ExamBro
ExamBro
TS EAMCET · Maths · Three Dimensional Geometry

Let \(A(4,3,5), B(1,-2,1), C(3,2,1)\) be the vertices of a triangle \(\mathrm{ABC}\). If the internal bisector of \(\angle \mathrm{BAC}\) meet the side \(\mathrm{BC}\) at \(\mathrm{D}\), then \(\mathrm{CD}=\)

  1. A \(\frac{\sqrt{5}}{4}\)
  2. B \(\frac{3 \sqrt{5}}{4}\)
  3. C \(2 \sqrt{5}\)
  4. D \(\frac{5 \sqrt{5}}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{3 \sqrt{5}}{4}\)

Step-by-step Solution

Detailed explanation

\[ \begin{aligned} A B= & \sqrt{3^2+5^2+4^2}=5 \sqrt{2} \\ A C= & \sqrt{1^2+1^2+4^2}=3 \sqrt{2} \\ & \frac{A B}{A C}=\frac{B D}{D C} \\ \Rightarrow & \frac{5 \sqrt{2}}{3 \sqrt{2}}=\frac{B D}{D C}=\frac{5}{3} \end{aligned} \] \(\therefore\) Co-ordinates of \(D\) are:…