Question

\(A(2,1,2), B(1,0,0), C(1+\sqrt{3},\sqrt{3},-\sqrt{6})\) are vertices of a triangle. If the length of the median drawn through \(A\) is \(\lambda\sqrt{9-2\sqrt{3}+2\sqrt{6}}\), then \(\lambda=\)

• A. \(4\)
• B. \(3\)
• C. \(2\)
• D. \(1\)

Hint:

To find the length of a median in coordinate geometry, first find the midpoint of the opposite side and then apply the distance formula between the vertex and that midpoint.

Answer 1

The Correct Option is D

Step 1: Find the midpoint of \(BC\).
Let the midpoint of \(BC\) be \(M\).
Since, \[ B(1,0,0) \] and \[ C(1+\sqrt{3},\sqrt{3},-\sqrt{6}) \] Therefore, \[ M=\left(\frac{1+1+\sqrt{3}}{2},\frac{0+\sqrt{3}}{2},\frac{0-\sqrt{6}}{2}\right) \] So, \[ M=\left(1+\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2},-\frac{\sqrt{6}}{2}\right) \]

Step 2: Use the distance formula for median from \(A\) to \(M\).
The median drawn through \(A\) is \(AM\).
Now, \[ A=(2,1,2) \] Hence, \[ AM=\sqrt{\left(1+\frac{\sqrt{3}}{2}-2\right)^2+\left(\frac{\sqrt{3}}{2}-1\right)^2+\left(-\frac{\sqrt{6}}{2}-2\right)^2} \] \[ AM=\sqrt{\left(\frac{\sqrt{3}}{2}-1\right)^2+\left(\frac{\sqrt{3}}{2}-1\right)^2+\left(-\frac{\sqrt{6}}{2}-2\right)^2} \]

Step 3: Simplify the expression.
First, \[ \left(\frac{\sqrt{3}}{2}-1\right)^2 = \frac{3}{4}-\sqrt{3}+1 \] \[ = \frac{7}{4}-\sqrt{3} \] Since this term occurs twice, \[ 2\left(\frac{7}{4}-\sqrt{3}\right)=\frac{7}{2}-2\sqrt{3} \] Now, \[ \left(-\frac{\sqrt{6}}{2}-2\right)^2 = \left(\frac{\sqrt{6}}{2}+2\right)^2 \] \[ = \frac{6}{4}+2\sqrt{6}+4 \] \[ = \frac{3}{2}+2\sqrt{6}+4 \] \[ = \frac{11}{2}+2\sqrt{6} \] Therefore, \[ AM^2=\frac{7}{2}-2\sqrt{3}+\frac{11}{2}+2\sqrt{6} \] \[ AM^2=9-2\sqrt{3}+2\sqrt{6} \] Thus, \[ AM=\sqrt{9-2\sqrt{3}+2\sqrt{6}} \]

Step 4: Compare with the given length.
Given length of median is \[ \lambda\sqrt{9-2\sqrt{3}+2\sqrt{6}} \] But we obtained \[ AM=\sqrt{9-2\sqrt{3}+2\sqrt{6}} \] Therefore, \[ \lambda=1 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{1} \]