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, `a = hati - hatk`
`b = xhati + hatj + (1-x)hatk`
and `ac = yhati + x hatj + (1+x -y)hatk`
`therefore [abc]=|{:(1,,0,,-1),(x,,1,,1-x),(y,,x,,1+x-y):}|`
Applying `C_(3) to C_(3) + C_(1)`
`=|{:(1,,0,,0),(x,,1,,1),(y,,x,,1+x):}|=1|{:(1,,1),(x,,1+x):}|`
`= 1(1 + x) - x = 0`
Thus, [abc] does not depends upon neither x nor y.