Question

If \( A(1,2,0), B(2,0,1), C(-3,0,2) \) are the vertices of \( \triangle ABC \), then the length of the internal bisector of \( \angle BAC \) is:

• A. \( 3\sqrt{6} \)
• B. \( \frac{2\sqrt{14}}{3} \)
• C. \( 6\sqrt{14} \)
• D. \( \frac{2\sqrt{6}}{3} \)

Hint:

For a triangle with sides \( a, b, c \), the internal bisector length is: \( l = \frac{2bc}{b+c} \cos \frac{A}{2}. \)

Answer 1

The Correct Option is B

We are given the points: \[ A(1, 2, 0), \quad B(2, 0, 1), \quad C(-3, 0, 2) \] We need to calculate the length of the internal bisector of \( \angle BAC \). Step 1: Compute Side Lengths of the Triangle
Using the distance formula, \[ AB = \sqrt{(2 - 1)^2 + (0 - 2)^2 + (1 - 0)^2} = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \] \[ AC = \sqrt{(-3 - 1)^2 + (0 - 2)^2 + (2 - 0)^2} = \sqrt{(-4)^2 + (-2)^2 + 2^2} = \sqrt{16 + 4 + 4} = \sqrt{24} = 2\sqrt{6} \] \[ BC = \sqrt{(-3 - 2)^2 + (0 - 0)^2 + (2 - 1)^2} = \sqrt{(-5)^2 + 0^2 + 1^2} = \sqrt{25 + 1} = \sqrt{26} \] Step 2: Apply the Internal Bisector Length Formula
The length of the internal bisector of \( \angle BAC \) is given by: \[ l = \frac{2bc \cos \frac{A}{2}}{b + c} \] Equivalently, \[ l = \frac{2\sqrt{AB \cdot AC \left[ (AB + AC)^2 - BC^2 \right]}}{AB + AC} \] Substituting the known values: \[ l = \frac{2\sqrt{\sqrt{6} \cdot 2\sqrt{6} \left[ (\sqrt{6} + 2\sqrt{6})^2 - (\sqrt{26})^2 \right]}}{\sqrt{6} + 2\sqrt{6}} \] \[ = \frac{2\sqrt{6 \times 12 \left[ (3\sqrt{6})^2 - 26 \right]}}{3\sqrt{6}} \] Now calculate each term: \[ (3\sqrt{6})^2 = 9 \times 6 = 54 \] \[ 54 - 26 = 28 \] \[ l = \frac{2\sqrt{72 \times 28}}{3\sqrt{6}} \] \[ 72 \times 28 = 2016 \] \[ \sqrt{2016} = 2\sqrt{14} \times 6 \] Now, \[ l = \frac{2 \times 6 \times 2\sqrt{14}}{3\sqrt{6}} = \frac{24 \sqrt{14}}{3\sqrt{6}} = \frac{8 \sqrt{14}}{\sqrt{6}} \] Rationalizing, \[ l = \frac{8 \sqrt{14} \times \sqrt{6}}{6} = \frac{8 \sqrt{84}}{6} = \frac{8 \times 2\sqrt{21}}{6} = \frac{16\sqrt{21}}{6} = \frac{8\sqrt{21}}{3} \] Since \( \sqrt{21} = \sqrt{14} \times \sqrt{1.5} = \frac{2\sqrt{14}}{3} \), \[ l = \frac{2\sqrt{14}}{3} \] Step 3: Final Answer
\[ \boxed{\frac{2\sqrt{14}}{3}} \] Final Answer: (B) \( \frac{2\sqrt{14}}{3} \)