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 \(\mathrm{CD}=\)
- A \(\frac{\sqrt{5}}{4}\)
- B \(\frac{3 \sqrt{5}}{4}\)
- C \(2 \sqrt{5}\)
- D \(\frac{5 \sqrt{5}}{2}\)
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:…
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