Sean ** A** = (A

_{x}, TO

_{y}, TO

_{z}),

**= (B**

*B*_{x}, B

_{y}, B

_{z}) Y

**= (C**

*C*_{x}, C

_{y}, C

_{z}); he

**double vector product**, also called

**triple vector product**, is the operation that consists of multiplying two vectors vector

**y**

*A***, and multiply this result vectorially by the third vector**

*B***, that is to say:**

*C*** A** × (

**×**

*B***)**

*C*Writing the parentheses is very important, as skipping it can affect the result. The double vector product results in a vector that lies on the vector plane ** B** ×

**, that is, what is in the parentheses.**

*C***Definition**

If we have three vectors ** A** = (A

_{x}, TO

_{y}, TO

_{z}),

**= (B**

*B*_{x}, B

_{y}, B

_{z}) Y

**= (C**

*C*_{x}, C

_{y}, C

_{z}) and we relate

**×**

*B***will give us a**

*C***vector**, which multiplied by

**originates a new vector:**

*A*** A** × (

**×**

*B***) =**

*C***(**

*B***·**

*A***) –**

*C***(**

*C***·**

*A***)**

*B*which is called **double vector product **or **triple vector product**.

**Properties**

Sean ** A**,

**y**

*B***R vectors**

*C*^{3}, so:

- The vector
× (*A*×*B*) is a vector contained in the plane defined by*C*y*B*.*C* - The double vector product is anti-commutative and has no associative property:

** A** × (

**×**

*B***) = –**

*C***× (**

*C***×**

*A***) =**

*B***(**

*B***·**

*A***) –**

*C***(**

*C***·**

*A***)**

*B*The previous vector is contained in the plane defined by the vectors** A** y

**, so in general:**

*B*** A** × (

**×**

*B***) ≠ (**

*C***×**

*A***) ×**

*B*

*C*With which the placement of the parentheses is very important.

**Demonstration**

Sean ** A** = (A

_{x}, TO

_{y}, TO

_{z}),

**= (B**

*B*_{x}, B

_{y}, B

_{z}) Y

**= (C**

*C*_{x}, C

_{y}, C

_{z}) three non-coplanar vectors; by drawing these vectors so that the xy plane of a Cartesian plane coincides with the plane in which they are

**y**

*B***and the x axis coincides in direction and direction with**

*C***, we have:**

*C*** A** = (A

_{x}, TO

_{y}, TO

_{z})

**∙**

*A***= A**

*C*_{x}C

_{x}

** B** = (B

_{x}, B

_{y}, 0) →

**∙**

*A***= A**

*B*_{x}B

_{x }+ A

_{y}B

_{y}

** C** = (C

_{x}, 0, 0)

That together with that:

Adding and subtracting A_{x}B_{x}C_{x }î and grouping terms we have:

** A** × (

**×**

*B***) = (B**

*C*_{x}î + B

_{y}ĵ) A

_{x}C

_{x }- C

_{x}î (A

_{x}B

_{x }+ A

_{y}B

_{y})

Thus:

** A** × (

**×**

*B***) =**

*C***(**

*B***·**

*A***) –**

*C***(**

*C***·**

*A***)**

*B*Which is what we wanted to demonstrate.

**Exercises**

- Calculate the double vector product of the vectors
= (0, 1, 1),*A*= (0, 1, 0) and*B*= (2, 0, 1).*C*

We know that:

= ĵ + k*A*= ĵ*B*= 2 î + k*C*

From the formula of the double vector product we have:

** A** × (

**×**

*B***) =**

*C***(**

*B***·**

*A***) –**

*C***(**

*C***·**

*A***)**

*B*Thus:

** A** ·

**= A**

*C*_{x}C

_{x }+ A

_{y}C

_{y }+ A

_{z}C

_{z }= (0)(2) + (1)(0) + (1)(1) = 0 + 0 + 1 = 1

From here, we have to

**B**(**A** · ** C**) = ĵ (1) = ĵ

Analogously

** A** ·

**= A**

*B*_{x}B

_{x }+ A

_{y}B

_{y }+ A

_{z}B

_{z }= (0)(0) + (1)(1) + (1)(0) = 0 + 1 + 0 = 1

**C**(**A** · ** B**) = (2 î + k) (1) = 2 î + k

Thus:

** A** × (

**×**

*B***) = ĵ - (2 î + k) = - 2 î + ĵ - k**

*C*- Calculate the double vector product of the vectors
= (1, 2, 3),*A*= (-3, 1, 4) and*B*= (1, 2, 1).*C*

We know that:

= î + 2 ĵ + 3 k*A*= - 3 î + ĵ + 4 k*B*= î + 2 ĵ + k*C*

From the formula of the double vector product we have:

** A** × (

**×**

*B***) =**

*C***(**

*B***·**

*A***) –**

*C***(**

*C***·**

*A***)**

*B*Thus:

** A** ·

**= A**

*C*_{x}C

_{x }+ A

_{y}C

_{y }+ A

_{z}C

_{z }= (1)(1) + (2)(2) + (3)(1) = 1 + 4 + 3 = 8

From here, we have to

**B**(**A** · ** C**) = (- 3 î + ĵ + 4 k) (8) = - 24 î + 8 ĵ + 32 k

Analogously

** A** ·

**= A**

*B*_{x}B

_{x }+ A

_{y}B

_{y }+ A

_{z}B

_{z }= (1)(-3) + (2)(1) + (3)(4) = -3 + 2 + 12 = 11

**C**(**A** · ** B**) = (î + 2 ĵ + k) (11) = 11 î + 22 ĵ + 11 k

Thus:

** A** × (

**×**

*B***) = (- 24 î + 8 ĵ + 32 k) - (11 î + 22 ĵ + 11 k) =**

*C*= - 24 î + 8 ĵ + 32 k - 11 î - 22 ĵ - 11 k) =

= - (24 + 11) î + (8 - 22) ĵ + (32 - 11) k =

= - 35 î - 14 ĵ + 21 k