The position vectors of the vertices `A ,B ,a n dC` of a triangle are ` hat i+ hat j , hat j+ hat ka n d hat i+ hat k` , respectively. Find the unite vector ` hat r` lying in the plane of `A B C` and perpendicular to `I A ,w h e r eI` is the incentre of the triangle.

Options

A. `(3pi)/(4)`
B. ` (pi)/(4)`
C. `(pi)/(2)`
D. `pi`

Answer

A

Explanation

Since `vec(a) xx (vec(b) xx vec(c )) = (vec(b) + vec(c ))/(sqrt(2))`
` rArr (vec(a) " ." vec(c )) vec(b) - (vec(a) ". " vec(b)) vec(c ) = (1)/(sqrt(2)) vec(b) + (1)/(sqrt(2)) vec( c)`
On equating the coefficient of `vec(c )` we get
`rArr |vec(a)||vec(b)|| cos 0 = (1)/(sqrt(2))`
` :. cos 0 = (1)/(sqrt(2)) rArr 0= (3pi)/(4)`