CROSS PRODUCT: PROPERTIES, APPLICATIONS AND EXERCISES - MATHS - 2025

The cross product or vector product is a way of multiplying two or more vectors. There are three ways to multiply vectors, but none of these is multiplication in the usual sense of the word. One of these forms is known as a vector product, which results in a third vector.

The cross product, which is also called the cross product or outer product, has different algebraic and geometric properties. These properties are very useful, especially in terms of the study of physics.

Definition

A formal definition of the vector product is the following: if A = (a1, a2, a3) and B = (b1, b2, b3) are vectors, then the vector product of A and B, which we will denote as AxB, is:

AxB = (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1)

Due to the AxB notation, it is read as "A cross B".

An example of how to use the outer product is that if A = (1, 2, 3) and B = (3, -2, 4) are vectors, then using the definition of a vector product we have:

AxB = (1,2,3) x (3, -2, 4) = (2 * 4 - 3 * (- 2), 3 * 3 - 1 * 4, 1 * (- 2) - 2 * 3)

AxB = (8 + 6, 9 - 4, - 2 - 6) = (14, 5, - 8).

Another way of expressing the vector product is given by the notation of determinants.

The calculation of a second order determinant is given by:

Therefore, the formula for the cross product given in the definition can be rewritten as follows:

This is usually simplified into a third-order determinant as follows:

Where i, j, k represent the vectors that form the basis of R ³.

Using this way of expressing the cross product, we have that the previous example can be rewritten as:

Properties

Some properties that the vector product possesses are the following:

Property 1

If A is any vector in R ³, we have:

- AxA = 0

- Ax0 = 0

- 0xA = 0

These properties are easy to check using just the definition. If A = (a1, a2, a3) we have:

AxA = (a2a3 - a3a2, a3a1 - a1a3, a1a2 - a2a1) = (0, 0, 0) = 0.

Ax0 = (a2 * 0 - a3 * 0, a3 * 0 - a1 * 0, a1 * 0 - a2 * 0) = (0, 0, 0) = 0.

If i, j, k represent the unit base of R ³, we can write them as follows:

i = (1, 0, 0)

j = (0, 1, 0)

k = (0, 0, 1)

So, we have that the following properties are true:

As a mnemonic rule, the following circle is often used to remember these properties:

There we must note that any vector with itself gives vector 0 as a result, and the rest of the products can be obtained with the following rule:

The cross product of two consecutive vectors in a clockwise direction gives the next vector; and when the counter-clockwise direction is considered, the result is the following vector with a negative sign.

Thanks to these properties we can see that the vector product is not commutative; for example, just note that ixj ≠ jx i. The following property tells us how AxB and BxA are related in general.

Property 2

If A and B are vectors of R ³, we have:

AxB = - (BxA).

Demonstration

If A = (a1, a2, a3) and B = (b1, b2, b3), by definition of external product we have:

AxB = (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1)

= (- 1) (a3b2 - a2b3, a1b3 - a3b1, a2b1 - a1b2)

= (- 1) (BxA).

We can also see that this product is not associative with the following example:

ix (ixj) = ixk = - j but (ixi) xj = 0xj = 0

From this we can see that:

ix (ixj) ≠ (ixi) xj

Property 3

If A, B, C are vectors of R ³ and r is a real number, the following is true:

- Ax (B + C) = AxB + AxC

- r (AxB) = (rA) xB = Ax (rB)

Thanks to these properties we can calculate the vector product using the laws of algebra, provided that the order is respected. For example:

If A = (1, 2, 3) and B = (3, -2, 4), we can rewrite them in terms of the canonical basis of R ³.

Thus, A = i + 2j + 3k and B = 3i - 2j + 4k. Then, applying the previous properties:

AxB = (i + 2j + 3k) x (3i - 2j + 4k)

= 3 (ixi) - 2 (ixj) + 4 (ixk) + 6 (jxi) - 4 (jxj) + 8 (jxk) + 9 (kxi) - 6 (kxj) +12 (kxk)

= 3 (0) - 2 (k) + 4 (- j) + 6 (- k) - 4 (0) + 8 (i) + 9 (j) - 6 (- i) +12 (0)

= - 2k - 4j - 6k + 8i + 9j + 6i = 14i + 5j - 4k

= (14, 5, - 8).

Property 4 (triple dot product)

As we mentioned at the beginning, there are other ways to multiply vectors besides the vector product. One of these ways is the scalar product or inner product, which is denoted as A ∙ B and whose definition is:

If A = (a1, a2, a3) and B = (b1, b2, b3), then A ∙ B = a1b1 + a2b2 + a3b3

The property that relates both products is known as the triple scalar product.

If A, B, and C are vectors of R ³, then A ∙ BxC = AxB ∙ C

As an example, let's see that, given A = (1, 1, - 2), B = (- 3, 4, 2) and C = (- 5, 1, - 4), this property is satisfied.

BxC = - 3k - 12j + 20k - 16i - 10j - 2i = - 18i - 22j + 17k

A ∙ BxC = (1, 1, - 2) ∙ (- 18, - 22, 17) = (1) (- 18) + (1) (- 22) + (- 2) (17) = - 74

On the other hand:

AxB = 4k - 2j + 3k + 2i + 6j + 8i = 10i + 4j + 7k

AxB ∙ C = (10, 4, 7) ∙ (- 5, 1, - 4) = (10) (- 5) + (4) (1) + (7) (- 4) = - 74

Another triple product is Ax (BxC), which is known as the triple vector product.

Property 5 (triple vector product)

If A, B and C are vectors of R ³, then:

Ax (BxC) = (A ∙ C) B - (A ∙ B) C

As an example, let's see that, given A = (1, 1, - 2), B = (- 3, 4, 2) and C = (- 5, 1, - 4), this property is satisfied.

From the previous example we know that BxC = (- 18, - 22, 17). Let's calculate Ax (BxC):

Ax (BxC) = - 22k - 17j + 18k + 17i + 36j - 44i = - 27i + 19j - 4k

On the other hand, we have to:

A ∙ C = (1, 1, - 2) ∙ (- 5, 1, - 4) = (1) (- 5) + (1) (1) + (- 2) (- 4) = - 5 + 1 + 8 = 4

A ∙ B = (1, 1, - 2) ∙ (- 3, 4, 2) = (1) (- 3) + (1) (4) + (- 2) (2) = - 3 + 4 - 4 = - 3

Thus, we have to:

(A ∙ C) B - (A ∙ B) C = 4 (- 3, 4, 2) + 3 (- 5, 1, - 4) = (- 12, 16, 8) + (- 15, 3, - 12) = (- 27,19, –4)

Property 6

It is one of the geometric properties of vectors. If A and B are two vectors in R ³ and ϴ is the angle formed between them, then:

--AxB-- = --A ---- B - sin (ϴ), where - ∙ - denotes the modulus or magnitude of a vector.

The geometric interpretation of this property is as follows:

Let A = PR and B = PQ. So the angle formed by vectors A and B is the angle P of triangle RQP, as shown in the following figure.

Therefore, the area of ​​the parallelogram that has PR and PQ as adjacent sides is --A ---- B - sin (ϴ), since we can take --A-- as a base and its height is given by --B - sin (ϴ).

Therefore, we can conclude that --AxB-- is the area of ​​said parallelogram.

Example

Given the following vertices of a quadrilateral P (1, –2,3), Q (4, 3, –1), R (2, 2,1) and S (5,7, -3), show that said quadrilateral is a parallelogram and find its area.

For this we first determine the vectors that determine the direction of the sides of the quadrilateral. This is:

A = PQ = (1 - 4, 3 + 2, - 1 - 3) = (3, 5, - 4)

B = PR = (2 - 1, 2 + 2, 1 - 3) = (1, 4, - 2)

C = RS = (5 - 2, 7 - 2, - 3 - 1) = (3, 5, - 4)

D = QS = (5 - 4, 7 - 3, - 3 + 1) = (1, 4, - 2)

As we can see, A and C have the same director vector, for which we have that both are parallel; the same happens with B and D. Therefore, we conclude that PQRS is a parallelogram.

To have the area of ​​this parallelogram, we calculate BxA:

BxA = (i + 4j - 2k) x (3i + 5j - 4k)

= 5k + 4j - 12k - 16i - 6j + 10i

= - 6i - 2j - 7k.

Therefore, the area squared will be:

--BxA-- ² = (- 6) ² + (- 2) ² + (- 7) ² = 36 + 4 + 49 = 89.

It can be concluded that the parallelogram area will be the square root of 89.

Property 7

Two vectors A and B are parallel in R ³ if and only if AxB = 0

Demonstration

It is clear that if A or B are the null vector, it is fulfilled that AxB = 0. Since the zero vector is parallel to any other vector, then the property is valid.

If neither of the two vectors is the zero vector, we have that their magnitudes are different from zero; that is, both --A-- ≠ 0 and --B-- ≠ 0, so we will have --AxB-- = 0 if and only if sin (ϴ) = 0, and this happens if and only if ϴ = π or ϴ = 0.

Therefore, we can conclude AxB = 0 if and only if ϴ = π or ϴ = 0, which only happens when both vectors are parallel to each other.

Property 8

If A and B are two vectors in R ³, then AxB is perpendicular to both A and B.

Demonstration

For this proof, let's remember that two vectors are perpendicular if A ∙ B is equal to zero. Furthermore, we know that:

A ∙ AxB = AxA ∙ B, but AxA is equal to 0. Therefore, we have:

A ∙ AxB = 0 ∙ B = 0.

By this we can conclude that A and AxB are perpendicular to each other. Analogously, we have to:

AxB ∙ B = A ∙ BxB.

Since BxB = 0, we have:

AxB ∙ B = A ∙ 0 = 0.

Therefore, AxB and B are perpendicular to each other and with this the property is demonstrated. This is very useful to us, since they allow us to determine the equation of a plane.

Example 1

Obtain an equation of the plane that passes through the points P (1, 3, 2), Q (3, - 2, 2) and R (2, 1, 3).

Let A = QR = (2 - 3.1 + 2, 3 - 2) and B = PR = (2 - 1.1 - 3, 3 - 2). Then A = - i + 3j + k and B = i - 2j + k. To find the plane formed by these three points, it is enough to find a vector that is normal to the plane, which is AxB.

AxB = (- i + 3j + k) x (i - 2j + k) = 5i + 2j - k.

With this vector, and taking the point P (1, 3, 2), we can determine the equation of the plane as follows:

(5, 2, - 1) ∙ (x - 1, y - 3, z - 2) = 5 (x - 1) + 2 (y - 3) - (z - 2) = 0

Thus, we have that the equation of the plane is 5x + 2y - z - 9 = 0.

Example 2

Find the equation of the plane that contains the point P (4, 0, - 2) and that is perpendicular to each of the planes x - y + z = 0 and 2x + y - 4z - 5 = 0.

Knowing that a normal vector to a plane ax + by + cz + d = 0 is (a, b, c), we have that (1, -1,1) is a normal vector of x - y + z = 0 y (2,1, - 4) is a normal vector of 2x + y - 4z - 5 = 0.

Therefore a normal vector to the sought plane must be perpendicular to (1, -1,1) and to (2, 1, - 4). This vector is:

(1, -1,1) x (2,1, - 4) = 3i + 6j + 3k.

Then, we have that the sought plane is the one that contains the point P (4,0, - 2) and has the vector (3,6,3) as a normal vector.

3 (x - 4) + 6 (y - 0) + 3 (z + 2) = 0

x + 2y + z - 2 = 0.

Applications

Calculation of volume of a parallelepiped

An application that has the triple scalar product is to be able to calculate the volume of a parallelepiped whose edges are given by the vectors A, B and C, as shown in the figure:

We can deduce this application in the following way: as we said before, the vector AxB is a vector that is normal to the plane of A and B. We also have that the vector - (AxB) is another vector normal to said plane.

We choose the normal vector that forms the smallest angle with vector C; Without loss of generality, let AxB be the vector whose angle with C is the smallest.

We have that both AxB and C have the same starting point. Furthermore, we know that the area of ​​the parallelogram that forms the base of the parallelepiped is --AxB--. Therefore, if the height of the parallelepiped is given by h, we have that its volume will be:

V = --AxB - h.

On the other hand, let's consider the dot product between AxB and C, which can be described as follows:

However, by trigonometric properties we have that h = --C - cos (ϴ), so we have:

In this way, we have that:

In general terms, we have that the volume of a parallelepiped is given by the absolute value of the triple scalar product AxB ∙ C.

Solved exercises

Exercise 1

Given the points P = (5, 4, 5), Q = (4, 10, 6), R = (1, 8, 7) and S = (2, 6, 9), these points form a parallelepiped whose edges they are PQ, PR and PS. Determine the volume of said parallelepiped.

Solution

If we take:

- A = PQ = (-1, 6, 1)

- B = PR = (-4, 4, 2)

- C = PS = (-3, 2, 2)

Using the triple scalar product property, we have:

AxB = (-1, 6, 1) x (-4, 4, 2) = (8, -2, 20).

AxB ∙ C = (8, -2, 20) ∙ (-3, 2, 2) = -24 -4 +80 = 52.

Therefore, we have that the volume of said parallelepiped is 52.

Exercise 2

Determine the volume of a parallelepiped whose edges are given by A = PQ, B = PR and C = PS, where the points P, Q, R and S are (1, 3, 4), (3, 5, 3), (2, 1, 6) and (2, 2, 5), respectively.

Solution

First we have that A = (2, 2, -1), B = (1, -2, 2), C = (1, -1, 1).

We calculate AxB = (2, 2, -1) x (1, -2, 2) = (2, -5, -6).

Then we calculate AxB ∙ C:

AxB ∙ C = (2, -5, -6) ∙ (1, -1, 1) = 2 + 5 - 6 = 1.

Thus we conclude that the volume of said parallelepiped is 1 cubic unit.

References

1. Leithold, L. (1992). The calculation with analytic geometry. HARLA, SA
2. Resnick, R., Halliday, D., & Krane, K. (2001). Physics Vol. 1. Mexico: Continental.
3. Saenz, J. (nd). Vector Calculus 1ed. Hypotenuse.
4. Spiegel, MR (2011). Vectorial Analysis 2ed. Mc Graw Hill.
5. Zill, DG, & Wright, W. (2011). Calculation of Several Variables 4ed. Mc Graw Hill.