Vector Triple Product

In vector algebra, there is a branch called Vector Triple Product. In a vector triple product, we learn about the cross product of three vectors. 

If we do the cross-product of a vector along with the cross product of the other two vectors, the amount of the vector triple product can be calculated.

This cross-product generates a vector quantity as a result. After the simplification of the vector triple product, BAC – CAB identity name can be obtained from the result.

Let, \[\vec{a}, \vec{b}, \vec{c}\]

are the three vectors. Their Vector triple product can be defined as the cross product of vector  a with the cross product of the vectors \[\vec{b}  and  \vec{c}\]

Mathematically \[\vec{a}\times (\vec{b}\times \vec{c})\]

In this vector triple product \[\vec{a}\times (\vec{b}\times \vec{c})\]

The vectors  \[\vec{b}  and  \vec{c}\]

 are being coplanar with the triple product.

The triple product is also perpendicular to  \[\vec{a}\]

In other methods, we can also write it as a linear combination of vectors  \[\vec{b}  and     \vec{c}\]

The mathematical form is: \[\vec{a}\times (\vec{b}\times \vec{c})=x \vec{b} + y \vec{c}\]

Do you Know About the Scalar Triple Product?

The definition for the scalar triple product can be explained as it is the dot product of one of the vectors with the cross product of the other two vectors. This is also termed as the box product or mixed product.

We can explain the scalar triple product geometrically

a. (b × c)

It is the volume of the parallelepiped distinct by the three vectors shown. 

Here, the parentheses may be omitted without causing uncertainty, since the dot product cannot be estimated first. 

If it were, it would leave the cross product of a scalar and a vector, which is not defined.

Vector Triple Product Proof

We need to prove that the vector triple product is the right result generated from the cross product of

\[\vec{a}, \vec{b}  and  \vec{c}\]

\[\vec{a}\times (\vec{b}\times \vec{c})\]

This product can be written as the linear combination of vectors \[\vec{a} and \vec{b}\]

The product can be written as \[(\vec{a}\times \vec{b})\times \vec{c}=x \vec{a} + y \vec{b}\]

\[\vec{c}.(\vec{a}\times \vec{b})\times \vec{c}=\vec{c}. (x\vec{a}+y\vec{b})\]

\[x.(\vec{c}.\vec{a}) + y.(\vec{c}.\vec{b})\]

\[x.(\vec{a}.\vec{c}) + y.(\vec{b}.\vec{c}=0\]

\[\frac{x}{\vec{b}.\vec{c}}=-\frac{y}{\vec{a}.\vec{c}}=\lambda \]

So, \[x = \lambda (\vec{b}.\vec{c})\] and \[y = \lambda (\vec{a}.\vec{c})\]

Now, we have  \[(\vec{a}\times \vec{b}) \times \vec{c} = x \vec{a} + y \vec{b}\]......(1)

Let’s just substitute the value of x and y in the above equation (1)

\[(\vec{a} \times \vec{b}) \times \vec{c} = (\lambda \vec{b}. \vec{c}) \vec{a} + (-\lambda \vec{a}. \vec{c}) \vec{b} = (\lambda \vec{b}. \vec{c}) \vec{a}-(\lambda \vec{a}. \vec{c}) \vec{b}\]

The product is for each value of \[\vec{a},\vec{b} and \vec{c}\]

The reason is each of them has an identity.

Put \[\vec{a} = i, \vec{b} = j, \vec{c} = i\]

\[(i \times j) \times = (\lambda j . i) i - (\lambda i . i) j\]

\[j = - \lambda j\]

\[\lambda = -1\]

Therefore, \[(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} . \vec{c}) \vec{b} - (\vec{b} . \vec{c}) \vec{a}\]

Vector Triple Product Example 

Let’s solve some vector triple product example problems.

Example – 1

There are three vectors known as \[\vec{a}, \vec{b} and \vec{c}\].

The magnitudes of these vectors are \[\left | \vec{a} \right |= 1, \left | \vec{b} \right |=2, \left | \vec{c} \right |=1\]

The equation is \[\vec{a} \times (\vec{a} \times \vec{b}) + \vec{c} = 0\].

Then, calculate the angle between \[\vec{a} and \vec{b}\].

Ans: Consider the angle between \[\vec{a} and \vec{b}\] = A

We know the relation that

\[\vec{a}.\vec{b} = \left | \vec{a} \right | \left | \vec{b} \right |cos A\]

\[= 1*2*cos A\]

\[= 2\]

Also, \[\vec{a} \times (\vec{a} \times \vec{b}) + \vec{c} = 0\]

\[(\vec{a} . \vec{b}) \vec{a} - (\vec{a} . \vec{a}) \vec{b} + \vec{c} = 0\]

2cosA \[\vec{a} - \vec{b} + \vec{c} = 0\]

2cosA \[\vec{a} - \vec{b} = - \vec{c}\]

Lets just square both the sides

2cosA \[[\vec{a} - \vec{b}]^{2} = [-\vec{c}]^{2}\]

\[4cos^{2}A\left | \vec{a} \right |^{2} - 4cosA \vec{a} . \vec{b} + \left | \vec{b} \right |^{2} = \left | \vec{c} \right |^{2}\]

\[4cos^{2}A - 4cosA.2cosA + 4 = 1\]

\[4(1-A)=1\]

\[1=4sin^{2}A\]

Lets just apply square root on both sides, we get;

\[1= 2sin A\]

\[Sin A\frac{1}{2}\]

\[A = Sin^{-1}\frac{1}{2} = \frac{\pi}{6}\]

Example – 2

Let’s assume that the vectors  \[\vec{a},\vec{b}, and \vec{c}\]

are coplanar. The question is to demonstrate that  \[\vec{a} \times \vec{b}, \vec{a} \times \vec{c}, \vec{b} \times \vec{c}\]

 are also coplanar.

Ans: As \[\vec{a}, \vec{b}  and \vec{c}\]

are coplanar we can write them as 

\[[\vec{a} \times \vec{b} \times \vec{c}]\]

\[[\vec{a} \times \vec{b} \times \vec{c}]=0\]

By squaring both the sides, we get,

\[[\vec{a} \times \vec{b} \times \vec{c}]^{2}=0\]

\[[(\vec{a} \times \vec{b})(\vec{b} \times \vec{c})(\vec{c} \times \vec{a})] = 0\]

After this result, it is proven that \[\vec{a}\times \vec{b}, \vec{a} \times \vec{c}, \vec{b} \times \vec{c}\] are also coplanar.

Example – 3

When both \[\vec{a} and \vec{c}\]

are collinear, then prove that \[(\vec{a} \times \vec{b}) \times \vec{c} = \vec{a} \times (\vec{b} \times \vec{c})\].

Ans: Now, we need to assume that

\[(\vec{a} \times \vec{b}) \times \vec{c} = \vec{a} \times (\vec{b} \times \vec{c})\]

\[\Rightarrow -\vec{c} \times (\vec{a} \times \vec{b}) = \vec{a} \times  (\vec{b} \times \vec{c})\]

\[\Rightarrow (-\vec{c}.\vec{b}) \vec{a} - (-\vec{c}.\vec{a}) \vec{b} = (\vec{a} - \vec{c}) \vec{b} - (\vec{a} . \vec{b}) \vec{c}\]

\[\Rightarrow (\vec{b} . \vec{c}) \vec{a} = (\vec{a} . \vec{b}) \vec{c}\]

\[\Rightarrow \vec{a} = (\frac{ \vec{a}.\vec{b}}{\vec{b}.\vec{c}})  \vec{c}\]

\[\Rightarrow \vec{a} = \lambda \vec{c}\]

Here, \[\lambda \epsilon R\]

Vector Triple Product Formula 

The vector triple product formula can be written as:

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

Here, \[\vec{a} \times (\vec{b} \times \vec{c}) \neq (\vec{a} \times \vec{b}) \times \vec{c}\]

Vector Triple Product Properties

Let’s assume that there are three vectors such as a, b, c.  The cross-product of the vectors such as a × (b × c) and (a × b) × c is known as the vector triple product of a, b, c.

Therefore, it can be written as, a × (b × c) = (a. c) b − (a. b) c

The vector triple product a × (b × c) is a linear combination of those two vectors which are within brackets.

The ‘r’ vector r=a×(b×c) is perpendicular to a vector and remains in the b and c plane.

The expression for the vector r = a1 + λb is factual only when the vector lies external to the bracket is on the leftmost side. 

When ‘r’ is not found as mentioned in the above theory, we first change it to the left via the properties of the cross-product and then put on the exact expression.

Therefore, (b × c) × a

= − {a × (b × c)} 

= − {(a. c) b − (a. b) c} 

= (a. b) c − (a. c) b

Vector triple product is recognized as a vector quantity.

a × (b × c) ≠ (a × b) × c

Example 5: 

Let a x b=c, b x c=a, and a, b, c be the moduli of the vectors a, b, c, then find a and b.

Solution: 

a = b × c and a × b = c

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

a, b, and c are perpendicular to each other

Now, a = b × c = b × (a × b) = (b . b) a − (b . a) b or 

a =b² a − (b.a) b= b² a, {because a⊥b}

⇒1= b,

Therefore, \[c = a × b = ab sin90^{0}ń\]

Taking the moduli of both sides, c = ab, but b = 1 ⇒ c = a.

Example 6: Given these simultaneous equations for two vectors x and y.

x + y = a …..(i)

x × y = b …..(ii)

x . a = 1 …..(iii)

Find the values of x and y.

Solution: 

By multiplying (i) scalarly by a, we get

a . x + a . y = a²

∴ a . y = a² − 1 ..(iv),

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

(a² − 1) x − y = a × b ..(v),

Adding and subtracting (i) and (v),

we get x = [a + (a × b)] / [a²] and y = a − x