The vertices of a triangle ABC are A(3, 2, 0), B(5, 3, 2) and C(-9, 6, -3). The bisector AD of A meets BC at D, find the fourth vertex D.

The given co-ordinates: A(3, 2, 0), B(5, 3, 2) and C(-9, 6, -3)


Now, AB = 3


Also, AC


Now, we have,


By the property of internal angle bisector,



Therefore,


Applying the section formula, we get,


D(x, y, z)


D(x, y, z)


Answer.


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