Scalar and Vector Triple Product

Q: How do I remember the BAC minus CAB expansion correctly?

Read a cross the bracket b cross c as BAC minus CAB: write b times the dot product of a and c, minus c times the dot product of a and b. The vector next to the opening bracket comes out first, and the outer vector always sits inside both dot products.

Q: Is there a scalar quadruple product I need to know?

The useful one states that the dot product of a cross b with c cross d equals the product of the dot products of a with c and b with d, minus the product of the dot products of a with d and b with c. Lagrange's identity is the special case where c equals a and d equals b.

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Scalar and Vector Triple Product

Maths · Vector Algebra · Concept 4

With three vectors there are only two sensible ways to combine dot and cross products. One gives a number that measures volume and detects coplanarity; the other gives a vector that always lies in a particular plane. Learn the determinant for the first and the BAC minus CAB rule for the second, and this entire topic collapses into two lines.

Scalar triple product (box product)

Definition

[a b c] = (a \times b) \cdot c

Writing $θ$ for the angle between $a$ and $b$ , and $ϕ$ for the angle between $a \times b$ and $c$ ,

[a b c] = ∣ a ∣ ∣ b ∣ ∣ c ∣ sin θ cos ϕ

The result is a scalar, and it is also written $a \cdot (b \times c)$ .

Geometrical meaning: volume

Build a parallelepiped on $a$ , $b$ , $c$ as coterminous edges. The base parallelogram spanned by $a$ and $b$ has area $a \times b$ , and the height is the projection of $c$ on the normal $a \times b$ . Multiplying the two gives

Volume of a parallelepiped

V = [a b c]

The box product is base area times height, so its absolute value is the volume of the parallelepiped built on the three vectors. Here

a \times b

and

c

point to the same side of the base, so

[a b c]

is positive and the triple is right-handed; interchanging any two vectors reverses the sign but not the volume.

Determinant form

Components

If $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ , $b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ and $c = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ , then

[a b c] = a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} c_{1} c_{2} c_{3}

More generally, if $a = a_{1} l + a_{2} m + a_{3} n$ and similarly for $b$ and $c$ in terms of non-coplanar $l, m, n$ , then

[a b c] = a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} c_{1} c_{2} c_{3} [l m n]

Every property below is a determinant property in disguise, which is the fastest way to remember them.

Properties of the scalar triple product

Key properties

Cyclic invariance: $[a b c] = [b c a] = [c a b]$
Swap changes sign: $[a b c] = - [a c b]$
Dot and cross can be interchanged: $(a \times b) \cdot c = a \cdot (b \times c)$
Repeated vector kills it: if any two of the three are equal or parallel, the product is $0$
Linearity: $[k a b c] = k [a b c]$ and $[(a + b) c d] = [a c d] + [b c d]$
$[\hat{i} \hat{j} \hat{k}] = 1$

Coplanarity test

a, b, c are coplanar ⟺ [a b c] = 0

Equivalently, they are linearly dependent. If $[a b c] \neq = 0$ they are non-coplanar and form a basis of space: positive for a right handed system, negative for a left handed system.

Two standard results

[a + b b + c c + a] = 2 [a b c]

[a \times b b \times c c \times a] = [a b c]^{2}

A third, worth recognising on sight:

[a - b b - c c - a] = 0

JEE Advanced

Gram determinant. The square of the box product can be written entirely in terms of dot products:

[a b c]^{2} = a \cdot a a \cdot b a \cdot c b \cdot a b \cdot b b \cdot c c \cdot a c \cdot b c \cdot c

This is the three vector version of Lagrange's identity, and it is the fastest route when a question supplies only magnitudes and mutual angles rather than components.

Volume of a tetrahedron

Tetrahedron

With one vertex at the origin and the other three at $a, b, c$ :

V = \frac{1}{6} [a b c]

With four general vertices $a, b, c, d$ :

V = \frac{1}{6} [a - d b - d c - d]

The centroid of the tetrahedron is $\frac{1}{4} (a + b + c + d)$ .

Example 1

Find the volume of the parallelepiped with edges $- 3 \hat{i} + 7 \hat{j} + 5 \hat{k}$ , $- 5 \hat{i} + 7 \hat{j} - 3 \hat{k}$ and $7 \hat{i} - 5 \hat{j} - 3 \hat{k}$ .

Solution.

[a b c] = - 375 - 57 - 3 7 - 5 - 3

Expanding along the first row:

$- 3 (- 21 - 15) - 7 (15 + 21) + 5 (25 - 49) = 108 - 252 - 120 = - 264$ .

Volume $= ∣ - 264∣ = 264$ cubic units. The negative sign only says the three edges form a left handed system.

Example 2

Simplify $[a - b b - c c - a]$ .

Solution. Expand the cross product first:

$(a - b) \times (b - c) = a \times b - a \times c - b \times b + b \times c = a \times b + b \times c + c \times a$ .

Dotting with $c - a$ and using the fact that a box product with a repeated vector vanishes:

$[a b c] - [b c a] = [a b c] - [a b c] = 0$ .

Example 3

Show that $a = \hat{i} + 2 \hat{j} - \hat{k}$ , $b = 3 \hat{i} + \hat{j} + 2 \hat{k}$ and $c = 5 \hat{i} + 5 \hat{j}$ are coplanar.

Solution.

$[a b c] = 12 - 1 312550 = 1 (0 - 10) - 2 (0 - 10) - 1 (15 - 5) = - 10 + 20 - 10 = 0$ .

Since the box product vanishes, the three vectors are coplanar.

The structural reason is visible directly: $2 a + b = (2 + 3) \hat{i} + (4 + 1) \hat{j} + (- 2 + 2) \hat{k} = 5 \hat{i} + 5 \hat{j} = c$ , so $c$ is a linear combination of the other two.

Vector triple product

The expression $a \times (b \times c)$ is a vector. Since $b \times c$ is perpendicular to the plane of $b$ and $c$ , crossing it again with $a$ produces a vector perpendicular to that normal, that is, a vector lying back in the plane of $b$ and $c$ , and also perpendicular to $a$ . So it must be expressible as $x b + y c$ , and working out the coefficients gives the central formula of this topic.

Expansion formula (BAC minus CAB)

a \times (b \times c) = (a \cdot c) b - (a \cdot b) c

(a \times b) \times c = (a \cdot c) b - (b \cdot c) a

Memory aid: the middle vector of the bracketed pair comes out first, multiplied by the dot product of the other two; then subtract, with the remaining vector multiplied by the dot product of the outer vector with the first one.

The vector triple product returns to the plane of the two bracketed vectors, and is perpendicular to the outer vector.

The bracket is not optional

$a \times (b \times c)$ lies in the plane of $b$ and $c$ , while $(a \times b) \times c$ lies in the plane of $a$ and $b$ . They are different vectors in general, so writing $a \times b \times c$ without brackets is meaningless.

The two agree exactly when $(a \cdot b) c = (b \cdot c) a$ , which happens when $a$ and $c$ are parallel, or when $b$ is perpendicular to both.

Jacobi identity

a \times (b \times c) + b \times (c \times a) + c \times (a \times b) = 0

Expanding all three by BAC minus CAB, every term appears twice with opposite signs.

Example 4

For any vector $a$ , prove that $\hat{i} \times (a \times \hat{i}) + \hat{j} \times (a \times \hat{j}) + \hat{k} \times (a \times \hat{k}) = 2 a$ .

Solution. By BAC minus CAB, $\hat{i} \times (a \times \hat{i}) = (\hat{i} \cdot \hat{i}) a - (\hat{i} \cdot a) \hat{i} = a - (a \cdot \hat{i}) \hat{i}$ .

Adding the three analogous terms:

$3 a - [(a \cdot \hat{i}) \hat{i} + (a \cdot \hat{j}) \hat{j} + (a \cdot \hat{k}) \hat{k}] = 3 a - a = 2 a$ ,

using the resolution identity from the dot product page.

Example 5

Prove that $a \times {b \times (c \times d)} = (b \cdot d) (a \times c) - (b \cdot c) (a \times d)$ .

Solution. Expand the inner bracket first:

$b \times (c \times d) = (b \cdot d) c - (b \cdot c) d$ .

Now cross with $a$ on the left, pulling the scalars out:

$a \times [(b \cdot d) c - (b \cdot c) d] = (b \cdot d) (a \times c) - (b \cdot c) (a \times d)$ .

JEE Advanced

Combining both triple products. Let $a = \hat{i} + 2 \hat{j} - 3 \hat{k}$ , $b = λ \hat{i} + \hat{j} + (λ - 2) \hat{k}$ and $c = 2 \hat{i} - \hat{j} + \hat{k}$ . Consider

$(a \times b) \times {(b \times c) \times (c \times a)}$ .

Using the identity $(b \times c) \times (c \times a) = [a b c] c$ , the whole expression becomes

$[a b c] {(a \times b) \times c} = [a b c] {(a \cdot c) b - (b \cdot c) a}$ .

It therefore vanishes if either $[a b c] = 0$ , or $(a \cdot c) b = (b \cdot c) a$ . The lesson is structural: in long products, hunt for a box product factor before touching components.

JEE Advanced

Solving for an unknown vector. To solve $x \times a + k x = b$ , first dot with $a$ , which kills the cross term:

$k (a \cdot x) = a \cdot b$ , so $a \cdot x = \frac{a \cdot b}{k}$ .

Next cross the original equation with $a$ on the left and expand the triple product:

$(a \cdot x) a - (a \cdot a) x + k (a \times x) = a \times b$ .

From the original equation $x \times a = b - k x$ , so $a \times x = k x - b$ . Substituting and collecting terms gives

x = \frac{1}{k ^{2} + ∣ a ∣ ^{2}} [k b + (a \times b) + \frac{a \cdot b}{k} a]

The technique matters more than the closed form: dot with the known vector to get the scalar unknown, then cross with it to get the vector unknown.

Mistakes that cost marks

Writing $[a b c] = [a c b]$ . Only cyclic rotation is free; any single swap flips the sign.
Treating the vector triple product as a scalar, or the scalar triple product as a vector.
Applying BAC minus CAB with the wrong vector coming out first. The vector adjacent to the bracket opening is the one paired with the outer vector in the second term.
Forgetting the modulus when reporting a volume. A box product can legitimately be negative.
Concluding from $[a b c] = 0$ that one of the vectors is zero. It only means the three are coplanar.
Using $[a b c]^{2}$ and $[a^{2} b^{2} c^{2}]$ interchangeably. The second is not a defined object.

Quick recap

$[a b c] = (a \times b) \cdot c =$ the $3 \times 3$ determinant of the components
$[a b c]$ is the volume of the parallelepiped; one sixth of it is the tetrahedron
Coplanar $⟺$ box product zero $⟺$ linearly dependent
Cyclic rotation keeps the sign; a swap reverses it; a repeat makes it zero
$a \times (b \times c) = (a \cdot c) b - (a \cdot b) c$
$[a \times b b \times c c \times a] = [a b c]^{2}$ and $[a + b b + c c + a] = 2 [a b c]$

Frequently asked questions

Why can the dot and the cross be interchanged in a scalar triple product?

Because both arrangements equal the same determinant of the nine components. Geometrically both compute the same signed volume, just taking a different face as the base, and the volume of a solid does not depend on which face you call the bottom.

What does a negative scalar triple product mean?

It means the three vectors, in the order given, form a left handed system. The magnitude is still the volume; only the orientation information sits in the sign. That is why a volume answer always carries a modulus.

How do I remember the BAC minus CAB expansion correctly?

Read $a \times (b \times c)$ as "a cross b cross c" and write $b (a \cdot c) - c (a \cdot b)$ , which reads as BAC minus CAB. The vector next to the opening bracket comes out first, and the outer vector always sits inside both dot products.

Does a zero box product mean the vectors are linearly dependent?

Yes, for three vectors in space the two statements are equivalent. A zero box product means zero volume, which means all three lie in one plane, which is exactly linear dependence.

Is there a scalar quadruple product I need to know?

The useful one is $(a \times b) \cdot (c \times d) = (a \cdot c) (b \cdot d) - (a \cdot d) (b \cdot c)$ , obtained by treating $c \times d$ as a single vector and applying the box product rules. Lagrange's identity is the special case with $c = a$ and $d = b$ .

Previous year questions on Scalar and Vector Triple Product

12 questions from past papers, each with a step-by-step solution.

Show all 12 questions

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