Let A(4,7,8) B(2,3,4) and C(2,5,7) be the position vectors of the vertices of a triangle ABC. Then the length of the internal angular bisector of angle A is

Let A (4, 7, 8), B (2, 3, 4) and C (2, 5, 7) be the position vectors of the vertices of a triangle ABC. The length of the internal bisector of the angle at A is

A(1,-1,-3), B(2, 1,-2) & C(-5, 2,-6) are the position vectors of the vertices of a triangle ABC. The length of the bisector of its internal angle at A is :

A(1,-1,-3), B(2, 1,-2) & C(-5, 2,-6) are the position vectors of the vertices of a triangle ABC. The length of the bisector of its internal angle at A is :

A(1,-1,-3), B(2, 1,-2) & C(-5, 2,-6) are the position vectors of the vertices of a triangle ABC. The length of the bisector of its internal angle at A is :

A(1,-1,-3), B(2, 1,-2) & C(-5, 2,-6) are the position vectors of the vertices of a triangle ABC. The length of the bisector of its internal angle at A is :

A(1,-1,-3),B(2,1,-2)&C(-5,2,-6) are the position vectors of the vertices of a triangle ABC.The length of the bisector of its internal angle at A is :

If A(3,1,-1),B(4,-1,1),C(5,2,1) be the vertices of a Delta ABC then the length of the internal bisector of angle A is

If 4i+7j+8k , 2i+3j+4k and 2i+5j+7k are the position vectors of the vertices A,B and C of triangle ABC ,The position vector of the point where the bisector of /_A meets BC is

A (1,3) and C (-2/5,-2/5) are the vertices of a triangle ABC and the equation of the internal angle bisector of angleABC is x+y =2 Equation of BC is