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Question: Answered & Verified by Expert
$\mathrm{A}(2,1,2), \mathrm{B}(1,0,0), C(1+\sqrt{3}, \sqrt{3},-\sqrt{6})$ are vertices of a triangle. If the length of the median drawn through $\mathrm{A}$ is $\lambda \sqrt{9-2 \sqrt{3}+2 \sqrt{6}}$ then $\lambda=$
MathematicsVector AlgebraAP EAMCETAP EAMCET 2022 (08 Jul Shift 1)
Options:
  • A $4$
  • B $3$
  • C $2$
  • D $1$
Solution:
2362 Upvotes Verified Answer
The correct answer is: $4$


$\mathrm{D}\left[\frac{2+\sqrt{3}}{2}, \frac{\sqrt{3}}{2}, \frac{-\sqrt{6}}{2}\right]$
$\begin{aligned} & \therefore \mathrm{AD}=\sqrt{\left[2-\frac{(2+\sqrt{3})}{2}\right]^2+\left(1-\frac{\sqrt{3}}{2}\right)^2+\left(2+\frac{\sqrt{6}}{2}\right)^2} \\ & \therefore \mathrm{AD}=\frac{1}{2} \sqrt{2(2-\sqrt{3})^2+(4+\sqrt{6})^2}\end{aligned}$
$\begin{aligned} & \Rightarrow \mathrm{AD}=\frac{1}{2} \sqrt{2(7-4 \sqrt{3})+22+8 \sqrt{6}} \\ & =\frac{1}{2} \times 2 \sqrt{9-2 \sqrt{3}+2 \sqrt{6}} \\ & =\sqrt{9-2 \sqrt{3}+2 \sqrt{6}} \\ & \Rightarrow \lambda=1\end{aligned}$

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