Let P(-2, -1, 1) and Q(56/17, 43/17, 111/17) be the vertices of the rhombus PRQS. If the direction r

Q: Wait, why do we care about the direction ratios of $PQ$ if we are looking for $RS$?

Great question! In a rhombus, the diagonals are perpendicular. By finding the direction ratios of $PQ$, we define the orientation of the first diagonal. Because $RS$ must be perpendicular to $PQ$, the dot product of their direction ratios must be zero. This gives us the constraint we need to solve for $\alpha$ and $\beta$.

Q: Why can we just scale the direction ratios of $PQ$ by $17$ and divide by $2$?

Direction ratios are not unique; they represent a direction, not a specific vector length. Any vector $k\vec{v}$ (where $k \neq 0$) represents the same line. Scaling simplifies the numbers, making the subsequent algebra much cleaner and less prone to calculation errors.

Q: How do I know if $\alpha = -15$ and $\beta = 15$ are the 'minimum absolute values'?

We derived the relationship $47k = 2 - 3\alpha$. By testing integer values for $k$, we look for the smallest resulting integers for $\alpha$ and $\beta$. Since $k=1$ yields a valid integer solution, and any larger $k$ would increase the magnitude of the variables, we have found the minimal set.

Q: Can I use the midpoint formula to find the center of the rhombus?

You absolutely could! The diagonals of a rhombus bisect each other at their midpoint. While you don't strictly need the midpoint to solve for the direction ratios, calculating it is a fantastic way to verify your understanding of the rhombus's symmetry.

Q: Is there any other way to solve $45\alpha + 47\beta = 30$ besides inspection?

Yes, you could use the Extended Euclidean Algorithm to find the general solution for this linear Diophantine equation. However, for JEE-level problems, recognizing the divisibility property—where $47\beta$ must be a multiple of $15$—is usually the fastest and most elegant path.

Question

Let $P(-2, -1, 1)$ and $Q\left(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\right)$ 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 ______.

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of Symmetry

Imagine you are floating in a three-dimensional coordinate space. Before you lies a rhombus, $PRQS$, suspended in the void.

It is a beautiful, symmetric figure, and its most defining characteristic is the relationship between its diagonals. In any rhombus, the diagonals are not merely lines; they are the axes of symmetry that bisect each other at a perfect $90^{\circ}$ angle.

This simple geometric truth is the key that will unlock our entire problem.

### Defining the Backbone

We are given the coordinates of two vertices, $P(-2, -1, 1)$ and $Q\left(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}ight)$. These two points define the diagonal $PQ$.

To understand the orientation of this diagonal, we calculate its direction ratios. The direction ratios of a line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are simply the differences in their coordinates: $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.

Calculating these, we get:

$$	ext{DRs of } PQ = \left(\frac{56}{17} - (-2), \frac{43}{17} - (-1), \frac{111}{17} - 1ight) = \left(\frac{90}{17}, \frac{60}{17}, \frac{94}{17}ight)$$

### The Elegance of Scaling

Those fractions may look intimidating, but remember that direction ratios define a direction, not a magnitude. We can scale them by any non-zero constant without changing the line's orientation.

Let us multiply by $17$ to clear the denominators, giving us $(90, 60, 94)$. We can simplify this further by dividing by $2$, resulting in the clean, manageable vector $\langle 45, 30, 47 angle$.

This vector represents the orientation of our diagonal $PQ$.

### The Perpendicularity Constraint

We are told the other diagonal, $RS$, has direction ratios $\langle \alpha, -1, \beta angle$. Because $PQ \perp RS$, the dot product of their direction ratios must vanish.

Mathematically, this is expressed as:

$$a_1a_2 + b_1b_2 + c_1c_2 = 0$$

Substituting our values, we get:

$$(45)(\alpha) + (30)(-1) + (47)(\beta) = 0$$

This simplifies to the linear equation:

$$45\alpha + 47\beta = 30$$

### Solving the Diophantine Puzzle

We are looking for integers $\alpha$ and $\beta$ with the minimum absolute values. Let us rearrange our equation to isolate the variables:

$$47\beta = 30 - 45\alpha = 15(2 - 3\alpha)$$

Since $47$ and $15$ are coprime, for the equality to hold, $\beta$ must be a multiple of $15$. Let $\beta = 15k$.

Substituting this back into our equation:

$$47(15k) = 15(2 - 3\alpha) \implies 47k = 2 - 3\alpha$$

To find the minimum absolute values, we test small integers for $k$. If we set $k = 1$, we get $47 = 2 - 3\alpha$, which leads to $3\alpha = -45$, or $\alpha = -15$.

Thus, our minimal integer pair is $(\alpha, \beta) = (-15, 15)$.

### Final Calculation

We have navigated the 3D space, respected the perpendicularity of the rhombus diagonals, and solved the Diophantine constraint. The final step is a simple calculation:

$$\alpha^2 + \beta^2 = (-15)^2 + (15)^2 = 225 + 225 = 450$$

Through logic, symmetry, and algebraic finesse, we have arrived at the final result: **450**.

Enter Numerical Value:

Visualized Solution

Visualizing Rhombus $P R QS$

The Diagonal Property

Direction Ratios of $P Q$

Calculating DRs of $P Q$

Simplifying the Ratios

Direction Ratios of $R S$

Applying Perpendicularity Condition

Forming the Linear Equation

Solving for Integers

Analyzing Divisibility

Finding Minimal Values

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Geometry of Symmetry

Defining the Backbone

The Elegance of Scaling

The Perpendicularity Constraint

Solving the Diophantine Puzzle

Final Calculation

Similar Questions

If the distance of the point $P (43, α, β), β < 0$ , from the line $r = 4 \hat{i} - \hat{k} + μ (2 \hat{i} + 3 \hat{k}), μ \in R$ along a line with direction ratios $3, - 1, 0$ is $1310$ , then $α^{2} + β^{2}$ is equal to ____

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Let a line pass through two distinct points $P (- 2, - 1, 3)$ and Q, and be parallel to the vector $3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ If the distance of the point Q from the point $R (1, 3, 3)$ is 5, then the square of the area of $△ P QR$ is equal to:

Enter Numerical Value:

Visualized Solution

Visualizing Rhombus .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PRQS

The Diagonal Property

Direction Ratios of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ

Calculating DRs of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ

Simplifying the Ratios

Direction Ratios of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RS

Applying Perpendicularity Condition

Forming the Linear Equation

Solving for Integers

Analyzing Divisibility

Finding Minimal Values

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Geometry of Symmetry

Defining the Backbone

The Elegance of Scaling

The Perpendicularity Constraint

Solving the Diophantine Puzzle

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a line having direction ratios 1,−4,2 intersect the lines 3x−7​=−1y−1​=1z+2​ and 2x​=3y−7​=1z​ at the point A and B. Then (AB)2 is equal to ____.

Visualizing Rhombus $P R QS$

Direction Ratios of $P Q$

Calculating DRs of $P Q$

Direction Ratios of $R S$

Let a line having direction ratios $1, - 4, 2$ intersect the lines $\frac{x - 7}{3} = \frac{y - 1}{- 1} = \frac{z + 2}{1}$ and $\frac{x}{2} = \frac{y - 7}{3} = \frac{z}{1}$ at the point $A$ and $B$ . Then $(A B)^{2}$ is equal to ____.