TS EAMCET · Maths · Three Dimensional Geometry
If \(\mathrm{A}(0,3,4), \mathrm{B}(1,5,6), \mathrm{C}(-2,0,-2)\) are the vertices of a triangle ABC and the bisector of angle A meets the side BC at D, then \(\mathrm{AD}=\)
- A \(\frac{\sqrt{21}}{5}\)
- B \(\frac{\sqrt{42}}{10}\)
- C \(10\)
- D \(4\)
Answer & Solution
Correct Answer
(B) \(\frac{\sqrt{42}}{10}\)
Step-by-step Solution
Detailed explanation
\( \vec{AB} = B - A = (1-0, 5-3, 6-4) = (1, 2, 2) \) \( |\vec{AB}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3 \) \( \vec{AC} = C - A = (-2-0, 0-3, -2-4) = (-2, -3, -6) \) \( |\vec{AC}| = \sqrt{(-2)^2 + (-3)^2 + (-6)^2} = \sqrt{4+9+36} = \sqrt{49} = 7 \) By angle…
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