Question

A(3,2,0), B(5,3,2), C(-9,6,-3) are three points forming a triangle. AD, the bisector of angle $BAC$ meets BC in D. Find the coordinates of D:

• A. $\left( \frac{19}{8}, \frac{57}{15}, \frac{57}{15} \right)$
• B. $\left( \frac{19}{8}, \frac{57}{16}, \frac{17}{16} \right)$
• C. (2,3,0)
• D. (4,5,6)

Hint:

The angle bisector theorem is a powerful tool for solving triangle-related coordinate geometry problems.

Answer 1

The Correct Option is B

Since \( AD \) is the bisector of \( \angle BAC \), we use the angle bisector theorem:
\[ \frac{BD}{DC} = \frac{AB}{AC} \quad {...(i)} \] Now, we calculate the lengths of \( AB \) and \( AC \).

Calculating \( AB \)

\[ AB = \sqrt{(5 - 3)^2 + (3 - 2)^2 + (2 - 0)^2} \] \[ = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

Calculating \( AC \)

\[ AC = \sqrt{(-9 - 3)^2 + (6 - 2)^2 + (-3 - 0)^2} \] \[ = \sqrt{144 + 16 + 9} \] \[ = \sqrt{169} = 13 \]

Finding the Ratio of \( BD : DC \)

From equation (i): \[ \frac{BD}{DC} = \frac{3}{13} \] Thus, \( D \) divides \( BC \) in the ratio \( 3:13 \).

Finding the Coordinates of \( D \)

Using the section formula: \[ D \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n} \right) \] where \( B(-9,6,-3) \) and \( C(5,3,2) \), and the ratio \( m:n = 3:13 \), \[ x = \frac{3(-9) + 13(5)}{3+13}, \quad y = \frac{3(6) + 13(3)}{3+13}, \quad z = \frac{3(-3) + 13(2)}{3+13} \] \[ = \left( \frac{-27 + 65}{16}, \frac{18 + 39}{16}, \frac{-9 + 26}{16} \right) \] \[ = \left( \frac{19}{8}, \frac{57}{16}, \frac{17}{16} \right) \] Thus, the coordinates of \( D \) are: \[ D \left( \frac{19}{8}, \frac{57}{16}, \frac{17}{16} \right) \]

Answer 2

Step 1: Understanding the Problem
We are given three points:
A(3, 2, 0), B(5, 3, 2), and C(–9, 6, –3).
AD is the angle bisector of angle BAC, and it meets BC at point D.
We are to find the coordinates of point D.

Step 2: Use of Internal Angle Bisector Theorem
According to the internal angle bisector theorem in 3D geometry, the angle bisector from point A to side BC divides BC in the ratio of the lengths of the adjacent sides AB and AC.

Step 3: Find Magnitudes of AB and AC
AB = B - A = (5 - 3, 3 - 2, 2 - 0) = (2, 1, 2)
|AB| = √(2² + 1² + 2²) = √(4 + 1 + 4) = √9 = 3

AC = C - A = (–9 - 3, 6 - 2, –3 - 0) = (–12, 4, –3)
|AC| = √(144 + 16 + 9) = √169 = 13

Therefore, point D divides BC in the ratio AB:AC = 3:13.

Step 4: Use Section Formula to Find Coordinates of D
Let D divide BC in the ratio 3:13. Then coordinates of D are:
\[ D = \left( \frac{13 \cdot 5 + 3 \cdot (-9)}{3 + 13}, \frac{13 \cdot 3 + 3 \cdot 6}{3 + 13}, \frac{13 \cdot 2 + 3 \cdot (-3)}{3 + 13} \right) \]
Now compute each component:
x = \( \frac{65 - 27}{16} = \frac{38}{16} = \frac{19}{8} \)
y = \( \frac{39 + 18}{16} = \frac{57}{16} \)
z = \( \frac{26 - 9}{16} = \frac{17}{16} \)

Step 5: Final Answer
Therefore, the coordinates of D are:
\[ \left( \frac{19}{8}, \frac{57}{16}, \frac{17}{16} \right) \]