If `veca=xhati + y hatj + zhatk, vecb= yhati + zhatj + xhatk ` and `vecc= zhati + xhatj+ yhatk `, `" then " vecaxx (vecbxx vecc) `is

Options

A. A. parallel to ` ( y- z) hati + (z - x) hatj + (x -y) hatk`
B. B. orthogonal to `hati + hatj + hatk`
C. C. orthogonal to ` ( y+z) hati + ( z + x) hatj + ( x + y) hatk`
D. D. orthogonal to `xhati + yhatj + zhatk`

Answer

a,b,c,d

Explanation

`veca xx (vecbxxvecc0 = (veca .vecc) vecb- (veca .vecb)vecc`
`= (yz+yx +zx) {(y-z)hati+(z-x)hatj +(x-y)hatk}`
Clearly vector is parallel to `(y-z) hati + (z-x) hatj + (x-y) hatk`
it is orthogonal to `hati+hatj +hatk as (y-z) (1) + (z-x) (1) + (x-y) (1) =0`
it is orhtogonal to `( y+z) hati+ (z+x) hatj + (x +y) hatk`
as (y-z) (y+z) + (z+x) + (x-y) (x+y)
` = y^(2) -z^(2) +z^(2)- x^(2) +x^(2) -y^(2) =0`
Also it is orthogonal to `xhati + yhatj + zhatk`