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

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Three Dimensional Geometry

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

α, - 1, β

, where both

α

and

β

are integers of minimum absolute values, then

α^{2} + β^{2}

is equal to ____.

Answer: 450

In a rhombus, the diagonals are perpendicular. The direction ratios of

P Q

are:

(\frac{56}{17} + 2, \frac{43}{17} + 1, \frac{111}{17} - 1) = (\frac{90}{17}, \frac{60}{17}, \frac{94}{17})

Simplifying:

(90, 60, 94) \sim (45, 30, 47)

.

Since

R S ⊥ P Q

90 α + 60 (- 1) + 94 β = 0

90 α + 94 β = 60

45 α + 47 β = 30

For minimum absolute integer values:

β = \frac{30 - 45 α}{47}

Trying

α = - 15

β = \frac{30 + 675}{47} = \frac{705}{47} = 15

.

Verify:

45 (- 15) + 47 (15) = - 675 + 705 = 30

✓

Therefore

α^{2} + β^{2} = 225 + 225 = 450

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Integer answer