Final Answer:
A) B⃗ ×C⃗ = (14, -7, -13)
B) C⃗ ×B⃗ = (-14, 7, 13)
C) (2B⃗ )×(3C⃗ ) = (-90, 30, 60)
D) A⃗ ×(B⃗ ×C⃗ ) = (-20, -19, -6)
E) A⃗ ⋅(B⃗ ×C⃗ ) = 0
F) |B⃗ ×C⃗ | = 27
G) |C⃗ ×B⃗ | = 27
Step-by-step explanation:
For A), we use the cross product formula, (a₁b₂ - a₂b₁, a₃b₁ - a₁b₃, a₂b₃ - a₃b₂), and find B⃗ ×C⃗ = (14, -7, -13).
For B), we swap the order of vectors in the cross product, resulting in C⃗ ×B⃗ = (-14, 7, 13).
For C), we apply the scalar multiplication before taking the cross product, obtaining (2B⃗ )×(3C⃗ ) = (-90, 30, 60).
For D), we use the vector triple product formula, A⃗ ×(B⃗ ×C⃗ ) = B⃗ (A⃗ ⋅C⃗ ) - C⃗ (A⃗ ⋅B⃗ ), leading to (-20, -19, -6).
For E), we find the dot product of A⃗ and the result from D), yielding 0.
For F) and G), we calculate the magnitude of the cross products |B⃗ ×C⃗ | and |C⃗ ×B⃗ |, both resulting in 27.
In summary, these calculations involve the application of vector operations such as cross products, dot products, and scalar multiplication. The order of vectors in cross products, scalar multiplication effects, and the use of vector triple product formula are key aspects in obtaining the correct results. The dot product being zero in part E) indicates orthogonality between vectors A⃗ and (B⃗ ×C⃗ ). The magnitudes in F) and G) represent the areas of parallelograms formed by the cross products, confirming their equality.