If `vec a` and `vec b` are two unit vectors and `theta` is the angle between them, then the unit vector along the angular bisector of `vec a` and `vec b` will be given by

Options

A. `[vecn_(2)vecn_(3)vecn_(4)](vecr.vecn_(1)-q_(1))=[vecn_(1)vecn_(3)vecn_(4)](vecr.vecn_(2)-q_(2))`
B. `[vecn_(1)vecn_(2)vecn_(3)](vecr.vecn_(4)-q_(4))=[vecn_(4)vecn_(3)vecn_(1)](vecr.vecn_(2)-q_(2))`
C. `[vecn_(4)vecn_(3)vecn_(1)](vecr.vecn_(4)-q_(4))=[vecn_(1)vecn_(2)vecn_(3)](vecr.vecn_(2)-q_(2))`
D. none of these

Answer

a

Explanation

`vecr*vec(n_(1))+lamdavecr*vec(n_(2))= q_(1)+ lamdaq_(2)" "` (i)
where `lamda` is a parameter.
So `vec(n_(1))+ lamda vec(n_2)` is normal to plane (i). Now, any plane parallel to the line of intersection of the planes `vecr`.
`vec(n_3)= q_(3) and vecr*vec(n_(4)) = q_(4)` is of form `vecr*(vec(n_(3))xxvec(n_(4)))=d`. Hence, we must have
`" "[vec(n_1)+lamdavec(n_2)]*[vec(n_3)xxvec(n_4)]=0`
or `" "[vec(n_1)vec(n_3)vec(n_4)] + lamda[vec(n_2)vec(n_3)vec(n_4)]=0`
or `" " lamda = (-[vec(n_1)vec(n_2)vec(n_4)])/([vec(n_1)vec(n_3)vec(n_4)] )`
On putting this value in Eq. (i), we have the equation of the required plane as
`" "vecr*vec(n_1)-q_1= ([vec(n_1)vec(n_3)vec(n_4)])/([vec(n_2)vec(n_3)vec(n_4)])(r*vec(n_2)-q_2)`
or `" "[vec(n_2) vec(n_3)vec(n_4)](vecr*vec(n_1)-q_1)`
`" "=[vec(n_1)vec(n_3)vec(n_4)] (vecr*vec(n_2) - q_2) `