c ⃗ \vec{c} c
In both formulas, the vector inside the bracket that is not dotted keeps the positive sign.
Useful results
Result [ a ⃗ + b ⃗ b ⃗ + c ⃗ c ⃗ + a ⃗ ] = 2 [ a ⃗ b ⃗ c ⃗ ] [\vec{a} + \vec{b}\ \ \vec{b} + \vec{c}\ \ \vec{c} + \vec{a}] = 2[\vec{a}\ \vec{b}\ \vec{c}] [ a + b b + c c + a ] = 2 [ a b c ] [ a ⃗ × b ⃗ b ⃗ × c ⃗ c ⃗ × a ⃗ ] = [ a ⃗ b ⃗ c ⃗ ] 2 [\vec{a} \times \vec{b}\ \ \vec{b} \times \vec{c}\ \ \vec{c} \times \vec{a}] = [\vec{a}\ \vec{b}\ \vec{c}]^2 [ a × b b × c c × a ] = [ a b c ] 2 i ^ × ( i ^ × a ⃗ ) + j ^ × ( j ^ × a ⃗ ) + k ^ × ( k ^ × a ⃗ ) = − 2 a ⃗ \hat{i} \times (\hat{i} \times \vec{a}) + \hat{j} \times (\hat{j} \times \vec{a}) + \hat{k} \times (\hat{k} \times \vec{a}) = -2\vec{a} i ^ × ( i ^ × a ) + j ^ × ( j ^ × a ) + k ^ × ( k ^ × a ) = − 2 a
Worked examples
Example 1: volume
Find the volume of the parallelepiped on a ⃗ = i ^ + j ^ \vec{a} = \hat{i} + \hat{j} a = i ^ + j ^ b ⃗ = j ^ + k ^ \vec{b} = \hat{j} + \hat{k} b = j ^ + k ^ c ⃗ = i ^ + k ^ \vec{c} = \hat{i} + \hat{k} c = i ^ + k ^
∣ 1 1 0 0 1 1 1 0 1 ∣ = 1 ( 1 − 0 ) − 1 ( 0 − 1 ) + 0 = 2 \begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix} = 1(1 - 0) - 1(0 - 1) + 0 = 2 1 0 1 1 1 0 0 1 1 = 1 ( 1 − 0 ) − 1 ( 0 − 1 ) + 0 = 2
Volume = 2 = 2 = 2
Example 2: coplanarity
Find λ so that i ^ + 2 j ^ + 3 k ^ \hat{i} + 2\hat{j} + 3\hat{k} i ^ + 2 j ^ + 3 k ^ 2 i ^ + j ^ + λ k ^ 2\hat{i} + \hat{j} + \lambda\hat{k} 2 i ^ + j ^ + λ k ^ i ^ − j ^ \hat{i} - \hat{j} i ^ − j ^
Step 1:
∣ 1 2 3 2 1 λ 1 − 1 0 ∣ = 0 \begin{vmatrix} 1 & 2 & 3 \\ 2 & 1 & \lambda \\ 1 & -1 & 0 \end{vmatrix} = 0 1 2 1 2 1 − 1 3 λ 0 = 0
Step 2: 1 ( 0 + λ ) − 2 ( 0 − λ ) + 3 ( − 2 − 1 ) = 3 λ − 9 = 0 1(0 + \lambda) - 2(0 - \lambda) + 3(-2 - 1) = 3\lambda - 9 = 0 1 ( 0 + λ ) − 2 ( 0 − λ ) + 3 ( − 2 − 1 ) = 3 λ − 9 = 0 λ = 3 \lambda = 3 λ = 3
Example 3: vector triple product
Find i ^ × ( j ^ × k ^ ) \hat{i} \times (\hat{j} \times \hat{k}) i ^ × ( j ^ × k ^ )
Directly: j ^ × k ^ = i ^ \hat{j} \times \hat{k} = \hat{i} j ^ × k ^ = i ^ i ^ × i ^ = 0 ⃗ \hat{i} \times \hat{i} = \vec{0} i ^ × i ^ = 0
By the formula: ( i ^ ⋅ k ^ ) j ^ − ( i ^ ⋅ j ^ ) k ^ = 0 ⃗ (\hat{i} \cdot \hat{k})\hat{j} - (\hat{i} \cdot \hat{j})\hat{k} = \vec{0} ( i ^ ⋅ k ^ ) j ^ − ( i ^ ⋅ j ^ ) k ^ = 0
Example 4: a tetrahedron
Find the volume of the tetrahedron with edges i ^ \hat{i} i ^ j ^ \hat{j} j ^ k ^ \hat{k} k ^
1 6 [ i ^ j ^ k ^ ] = 1 6 \frac{1}{6}[\hat{i}\ \hat{j}\ \hat{k}] = \frac{1}{6} 6 1 [ i ^ j ^ k ^ ] = 6 1
Summary table
Concept Key result Scalar triple product Determinant of components Volume ∣ [ a ⃗ b ⃗ c ⃗ ] ∣ \lvert[\vec{a}\ \vec{b}\ \vec{c}]\rvert ∣[ a b c ]∣ Tetrahedron One-sixth of that Coplanar [ a ⃗ b ⃗ c ⃗ ] = 0 [\vec{a}\ \vec{b}\ \vec{c}] = 0 [ a b c ] = 0 a ⃗ × ( b ⃗ × c ⃗ ) \vec{a} \times (\vec{b} \times \vec{c}) a × ( b × c ) ( a ⃗ ⋅ c ⃗ ) b ⃗ − ( a ⃗ ⋅ b ⃗ ) c ⃗ (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} ( a ⋅ c ) b − ( a ⋅ b ) c
Exam tips
Treat any coplanarity question as "determinant = 0".
Keep cyclic order; a single swap flips the sign.
Use [ a ⃗ × b ⃗ b ⃗ × c ⃗ c ⃗ × a ⃗ ] = [ a ⃗ b ⃗ c ⃗ ] 2 [\vec{a} \times \vec{b}\ \ \vec{b} \times \vec{c}\ \ \vec{c} \times \vec{a}] = [\vec{a}\ \vec{b}\ \vec{c}]^2 [ a × b b × c c × a ] = [ a b c ] 2
See all topics in the NIMCET Coordinate Geometry and Vectors chapter .
Common mistakes
Assuming a ⃗ × ( b ⃗ × c ⃗ ) = ( a ⃗ × b ⃗ ) × c ⃗ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \times \vec{c} a × ( b × c ) = ( a × b ) × c
Forgetting the 1 6 \frac{1}{6} 6 1
Reporting a negative volume.
Mixing up the two dot products in the vector triple product expansion.
Frequently Asked Questions
What is the scalar triple product?
[a b c] = a · (b × c). It equals the determinant formed by the components of the three vectors.
When are three vectors coplanar?
When their scalar triple product is zero.
a × (b × c) = (a · c)b − (a · b)c.
How is the volume of a parallelepiped found using vectors?
It is the absolute value of the scalar triple product of the three edge vectors.
Is the cross product associative?
No. In general a × (b × c) is not equal to (a × b) × c.
