Question

tính: \(B=a\left(bz-cy\right)+b\left(cx-az\right)+c\left(ay-bx\right)\)

Answer 1

\(B=a\left(bz-cy\right)+b\left(cx-az\right)+c\left(ay-bx\right)=a.b.z-a.c.y+b.c.x-a.b.z+a.c.y-b.c.x=\left(a.b.z-a.b.z\right)-\left(a.c.y-a.c.y\right)+\left(b.c.x-b.c.x\right)=0-0+0=0\)

Answer 2

B = a(bz - cy) + b(cx - az) + c(ay - bx) = abz - acy + bcx - baz + cay - cbx = 0