A vector `(vecd)` is equally inclined to three vectors `veca= hati - hatj + hatk, vecb = 2 hati + hatj and vecc = 3hatj - 2hatk` let `vecx ,vecy, vecz` be three in the plane of `veca, vecb , vecb, vecc, vecc, veca` respectively, then

Options

A. `vecz.vecd =0`
B. `vecx.vecd=1`
C. `vecy.vecd= 32`
D. `vecr.vecd= 0 , " where " vecr= lambdavecx+ muvecy+ gammavecz`

Answer

a,d

Explanation

` veca =hati -hatj +hatk`
` vecb = 2hati +hatj`
`and vecc = 3hatj -2hatk`
since, `[veca vecb vecc]= |{:(1,-1,1),(2,1,0),(0,3,-2):}|=0`
therefore, `veca,vecb and vecc` are coplanar vectors, furthere, sicne `vecd` is equally inclined to `veca, vecb anv vecc` we have
`vecd. veca=vecd.vecb=vecd.vecc=0`
` vecd.vecx=vecd.vecy =vecd.vecz=0`,
`vecd.vecr =0`