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Multiplication of Vectors

−→
−→
PROOF. If a and b are colinear or λ = 0, then this property is obvious.
−→
−→
Let a and b are non-colinear and λ ̸= 0.

−→ −→ −→
According to the definition of a cross product, on the one side: λ · ( a × b ) = |λ| · | a | ·

−→ −→ −→ −→ −→
| b | · sin φ, on another side: (λ a ) × b ) = |λ| · | a | · | b | · sin φ. That is, the length of
−→ − →
−→
−→
vectors λ · ( a × b ) and (λ a ) × b is the same. Moreover, these vectors are perpendicular
−→
−→
to every vector a and b , so they are colinear. Finally, these vectors have the same direction,
notwithstanding the sign of digit λ (it is suggested to check it individually).
− → −→ −→
−→
Thus, λ · ( a × b ) = (λ a ) × b . 2
−→ −→
−→
−→
−→
−→
−→
5. ( a + b ) × c = a × c + b × c — a distributive property related to the
sum of vectors.
Let’s consider the cross product in a coordinates form. Let’s have two non-zero vectors in
−→
−→
their coordinates: a = (x 1 ; y 1 ; z 1 ), b = (x 2 ; y 2 ; z 2 ).
Theorem 5.2.
A cross product of two vectors can be calculated following the formula:
−→ −→ −→

i j k ( )
− → y z z x x y
−→ −→ 1 1 1 1 1 1
;
;
c = a × b = x y z = (5.9)

1 1 1 y z z x x y
x 2 y 2 z 2
2 2 2 2 2 2 ⋆
−→ −→ −→ −→ − → −→
−→
PROOF. Expansions of vectors a and b through basis vectors i , j , k are: a = x 1 i +
−→ −→ −→ −→ − → −→
y 1 j + z 1 k , b = x 2 i + y 2 j + z 2 k . Using properties 4 and 5, find the cross product of vectors
−→
−→
a and b :
− → −→ −→ −→ −→ −→ −→ − → −→
−→
a × b = x 1 x 2 · ( i × i ) + x 1 y 2 · ( i × j ) + x 1 z 2 · ( i × k ) + y 1 x 2 · ( j × i )+
−→ −→ −→ −→ −→ − → −→ −→ −→ −→
+y 1 y 2 · ( j × j ) + y 1 z 2 · ( j × k )+z 1 x 2 ·( k × i ) + z 1 y 2 ·( k × j ) + z 1 z 2 ·( k × k ).
− → −→ − → −→ −→ −→ −→ −→ −→ −→ −→ −→
Accordingtoproperties1and3: i × i = j × j = k × k = 0, i × j = k , i × k = − j ,
−→ − → −→ −→ −→ −→ −→ −→ −→ −→ −→ −→ −→ −→
j × i =− k , j × k = i , k × i = j , k × j = − i , we will get: a × b = (y 1 z 2 −
− → −→ −→
y 2 z 1 ) · i + (x 2 z 1 − x 1 z 2 ) · j + (x 1 y 2 − x 2 y 1 ) · k or
− → −→ −→

i j k
− → −→ −→ −→
−→ y 1 z 1 z 1 x 1 x 1 y 1
a × b = · i + · j + · k = x 1 y 1 (5.10)

z 1
y 2 z 2 z 2 x 2 x 2 y 2
2
x 2 y 2 z 2
Formula (5.10) is very often used in practice.
Example 5.3. Find the area of the triangle △ ABC with given coordinates
of its vertices: A(2; 1; 0), B(3; 4; −1), C(4; 3; 1). ,

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