Question

The length of the internal bisector of angle A in \( \triangle ABC \) with vertices \( A(4,7,8) \), \( B(2,3,4) \), and \( C(2,5,7) \) is:

• A. \( \frac{1}{3} \sqrt{29} \)
• B. \( \frac{2}{3} \sqrt{29} \)
• C. \( \frac{2}{3} \sqrt{34} \)
• D. \( \frac{4}{3} \sqrt{34} \)

Hint:

For 3D distance calculations, use the Euclidean formula \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} \) and apply the angle bisector formula.

Answer 1

The Correct Option is C

To find the length of the internal bisector of angle \( A \) in \( \triangle ABC \) with vertices \( A(4,7,8) \), \( B(2,3,4) \), and \( C(2,5,7) \), follow these steps: 1. Calculate the lengths of sides \( AB \) and \( AC \):
- Length of \( AB \): \[ AB = \sqrt{(4
-2)^2 + (7
-3)^2 + (8
-4)^2} = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 \]
- Length of \( AC \): \[ AC = \sqrt{(4
-2)^2 + (7
-5)^2 + (8
-7)^2} = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \] 2. Use the Angle Bisector Theorem: The internal bisector of angle \( A \) divides the opposite side \( BC \) in the ratio of the adjacent sides \( AB \) and \( AC \). Let the bisector intersect \( BC \) at point \( D \). Then: \[ \frac{BD}{DC} = \frac{AB}{AC} = \frac{6}{3} = 2 \] Therefore, \( BD = 2 \cdot DC \). 3. Find the coordinates of point \( D \): The coordinates of \( D \) can be found using the section formula. Given \( B(2,3,4) \) and \( C(2,5,7) \), and \( BD:DC = 2:1 \): \[ D = \left( \frac{2 \cdot 2 + 1 \cdot 2}{2 + 1}, \frac{2 \cdot 3 + 1 \cdot 5}{2 + 1}, \frac{2 \cdot 4 + 1 \cdot 7}{2 + 1} \right) = \left( \frac{4 + 2}{3}, \frac{6 + 5}{3}, \frac{8 + 7}{3} \right) = \left( 2, \frac{11}{3}, 5 \right) \] 4. Calculate the length of the angle bisector \( AD \): The length of \( AD \) is the distance between \( A(4,7,8) \) and \( D(2, \frac{11}{3}, 5) \): \[ AD = \sqrt{(4
-2)^2 + \left(7
- \frac{11}{3}\right)^2 + (8
-5)^2} = \sqrt{2^2 + \left(\frac{21}{3}
- \frac{11}{3}\right)^2 + 3^2} = \sqrt{4 + \left(\frac{10}{3}\right)^2 + 9} \] \[ AD = \sqrt{4 + \frac{100}{9} + 9} = \sqrt{\frac{36}{9} + \frac{100}{9} + \frac{81}{9}} = \sqrt{\frac{217}{9}} = \frac{\sqrt{217}}{3} \] However, let's verify the calculation: \[ AD = \sqrt{(4
-2)^2 + \left(7
- \frac{11}{3}\right)^2 + (8
-5)^2} = \sqrt{4 + \left(\frac{10}{3}\right)^2 + 9} = \sqrt{4 + \frac{100}{9} + 9} = \sqrt{\frac{36 + 100 + 81}{9}} = \sqrt{\frac{217}{9}} = \frac{\sqrt{217}}{3} \] But \( \sqrt{217} \) is not a perfect square, so let's re
-examine the problem. 5. Re
-evaluate the calculation: The correct approach is to use the formula for the length of the angle bisector in a triangle: \[ AD = \frac{2 \cdot AB \cdot AC \cdot \cos\left(\frac{A}{2}\right)}{AB + AC} \] However, a simpler formula for the length of the internal bisector is: \[ AD = \sqrt{AB \cdot AC \left(1
- \frac{BC^2}{(AB + AC)^2}\right)} \] First, calculate \( BC \): \[ BC = \sqrt{(2
-2)^2 + (3
-5)^2 + (4
-7)^2} = \sqrt{0 + 4 + 9} = \sqrt{13} \] Then: \[ AD = \sqrt{6 \cdot 3 \left(1
- \frac{13}{(6 + 3)^2}\right)} = \sqrt{18 \left(1
- \frac{13}{81}\right)} = \sqrt{18 \cdot \frac{68}{81}} = \sqrt{\frac{1224}{81}} = \sqrt{\frac{136}{9}} = \frac{\sqrt{136}}{3} = \frac{2\sqrt{34}}{3} \] Final Answer: \boxed{\frac{2}{3} \sqrt{34}}