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: Use the angle bisector theorem.
Since the bisector of \(\angle BAC\) meets \(BC\) at \(D\), by the angle bisector theorem, \[ \frac{BD}{DC}=\frac{AB}{AC}. \]

Step 2: Find \(AB\).
\[ A(7,6,4),\quad B(5,4,6). \] \[ AB=\sqrt{(5-7)^2+(4-6)^2+(6-4)^2}. \] \[ AB=\sqrt{(-2)^2+(-2)^2+2^2}. \] \[ AB=\sqrt{4+4+4}. \] \[ AB=2\sqrt3. \]

Step 3: Find \(AC\).
\[ A(7,6,4),\quad C(3,2,0). \] \[ AC=\sqrt{(3-7)^2+(2-6)^2+(0-4)^2}. \] \[ AC=\sqrt{(-4)^2+(-4)^2+(-4)^2}. \] \[ AC=\sqrt{16+16+16}. \] \[ AC=4\sqrt3. \]

Step 4: Find the ratio \(BD:DC\).
\[ \frac{BD}{DC}=\frac{AB}{AC}. \] \[ BD:DC=2\sqrt3:4\sqrt3. \] \[ BD:DC=1:2. \]

Step 5: Use section formula.
Since \(D\) divides \(BC\) internally in the ratio \[ 1:2, \] where \[ B(5,4,6),\quad C(3,2,0), \] the coordinates of \(D\) are \[ D=\left(\frac{1\cdot 3+2\cdot 5}{1+2},\frac{1\cdot 2+2\cdot 4}{1+2},\frac{1\cdot 0+2\cdot 6}{1+2}\right). \] \[ D=\left(\frac{3+10}{3},\frac{2+8}{3},\frac{12}{3}\right). \] \[ D=\left(\frac{13}{3},\frac{10}{3},4\right). \]

Step 6: Final conclusion.
Therefore, \[ \boxed{\left(\frac{13}{3},\frac{10}{3},4\right)} \]