Let `veca=hati-hatk, vecb=xhati+hatj+(1-x)hatk` and `vecc=yhati+xhatj+(1+x-y)hatk`, then `[veca vecb vecc]` depends on

Options

A. neither `x` nor `y`
B. both `x` and `y`
C. only `x`
D. only `y`

Answer

A

Explanation

We have
`[veca vecb vecc]=|(1,0,-1),(x,1,1-x),(y,x,1+x-y)|`
`implies[veca vecb vecc]=|(1,0,0),(x,1,1),(y,x,1+x)|` Appling `C_(3)toC_(3)+C_(1)`
`implies[veca vecb vecc]=1`, which is independent of both `x` and `y`