Question

Let $P(-2, -1, 1)$ and $Q(\frac{56}{17}, \frac{43}{17}, \frac{111}{17})$ be the vertices of the rhombus $PRQS$. If the direction ratios of the diagonal $RS$ are $\alpha, -1, \beta$, where both $\alpha$ and $\beta$ are integers of minimum absolute values, then $\alpha^2 + \beta^2$ is equal to

Answer 1

450

Answer 2

Here's how to solve the problem:

1. Midpoint: Find the midpoint $M$ of $PQ$, which is also the midpoint of $RS$: $M = \left(\frac{-2 + \frac{56}{17}}{2},\, \frac{-1 + \frac{43}{17}}{2},\, \frac{1 + \frac{111}{17}}{2}\right) = \left(\frac{11}{17},\, \frac{13}{17},\, \frac{64}{17}\right).$

2. Direction Vector: Compute the direction vector of $PQ$: $\vec{PQ} = Q-P = \left(\frac{56}{17} - (-2),\, \frac{43}{17} - (-1),\, \frac{111}{17} - 1\right) = \left(\frac{90}{17},\, \frac{60}{17},\, \frac{94}{17}\right).$

3. Perpendicularity: Since $RS$ (with direction ratios $(\alpha, -1, \beta)$) is perpendicular to $PQ$, their dot product is zero: $(\alpha,-1,\beta) \cdot \left(\tfrac{90}{17},\tfrac{60}{17},\tfrac{94}{17}\right) = 0.$ This simplifies to: $90\alpha - 60 + 94\beta = 0 \implies 45\alpha + 47\beta = 30.$

4. Diophantine Equation: Solve the Diophantine equation $45\alpha + 47\beta = 30$ for integers $\alpha$ and $\beta$. Using the Extended Euclidean algorithm, we find a particular solution. The general solution is: $\alpha = 690 + 47t, \quad \beta = -660 - 45t, \quad t\in\mathbb{Z}.$ Choosing $t=-15$ yields: $\alpha = 690 - 705 = -15, \quad \beta = -660 + 675 = 15.$

5. Final Calculation: Compute $\alpha^2 + \beta^2$: $\alpha^2+\beta^2 = (-15)^2 + 15^2 = 225+225 = 450.$

Therefore, $\alpha^2 + \beta^2 = 450$.