If the position vector of a point p with respect to the origin O is `hat(i) + 3hat(j) -2hat(k)` and that of a point Q is `3hat(i) + hat(j)- 2hat(k)`, then what is the position vector of the bisector of the `anglePOQ` ?

Options

A. `hat(i)-hat(j)-hat(k)`
B. `hat(i)+hat(j)-hat(k)`
C. `hat(i)+hat(j)+hat(k)`
D. None of these

Answer

B

Explanation

Let `vec(OP)=hat(i)+3hat(j)-2hat(k)and vec(OQ)=3hat(i)+hat(j)-2hat(k)`
Let `hat(i)+hat(j)-hat(k)` be required position vector of the bisector of the angle POQ since, it is the bisector of `anglePOQ` therefore. It will make equal angles with `vec(OP)` and `vec(OQ)`.
Let Angle between `hat(i)+3hat(j)-2hat(k) and hat(i)+hat(j)-hat(k)` is
`theta =cos^(-1)((1+3+2)/(sqrt(1+9+4)sqrt(1+1+1)))`
`=cos^(-1)((6)/(sqrt(14)sqrt(3)))`
and angle between `3hat(i)+hat(j)-2hat(k) and hat(i)+hat(j)-hat(k)`, is
`phi = cos ^(-1)((1+3+2)/(sqrt(9+1+4)sqrt(1+1+1)))=cos^(-1)((6)/(sqrt(14)sqrt(3)))`
Hence, `theta= phi`