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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.

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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).

The coordinates of C is given by,

 \(\Big(\frac{nx+ma}{m+n},\frac{ny+mb}{m+n},\frac{nz+mc}{m+n}\Big)\)  

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,

 = \(\sqrt{9}\)

= 3

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

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 \(\Big(\frac{-3}{10}, \frac{5}{2}, \frac{-1}{5}\Big)\) 

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