The vertices of a triangle are A(5,4,6) ,B(1,-1,3) C(4,3,2) The internal bisector of angleA meetsBC at D Find - Maths - Introduction to Three Dimensional Geometry

Question

The vertices of a triangle are A(5,4,6)
,B(1,-1,3)&C(4,3,2) The internal bisector of angleA meetsBC at
D Find the coordinates of D and the length AD

Aakash Sharma · Accepted Answer

Dear Student!

Here is the answer to your query.

Given : The coordinates of A, B, C are (5, 4, 6), (1, –1, 3), (4, 3, 2) respectively and angle bisector of A meets BC at D.

Then $\frac{BD}{CD} = \frac{AB}{AC}$

$\Rightarrow \frac{BD}{CD} = \frac{\sqrt{(1 - 5)^{2} + (- 1 - 4)^{2} + (3 - 6)^{2}}}{\sqrt{(4 - 5)^{2} + (3 - 4)^{2} + (2 - 6)^{2}}} = \frac{\sqrt{4^{2} + (- 5)^{2} + (- 3)^{2}}}{\sqrt{(- 1)^{2} + (- 1)^{2} + (- 4)^{2}}} = \frac{\sqrt{16 + 25 + 9}}{\sqrt{1 + 1 + 16}} = \frac{\sqrt{50}}{\sqrt{18}} = \frac{5 \sqrt{2}}{3 \sqrt{2}} = \frac{5}{3}$

∴ BD : CD = 5 : 3

Now using section formulae coordinates of D are

$(\frac{5 \times 4 + 3 \times 1}{5 + 3}, \frac{5 \times 3 + 3 \times (- 1)}{5 + 3}, \frac{5 \times 2 + 3 \times 3}{5 + 3}) = (\frac{20 + 3}{8}, \frac{15 - 3}{8}, \frac{10 + 9}{8}) = (\frac{23}{8}, \frac{3}{2}, \frac{19}{8}) ∴ The coordinates of D are (\frac{23}{8}, \frac{3}{2}, \frac{19}{8}) and AD = \sqrt{(\frac{23}{8} - 5)^{2} + (\frac{3}{2} - 4)^{2} + (\frac{19}{8} - 6)^{2}} = \sqrt{(\frac{23 - 40}{8})^{2} + (\frac{3 - 8}{2})^{2} + (\frac{19 - 48}{8})^{2}} = \sqrt{(\frac{- 17}{8})^{2} + (\frac{- 5}{2})^{2} + (\frac{- 29}{8})^{2}} = \sqrt{\frac{289}{64} + \frac{25}{4} + \frac{841}{64}} = \sqrt{\frac{289 + 400 + 841}{64}} = \sqrt{\frac{1530}{64}} = \frac{3 \sqrt{170}}{8} units H ence the length of AD is \frac{3 \sqrt{170}}{8} units$

Cheers!

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The vertices of a triangle are A(5,4,6) ,B(1,-1,3)&C(4,3,2). The internal bisector of angleA meetsBC at D. Find the coordinates of D and the length AD