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Question
A(1, 2, 3), B(0, 4, 1), C(–1, –1, –3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC.
Solution
AB =\[\sqrt{\left( - 1 \right)^2 + 2^2 + \left( - 2 \right)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3\]
AC =\[\sqrt{\left( - 2 \right)^2 + \left( - 3 \right)^2 + \left( - 6 \right)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\]
AD is the internal bisector of\[∠BAC\]
\[\therefore \frac{BD}{DC} = \frac{AB}{AC} = \frac{3}{7}\]
Thus, D divides BC internally in the ratio 3:7.
\[\therefore D \equiv \left( \frac{3\left( - 1 \right) + 3 \times 0}{3 + 7}, \frac{3\left( - 1 \right) + 7\left( 4 \right)}{3 + 7}, \frac{3\left( - 3 \right) + 7 \times 1}{3 + 7} \right)\]
\[ \Rightarrow D \equiv \left( \frac{- 3}{10}, \frac{25}{10}, \frac{- 2}{10} \right)\]
\[ \Rightarrow D \equiv \left( \frac{- 3}{10}, \frac{5}{2}, \frac{- 1}{5} \right) \]
\[\]
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