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Cross product of two vectors

Definition. Cross product (vector product) of vector a by the vector b is the vector c, the length of which is numerically equal to the area of the parallelogram constructed on the vectors a and b, perpendicular to the plane of this vectors and the direction so that the smallest rotation from a to b around the vector c was carried out counter-clockwise when viewed from the terminal point of c (Fig. 1).
Cross product of two vectors
Fig. 1

Cross product formulas

Cross product of two vectors a = {ax; ay; az} and b = {bx; by; bz} in Cartesian coordinates is a vector whose value can be calculated using the following formulas:

a × b =     i      j      k     = i(aybz - azby) - j(axbz - azbx) + k(axby - aybx)
 ax  ay  az 
 bx  by  bz 
a × b = {aybz - azby; azbx - axbz; axby - aybx}

Cross product properties

  • Geometric interpretation.
    The magnitude of the cross product of two vectors a and b is equal to the area of the parallelogram constructed on these vectors:

    Ap = |a × b|

  • Geometric interpretation.
    The area of the triangle constructed on the vectors a and a is equal to half the magnitude of the cross product of this vectors:
    AΔ1|a × b|
    2
  • Cross product of two non-zero vectors a and b is equal to zero if and only if the vectors are collinear.
  • The vector c that is equal to the cross product of non-zero vectors a and b, is perpendicular to these vectors.

    c = a × b    =>    c a and c b

  • a × b = -b × a
  • (k a) × b = a × (k b) = k (a × b)
  • (a + b) × c = a × c + b × c

Cross product examples

Example 1. Find the cross product of a = {1; 2; 3} and b = {2; 1; -2}.

Solution:

a × b  i   j   k   =
 1   2   3 
 2   1   -2 

= i(2 · (-2) - 3 · 1) - j(1 · (-2) - 2 · 3) + k(1 · 1 - 2 · 2) =

= i(-4 - 3) - j(-2 - 6) + k(1 - 4) = -7i + 8j - 3k = {-7; 8; -3}
triangle constructed on the vectors
Example 2. Find the area of a triangle formed by vectors a = {-1; 2; -2} and b = {2; 1; -1}.

Solution: Calculate the cross product of these vectors:

a × b  i   j   k   =
 -1   2   -2 
 2   1   -1 

= i(2 · (-1) - (-2) · 1) - j((-1) · (-1) - (-2) · 2) + k((-1) · 1 - 2 · 2) =

= i(-2 + 2) - j(1 + 4) + k(-1 - 4) = -5j - 5k = {0; -5; -5}

From the properties of the cross product:

AΔ1|a × b| = 102 + 52 + 52125 + 251505√2
22222

Answer: AΔ = 2.5√2.

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