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TS EAMCET · Maths · Three Dimensional Geometry

If \(A(4,3,2), B(5,4,6), C(-1,-1,5)\) are the vertices of a triangle, then the coordinates of the point in which the bisector of the angle \(A\) meet the side \(B C\) is

  1. A \(\left(\frac{22}{8}, \frac{17}{8}, \frac{45}{8}\right)\)
  2. B \(\left(\frac{17}{8}, \frac{22}{8}, \frac{45}{8}\right)\)
  3. C \(\left(\frac{-22}{8}, \frac{-17}{8}, \frac{45}{8}\right)\)
  4. D \(\left(\frac{-17}{8}, \frac{22}{8}, \frac{45}{8}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\left(\frac{22}{8}, \frac{17}{8}, \frac{45}{8}\right)\)

Step-by-step Solution

Detailed explanation

Given points are \(A(4,3,2), B(5,4,6), \mathrm{C}(-1,-1,5)\) \(A B=\sqrt{(5-4)^2+(4-3)^2+(6-2)^2}\) \(=\sqrt{(1)^2+(1)^2+(4)^2}=\sqrt{18}\) \(A B=3 \sqrt{2}\) \(A C=\sqrt{(-1-4)^2+(-1-3)^2+(5-2)^2}\) \(=\sqrt{(-5)^2+(-4)^2+(3)^2}=\sqrt{25+16+9}=\sqrt{50}\) \(A C=5 \sqrt{2}\)…