Double vector product

double vector productSean A = (Ax, TOy, TOz), B = (Bx, By, Bz) Y C = (Cx, Cy, Cz); he double vector product, also called triple vector product, is the operation that consists of multiplying two vectors vector A y B, and multiply this result vectorially by the third vector C, that is to say:

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 × C, that is, what is in the parentheses.

Definition

If we have three vectors A = (Ax, TOy, TOz), B = (Bx, By, Bz) Y C = (Cx, Cy, Cz) and we relate B × C will give us a vector, which multiplied by A originates a new vector:

A × (B × C) = B (A · C) – C (A · B)

which is called double vector product or triple vector product.

Properties

Sean A, B y C R vectors3, so:

  1. The vector A × (B × C) is a vector contained in the plane defined by B y C.
  2. 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 B, so in general:

A × (B × C) ≠ (A × B) × C

With which the placement of the parentheses is very important.

Demonstration

demonstration of the double vector productSean A = (Ax, TOy, TOz), B = (Bx, By, Bz) Y C = (Cx, Cy, Cz) 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 B y C and the x axis coincides in direction and direction with C, we have:

A = (Ax, TOy, TOz)                                  AC = Ax Cx

B = (Bx, By, 0) →                   AB = Ax Bx + Ay By

C = (Cx, 0, 0)

That together with that:

double vector product

Adding and subtracting AxBxCî and grouping terms we have:

A × (B × C) = (Bx î + By ĵ) AxCx - Cx î (Ax Bx + Ay By)

Thus:

A × (B × C) = B (A · C) – C (A · B)

Which is what we wanted to demonstrate.

Exercises

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

We know that:

  • A = ĵ + k
  • B = ĵ
  • C = 2 î + k

From the formula of the double vector product we have:

A × (B × C) = B (A · C) – C (A · B)

Thus:

A · C = AxCx + AyCy + AzCz = (0)(2) + (1)(0) + (1)(1) = 0 + 0 + 1 = 1

From here, we have to

B(A · C) = ĵ (1) = ĵ

Analogously

A · B = AxBx + AyBy + AzBz = (0)(0) + (1)(1) + (1)(0) = 0 + 1 + 0 = 1

C(A · B) = (2 î + k) (1) = 2 î + k

Thus:

A × (B × C) = ĵ - (2 î + k) = - 2 î + ĵ - k

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

We know that:

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

From the formula of the double vector product we have:

A × (B × C) = B (A · C) – C (A · B)

Thus:

A · C = AxCx + AyCy + AzCz = (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 · B = AxBx + AyBy + AzBz = (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 × C) = (- 24 î + 8 ĵ + 32 k) - (11 î + 22 ĵ + 11 k) =

= - 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