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# The Cross Product of Vectors in 3-Dimensional Space

the use of a vector in problems of geometry, physics (mechanics, electromagnetic theory, etc.). Very helpful, if we have an easy way to determine a vector that is perpendicular to the two given vectors.
Therefore, in this article we will discuss the concept of cross product by first defining the definitions for cross product vector u and v.

Definition:
If u = (u1, u2, u3) and v = (v1, v2, v3) are vectors in space-3, then the result of sialng u x v is a vector defined by,

u x v = (u2v3 – u3v2, u3v1-u1v3, u1v2-u2v1)

Or in determinant notation,

Example 6.
Calculate u x v, where u = (2, -1, 1) and v = (0, 1, 4).
Solution:
Pay attention to the matrix

Although the product of the two vectors is scalar, the cross product of two vectors is another vector.

The following theorem gives the relationship between dot product and cross product, and shows that u x v is orthogonal for both u and v.

If u and v are vectors in space-3, then:
1. u . ( u x v) = 0                       (u x v orthogonal to u)
2. v . (u x v) = 0                        (u x v orthogonal to v)
3. ||u x v||2 = ||u||2||v||2 – ( u v)2          (Lagrange's identity)
Proof:
For example u = (u1, u2, u3) and v = ( v1, v2, v3)

1.   u . u x v = (u1, u2, u3). (u2v3 – u3v2, u3v2 – u1v3, u1v2 – u2v1)
u . u x v = u1(u2v3 – u3v2) + u2(u3v2 – u1v3) + u3(u1v2 – u2v1)
u . u x v = 0
2.   Do it yourself as your practice.
3.   ||u x v||2 = (u2v3 – u3v2)2 + (u3v2 – u1v3)2 + (u1v2 – u2v1)2  ................(*)
and
||u||2||v||2 – (u . v )2 = (u12 + u22 + u32)( v12 + v22 + v32) – (u1v1 + u2v2 + u3v3)2 ......(**)

Lagrange identity can be generated by "Writing the Product results" of the right side (*) and (**) and proving their similarity.

Example 7.
Review the vectors u = (0, 1, -2) and v = (2, 0, 1)
Solution:
u x v = (1, 4, -2)

because
u . (u x v) = (0)(1) + (1)(-4) + (-2)(-2) = 0
and     v . (u x v) = (2)(1) + (0)(-4) + (1)(-2) = 0

then u x v is orthogonal for both u and v.
The important arithmetic properties of cross product are stated in the following theorem.

Theorem 5,
If u, v, and w are any vectors in space-2 and k is any scalar, then:
• u x v = -(v x u)
• u x (v + w) = (u x v) + (u x w)
• (u + v) x w = (u x w) + (v x w)
• k (x) = (k u) x v = u x (k v)
• u x 0 = 0 x u = 0
• u x u = 0

The properties (a) of the above theorem show that cross product is not commutative.
Look at vectors i = (1, 0, 0)    j = (0, 1, 0)    k = (0, 0, 1).
Each of these vectors has a length of 1 and is located along the coordinate axis, which is illustrated by the image.

These vectors are called standard unit vectors (standard unit vectors) in space-3. Each vector v = (v1, v2, v3) in space-3 can be expressed with i, j, k because
v = (v1, v2, v3) = v1 (1, 0, 0) + v2 (0, 1, 0) + v3 (0, 0, 1) = v1i + v2j + v3k

For example,
(4, -3, 2) = 4i  + (-3)j + 2k

Now we will look at the properties and relationships between the standard unit vectors.
• i x j = j x j = k x k = 0
• i x j = k, j x k = i, k x i = j
• j x i = -k, k x j = -i, i x k = -j
To remember the properties above the following image is very useful.

Two key sentences are:
1. The cross product of two vectors which are sorted in a clockwise direction in the image is the same as the next vector.
2. The cross product of two vectors which are ordered in the clockwise direction of the image is the same as the negative of the next vector.

Also note that cross product can also be given symbolically as a determinant of the following 3 x 3:

Example 8.
For example u = (1, -1, 2) and v = (2, 1, 0) then

The length of vector u x v has a useful geometric meaning. According to the 4 part c theorem, the so-called Lagrange identity states that

||u x v||2 = ||u||2||v||2 – (u . v)2

If Ɵ expresses the angle between u and v, then u.v = ||u||.||v|| cos Ɵ. Then,

||u x v||2 = ||u||2||v||2 - ||u||2.||v||2 cos2Ɵ.
||u x v||2 = ||u||2.||v||2 (1 - cos2Ɵ)
||u x v||2 = ||u||2.||v||2 sin2Ɵ

with the result that,
||u x v|| = ||u||.||v|| sin Ɵ

Because ||v|| sin Ɵ is the parallelogram height determined by u and v (Figure 3), it is clear that ||u x v|| state the width of the parallelogram stated by
A = (base)(height) = ||u||.||v|| sin Ɵ = ||u x v||

In other words, the norm (length) u x v is equal to the width of the parallelogram specified by u and v.

Example 16.
Calculate the area of a triangle determined by points P1 (2, 1, 0), P2 (-1, 0, 2) and P3 (0, 4, 3).
Settlement:
The area of triangle A is ½ the width of the parallelogram determined by vectors

Based on Figure 4, we can determine the distance between the two points as follows,

and

so the area of P1P2P3 is