Question

If the coordinates of the vertices of a triangle \(ABC\) are \[ A(7,6,4),\quad B(5,4,6),\quad C(3,2,0) \] and the bisector of \(\angle BAC\) meets the side \(BC\) at \(D\), then the coordinates of \(D\) are:

• A. \(\left(\frac{13}{3},\frac{10}{3},4\right)\)
• B. \(\left(\frac{11}{3},\frac{8}{3},2\right)\)
• C. \((9,8,6)\)
• D. \((7,5,3)\)

Hint:

For an internal angle bisector from \(A\) meeting \(BC\) at \(D\), use \(\frac{BD}{DC}=\frac{AB}{AC}\), then apply the internal section formula.

Answer 1

The Correct Option is A

Step 1: Apply the angle bisector theorem.
In triangle ABC, the internal bisector of ∠BAC meets BC at D, so BD/DC = AB/AC.

Step 2: Compute the lengths AB and AC.
A(7, 6, 4), B(5, 4, 6): AB = √[(5-7)² + (4-6)² + (6-4)²] = √(4 + 4 + 4) = 2√3. A(7, 6, 4), C(3, 2, 0): AC = √[(3-7)² + (2-6)² + (0-4)²] = √(16 + 16 + 16) = 4√3.

Step 3: Determine the division ratio.
BD : DC = 2√3 : 4√3 = 1 : 2.

Step 4: Use the section formula to find D.
D divides BC internally in the ratio 1:2. D = [(1·3 + 2·5)/3, (1·2 + 2·4)/3, (1·0 + 2·6)/3] = (13/3, 10/3, 4).

Step 5: Final conclusion.
The coordinates of D are (13/3, 10/3, 4).