# Proof of the Vector Triple Product

Vector Triple Product is a branch in vector algebra where we deal with the cross product of three vectors. The value of the vector triple product can be found by the cross product of a vector with the cross product of the other two vectors. It gives a vector as a result. When we simplify the vector triple product it gives us an identity name as BAC – CAB identity.

## Vector Triple Product Definition

Vector triple product of three vectors $\vec a, \vec b, \vec c$ is defined as the cross product of vector $\vec a$with the cross product of vectors $\vec b\ and\ \vec c$, i.e. $\vec a \times (\vec b \times \vec c)$.

Here $\vec a \times (\vec b \times \vec c)$ is coplanar with the vectors $\vec b\ and\ \vec c$ and perpendicular to $\vec a$ .

Hence we can write $\vec a \times (\vec b \times \vec c)$ as linear combination of vectors $\vec b\ and\ \vec c$,

That is, $\vec a \times (\vec b \times \vec c) = x \vec b + y \vec c$

## Vector Triple Product Formula

$\vec a \times (\vec b \times \vec c) = (\vec a . \vec c) \vec b\ – (\vec a . \vec b) \vec c$ $(\vec a \times \vec b) \times \vec c = (\vec a . \vec c) \vec b\ – (\vec b . \vec c) \vec a$

In general, $\vec a \times (\vec b \times \vec c) \neq (\vec a \times \vec b) \times \vec c$

## Vector Triple Product Proof

We can write $(\vec a \times \vec b) \times \vec c$ as linear combination of vectors $\vec a\ and\ \vec b$.

So, $(\vec a \times \vec b) \times \vec c = x \vec a + y \vec b$

=> $\vec c . (\vec a \times \vec b) \times \vec c = \vec c . (x \vec a + y \vec b)$

= $x . (\vec c . \vec a + y (\vec c . \vec b)$

=> 0 = $x . (\vec a . \vec c + y (\vec b . \vec c)$

=> $\frac{x}{\vec b . \vec c} = \frac{-y}{\vec a . \vec c} = \lambda$

Substituting value of x and y in = $(\vec a \times \vec b) \times \vec c = x \vec a + y \vec b$ we have,

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

= $(\lambda \vec b . \vec c) \vec a – (\lambda \vec a . \vec c) \vec b$

It is valid for every value of $\vec a, \vec b, \vec c$ because it is an identity

Put $\vec a = \hat i , \vec b = \hat j, \vec c = \hat i$

=> $(\hat i \times \hat j) \times \hat i = (\lambda \hat j . \hat i) \hat i – (\lambda \hat i . \hat i) \hat j$

=> $\hat j = – \lambda \hat j$

=> $\lambda = -1$

Hence, $(\vec a \times \vec b) \times \vec c = (\vec a . \vec c)\vec b – (\vec b . \vec c)\vec a$

## Properties

1. Vector triple product is a vector quantity.

2. Unit vector coplanar with $\vec a\ and\ \vec b$ and perpendicular to $\vec c$ is $\pm \frac{(\vec a \times \vec b)\times \vec c}{|(\vec a \times \vec b)\times \vec c|}$.

3. $\vec a \times (\vec b \times \vec c) \neq (\vec a \times \vec b) \times \vec c$.

4.

Note that if $\vec a, \vec b, \vec c$ are non coplanar vector then $\vec a \times \vec b, \vec b \times \vec c\ and\ \vec c \times \vec a$ are also non coplanar.

Some other useful results:

## Solved Examples

Example 1: Find the value of $\hat i \times (\hat j \times \hat k) + \hat j \times (\hat k \times \hat i)$

Solution: $\hat i \times (\hat j \times \hat k) + \hat j \times (\hat k \times \hat i) + \hat k \times (\hat i \times \hat j) = \hat i \times \hat i + \hat j \times \hat j = 0$

Example 2: If $\vec a, \vec b, \vec c$ are three vectors such that $| \vec a|$ = 1, $| \vec b|$ = 2, $| \vec c|$ = 1 and $\vec a \times (\vec a \times \vec b) + \vec c$ = 0 then find the acute angle between $\vec a\ and\ \vec b$.

Solution: let angle between $\vec a\ and \vec b$ is A, then

$\vec a . \vec b = |\vec a||\vec b|cos\ A$

= 1.2. cos A

= 2 cos A

But $\vec a \times (\vec a \times \vec b) + \vec c$ = 0

=> $(\vec a . \vec b)\vec a – (\vec a . \vec a)\vec b + \vec c = 0$

=> 2 cos A $\vec a – \vec b + \vec c = 0$

=> 2 cos A $\vec a – \vec b =- \vec c$

Squaring both side we have

=> [2 cos A $\vec a – \vec b]^2 = [- \vec c]^2$

=> $4 cos^2 A |\vec a|^2 – 4 cos A \vec a . \vec b + |\vec b|^2 = |\vec c|^2$

=> 4 cos2 A – 4 cos A . 2 cos A + 4 = 1

=> 4 cos2 A – 8 cos2 A + 4 = 1

=> $4 (1 – cos^2 A) = 1$

=> 4 sin2 A = 1

or sin A = 1/2

or A = π/6

Example 3: If $\vec a, \vec b, \vec c$ are coplanar then prove that $\vec a \times \vec b, \vec b \times \vec c, \vec c \times \vec a$ are also coplanar.

Solution: If $\vec a, \vec b, \vec c$ are coplanar then $[\vec a\;\; \vec b\;\; \vec c]$

=> $[\vec a\;\; \vec b\;\; \vec c]^2$ = 0

=> $[\vec a \times \vec b\;\; \vec b \times \vec c\;\; \vec c \times \vec a]$ = 0

So we can say that $\vec a \times \vec b, \vec b \times \vec c, \vec c \times \vec a$ are coplanar.

Example 4: Let $\vec c = 2 \hat i + \hat j – 2 \hat k\ and\ \vec b = \hat i + \hat j$ and if vector $\vec a$ is such that $\vec c . \vec a = |\vec a|, |\vec a – \vec c|$ = 2√2 and angle between $(\vec c \times \vec b)$ and $\vec a$ is π/6 then find the value of $|(\vec c \times \vec b) \times \vec a|$.

Solution: $\vec c = 2 \hat i + \hat j – 2 \hat k\ and\ \vec b = \hat i + \hat j$ $|\vec c|$ = 3 and $|\vec b|$ = √2

$|\vec a – \vec c|$ = 2√2 (Given)

Squaring both side we have

Example 5: If $\mathbf{a}\times \mathbf{b}=\mathbf{c},\,\,\mathbf{b}\times \mathbf{c}=\mathbf{a}$ and a, b, c be moduli of the vectors a, b, c respectively, then find the values of a and b.

Solution:

a = b × c and a × b = c

∴ a is perpendicular to both b and c and c is perpendicular to both a and b.

a, b, c are mutually perpendicular

Now, a = b × c = b × (a × b) = (b . b) a − (b . a) b or $\mathbf{a}={{b}^{2}}\mathbf{a}-(\mathbf{b}\,.\,\mathbf{a})\mathbf{b}={{b}^{2}}\mathbf{a}, \left\{ \text because \,\mathbf{a}\,\bot \,\mathbf{b} \right\} \\\Rightarrow 1={{b}^{2}}, \\\text therefore \,\mathbf{c}=\mathbf{a}\times \mathbf{b}=ab\sin 90{}^\circ \,\mathbf{\hat{n}}$

Take moduli of both sides, then c = ab, but b = 1 ⇒ c = a.

Example 6: Given the following simultaneous equations for vectors x and y.

x + y = a …..(i)

x × y = b …..(ii)

x . a = 1 …..(iii)

Then find the values of x and y.

Solution:

Multiplying (i) scalarly by a, we get

a . x + a . y = a^2

∴ a . y = a^2 − 1 ..(iv),

{By (iii)} Again a × (x × y) = a × b or (a . y) x − (a . x) y = a × b

(a^2 − 1) x − y = a × b ..(v),

Adding and subtracting (i) and (v),

we get x = [a + (a × b)] / [a^2] and y = a − x