A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle BAC meets BC.

Given: The vertices of the triangle are A(1, 2, 3), B(0, 4, 1) and C(-1, -1, -3)


To find: the coordinates of D


Formula used:


Distance Formula:


The distance between any two points (a, b, c) and (m, n, o) is given by,



Section Formula:


A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).


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The coordinates of C is given by,



We know angle bisector divides opposite side in the ratio of the other two sides.


As AD is angle bisector of A and meets BC at D


BD : DC = AB : BC



Distance between A(1, 2, 3) and B(0, 4, 1) is AB,






= 3


Distance between A(1, 2, 3) and C(-1, -1, -3) is AC,






= 7



AB : AC = 3:7


BD: DC = 3:7


Therefore, m = 3 and n = 7


B(0, 4, 1) and C(-1, -1, -3)


Coordinates of D using section formula:






Hence, Coordinates of D are


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