The points \left(\right. 3 , \textrm{ } 2 , \textrm{ } 0 \left.\right) , \textrm{ } \left(\right. 5 , \textrm{ } 3 , \textrm{ } 2 \left.\right) and \left(\right. - 9 , \textrm{ } 6 , \textrm{ } - 3 \left.\right), are the vertices of a triangle A B C . A D is the internal bisector of \angle \textrm{ } B A C which meets B C at D. Then the co-ordinates of D, are

Engineering

Mathematics

Line of Greatest Slope

Question

The points

(3, 2, 0), (5, 3, 2)

and

(- 9, 6, - 3)

, are the vertices of a triangle

A B C . A D

is the internal bisector of

∠ B A C

which meets

B C

D

. Then the co-ordinates of

D

, are

[\frac{17}{16}, \frac{57}{16}, \frac{19}{8}]

[\frac{19}{8}, \frac{57}{16}, \frac{17}{16}]

[0, 0, \frac{17}{16}]

[\frac{17}{16}, 0, 0]

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A D

is bisector of

∠ B A C

\Rightarrow

Ratio at

D

c : b

where

c = A B = \sqrt{(3 - 5)^{2} + (2 - 3) + (0 - 2)^{2}}

= \sqrt{4 + 1 + 4} = \sqrt{9} = 3

b = A C = \sqrt{(3 + 9)^{2} + (2 - 6)^{2} + (0 + 3)^{2}}

= \sqrt{144 + 16 + 9} = \sqrt{169} = 13

for point

D

x = \frac{c (- 9) + b (6)}{c + b} = \frac{3 (- 9) + 13 (5)}{3 + 13} = \frac{38}{16} = \frac{19}{8}

y = \frac{c (6) + b (3)}{c + b} = \frac{3 (6) + 13 (3)}{3 + 13} = \frac{57}{16}

z = \frac{c (- 3) + b (2)}{c + b} = \frac{3 (- 3) + 13 (2)}{3 + 13} = \frac{17}{16}

Hence, point

D

[\frac{19}{8}, \frac{57}{16}, \frac{17}{6}]

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