Task. Given that A × B = O A\times B=O A × B = O , B × C = O B\times C=O B × C = O , and A ≠ 0 A\neq 0 A = 0 , B ≠ 0 B\neq 0 B = 0 , C ≠ 0 C\neq 0 C = 0 . Find the value of A × C A\times C A × C .
Solution. Recall that for any two vectors A A A and B B B in R 3 \mathbb{R}^{3} R 3 the absolute value of their cross product ∣ A × B ∣ |A\times B| ∣ A × B ∣ is equal to
∣ A × B ∣ = ∣ A ∣ ⋅ ∣ B ∣ ⋅ sin A B ^ , |A\times B|=|A|\cdot|B|\cdot\sin\widehat{AB}, ∣ A × B ∣ = ∣ A ∣ ⋅ ∣ B ∣ ⋅ sin A B ,
where A B ^ \widehat{AB} A B is the angle between vectors A A A and B B B .
By assumption ∣ A ∣ ≠ 0 |A|\neq 0 ∣ A ∣ = 0 , ∣ B ∣ ≠ 0 |B|\neq 0 ∣ B ∣ = 0 , ∣ C ∣ ≠ 0 |C|\neq 0 ∣ C ∣ = 0 , and
∣ A ∣ ⋅ ∣ B ∣ ⋅ sin A B ^ = 0 , ∣ B ∣ ⋅ ∣ C ∣ ⋅ sin C B ^ = 0. |A|\cdot|B|\cdot\sin\widehat{AB}=0,\qquad|B|\cdot|C|\cdot\sin\widehat{CB}=0. ∣ A ∣ ⋅ ∣ B ∣ ⋅ sin A B = 0 , ∣ B ∣ ⋅ ∣ C ∣ ⋅ sin CB = 0.
Therefore
sin A B ^ = sin B C ^ = 0. \sin\widehat{AB}=\sin\widehat{BC}=0. sin A B = sin BC = 0.
This means that vectors A A A and B B B are collinear, and similarly B B B and C C C are also collinear. Hence, A A A and C C C are also collinear, and so
sin A C ^ = 0. \sin\widehat{AC}=0. sin A C = 0.
Therefore
∣ A × C ∣ = ∣ A ∣ ⋅ ∣ C ∣ ⋅ sin A C ^ = 0 , |A\times C|=|A|\cdot|C|\cdot\sin\widehat{AC}=0, ∣ A × C ∣ = ∣ A ∣ ⋅ ∣ C ∣ ⋅ sin A C = 0 ,
which means that A × C = 0 A\times C=0 A × C = 0 .
Answer. A × C = 0 A\times C=0 A × C = 0 .