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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, \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] 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 \[\overrightarrow{b}\] and \[\overrightarrow{c}\].

Mathematically = \[\overrightarrow{a} × (\overrightarrow{b} × \overrightarrow{c})\]

In this vector triple product, \[\overrightarrow{a} × (\overrightarrow{b} × \overrightarrow{c})\]

The vectors \[\overrightarrow{b}\] and \[\overrightarrow{c}\] are being coplanar with the triple product.

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

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

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

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.

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We need to prove that the vector triple product is the right result generated from the cross product of \[\overrightarrow{a}, \overrightarrow{b}, and \overrightarrow{c}\].

\[\overrightarrow{a} × (\overrightarrow{b} × \overrightarrow{c})\]

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

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

\[\overrightarrow{c} .(\overrightarrow{a} × \overrightarrow{b}) × \overrightarrow{c}\] = \[\overrightarrow{c}.(x\overrightarrow{a} + y\overrightarrow{b})\]

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

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

\[\frac{x}{\overrightarrow{b}.\overrightarrow{c}}\] = - \[\frac{y}{\overrightarrow{a}.\overrightarrow{c}}\] = λ

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

Now, we have \[(\overrightarrow{a} × \overrightarrow{b}) × \overrightarrow{c}\] = \[x\overrightarrow{a} + y\overrightarrow{b}\]…… (1)

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

\[(\overrightarrow{a} × \overrightarrow{b}) × \overrightarrow{c}\] = \[(λ\overrightarrow{b} . \overrightarrow{c}) \overrightarrow{a} + (-λ\overrightarrow{a} . \overrightarrow{c}) \overrightarrow{b}\]

= \[(λ\overrightarrow{b} . \overrightarrow{c}) \overrightarrow{a} - (λ\overrightarrow{a} . \overrightarrow{c}) \overrightarrow{b}\]

The product is for each value of \[\overrightarrow{a}, \overrightarrow{b}, and \overrightarrow{c}\]. The reason is each of them has an identity.

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

(i × j) × i =(λj . i) i - (λi . i) j

j = -λj

λ = -1

Therefore, \[(\overrightarrow{a} × \overrightarrow{b}) × \overrightarrow{c}\] = \[(\overrightarrow{a} . \overrightarrow{c})\overrightarrow{b} - (\overrightarrow{b} . \overrightarrow{c})\overrightarrow{a}\]

Let’s solve some vector triple product example problems.

Example – 1

There are three vectors known as \[\overrightarrow{a}, \overrightarrow{b}, and \overrightarrow{c}\]. The magnitudes of these vectors are \[|\overrightarrow{a}|=1, |\overrightarrow{b}|=2, |\overrightarrow{c}|=1\]. The equation is \[\overrightarrow{a} × (\overrightarrow{a} × \overrightarrow{b}) + \overrightarrow{c}\]=0. Then, calculate the angle between \[\overrightarrow{a}, and \overrightarrow{b}\].

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

We know the relation that

\[\overrightarrow{a}.\overrightarrow{b}\] = \[|\overrightarrow{a}||\overrightarrow{b}|cosA\]

= 1 * 2 * cos A

= 2

Also, \[\overrightarrow{a} × (\overrightarrow{a} × \overrightarrow{b}) + \overrightarrow{c}\] = 0

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

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

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

Let’s just square both the sides

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

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

4cos2A - 4cosA.2cosA + 4 = 1

4(1-A) = 1

1 = 4 sin2 A

Let’s just apply square root on both sides, we get;

1 = 2 sin A

Sin A = ½

A = Sin-1(1/2) = π/6

Example – 2

Let’s assume that the vectors \[\overrightarrow{a}, \overrightarrow{b}, and \overrightarrow{c}\] are coplanar. The question is to demonstrate that \[\overrightarrow{a} × \overrightarrow{b}\], \[\overrightarrow{a} × \overrightarrow{c}\], \[\overrightarrow{b} × \overrightarrow{c}\] are also coplanar.

Ans: As \[\overrightarrow{a}, \overrightarrow{b}, and \overrightarrow{c}\] are coplanar we can write them as \[[\overrightarrow{a} × \overrightarrow{b} × \overrightarrow{c}]\]

\[[\overrightarrow{a} × \overrightarrow{b} × \overrightarrow{c}]\] = 0

By squaring both the sides, we get,

\[[\overrightarrow{a} × \overrightarrow{b} × \overrightarrow{c}]^2\] = 0

\[[(\overrightarrow{a} × \overrightarrow{b})(\overrightarrow{b} × \overrightarrow{c})(\overrightarrow{c} × \overrightarrow{a})]\] = 0

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

Example – 3

When both \[\overrightarrow{a} and \overrightarrow{c}\] are collinear, then prove that \[(\overrightarrow{a} × \overrightarrow{b}) × \overrightarrow{c}\] = \[\overrightarrow{a} × (\overrightarrow{b} × \overrightarrow{c})\].

Ans: Now, we need to assume that

\[(\overrightarrow{a} × \overrightarrow{b}) × \overrightarrow{c}\] = \[\overrightarrow{a} × (\overrightarrow{b} × \overrightarrow{c})\]

⇒ \[-\overrightarrow{c} × (\overrightarrow{a} × \overrightarrow{b})\] = \[\overrightarrow{a} × (\overrightarrow{b} × \overrightarrow{c})\]

⇒ \[(-\overrightarrow{c}.\overrightarrow{b}) \overrightarrow{a}\ - (-\overrightarrow{c}.\overrightarrow{a}) \overrightarrow{b}\] = \[(\overrightarrow{a} - \overrightarrow{c})\overrightarrow{b} - (\overrightarrow{a}.\overrightarrow{b})\overrightarrow{c}\]

⇒ \[(\overrightarrow{b}.\overrightarrow{c})\overrightarrow{a}\] = \[(\overrightarrow{a}.\overrightarrow{b})\overrightarrow{c}\]

⇒ \[\overrightarrow{a}\] = \[(\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{b}.\overrightarrow{c}})\overrightarrow{c}\]

⇒ \[\overrightarrow{a} = λ\overrightarrow{c}\]

Here, λ ∊ R

The vector triple product formula can be written as:

\[\overrightarrow{a} × (\overrightarrow{b} × \overrightarrow{c})\] = \[(\overrightarrow{a}.\overrightarrow{c})\overrightarrow{b} - (\overrightarrow{a}.\overrightarrow{b})\overrightarrow{c}(\overrightarrow{a} × \overrightarrow{b}) × \overrightarrow{c} \] = \[(\overrightarrow{a}.\overrightarrow{c})\overrightarrow{b} - (\overrightarrow{b}.\overrightarrow{c})\overrightarrow{a}\]

Here,\[\overrightarrow{a} × (\overrightarrow{b} × \overrightarrow{c})\] ≠ \[(\overrightarrow{a} × \overrightarrow{b}) × \overrightarrow{c}\]

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

FAQ (Frequently Asked Questions)

Q1. How to calculate the product of two Vectors?

Ans: There are two ways of multiplying vectors which can be explained as the vector product and the scalar product. The vector product has a huge application in physics and astronomy.

The product of two vectors implies a vector that is perpendicular to each other. The result can be acquired by its magnitude i.e., by the product of their magnitudes by the sine of the angle in between these vectors.

Q2. Proof that cross product is distributive over subtraction.

Ans: There is a best and quite easy way to prove that cross product is distributive over the subtraction of two vectors.

It can also be completed by negating the components of either vector. Therefore, the cross-product is distributive over subtraction.

Q3. What is the result of a Cross-product; Scalar or Vector?

Ans: The first one is known as the dot product. This kind of product is a scalar product. The outcome of the dot product of two vectors is also a scalar quantity.

Another kind of product is known as the cross product. It is a vector product as it generates another vector instead of a scalar.

Q4. What do you mean by a Position Vector?

Ans: A position vector can be defined as a straight line having one end fixed to a body, and the other end fixed with a moving point.

A point vector is used to define the location of the point comparative to the body.