The position vectors of the vertices `A, B and C` of triangle are `hati+hatj, hatj+hatk and hati+hatk`, respectively. Find the unit vectors `hatr` lying in the plane of `ABC` and perpendicular to `IA`, where I is the incentre of the triangle.

Answer

`vecr = pm (1)/(sqrt2)(hati +hatj)`

Explanation

Since `|vec(AB)|= |vec(BC)|= |vec(CA)|`, the incentre is same as the circumcentre, and hence IA is perpendicular of BC. Therefore, `vecr` is parallel to BC.
`vecr= lamda (hati-hatj)`
Hence, unit vector `vecr=pm (1)/(sqrt2)(hati-hatj)`