Question

Let \[ O(0,0,0) \] and \[ A(2,1,-3) \] be vertices of a triangle \(OAB\). If \[ (-1,2,1) \] is the midpoint of side \(AB\), and the perimeter of the triangle is \[ \sqrt2\,(k+l\sqrt7+m\sqrt{13}), \] then the value of \[ k+l+m \] is:

• A. \(7\)
• B. \(8\)
• C. \(5\)
• D. \(6\)

Hint:

In three-dimensional geometry, if a midpoint and one endpoint are known, always find the missing endpoint first. This usually converts a difficult coordinate geometry problem into a straightforward distance-formula calculation.

Answer 1

The Correct Option is B

Step 1: List what we know.
The triangle is $OAB$ with $O(0,0,0)$ and $A(2,1,-3)$. The midpoint of side $AB$ is $(-1,2,1)$. We must find the perimeter and then read off $k,l,m$.

Step 2: Find point $B$.
The midpoint of $AB$ is the average of $A$ and $B$. So $B=2M-A$: \[ B=(2(-1)-2,\,2(2)-1,\,2(1)-(-3))=(-4,3,5). \]

Step 3: Find $OA$.
\[ OA=\sqrt{2^2+1^2+(-3)^2}=\sqrt{4+1+9}=\sqrt{14}=\sqrt2\,\sqrt7. \]

Step 4: Find $OB$.
\[ OB=\sqrt{(-4)^2+3^2+5^2}=\sqrt{16+9+25}=\sqrt{50}=5\sqrt2. \]

Step 5: Find $AB$.
\[ AB=\sqrt{(-4-2)^2+(3-1)^2+(5+3)^2}=\sqrt{36+4+64}=\sqrt{104}=2\sqrt2\,\sqrt{13}. \]

Step 6: Add and match.
Perimeter $=5\sqrt2+\sqrt2\sqrt7+2\sqrt2\sqrt{13}=\sqrt2\,(5+\sqrt7+2\sqrt{13})$. So $k=5,\,l=1,\,m=2$, giving $k+l+m=8$, which is option 2.
\[ \boxed{8} \]