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3D Geometry question

2022 · 28 Jul · Shift 1 · Q39
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  5. /2022 · 28 Jul · Shift 1 · Q39

3D Geometry question

2022 · 28 Jul · Shift 1 · Q39

JEE MainMathematics3D GeometryNumerical+4 / −1
Let P(−2,−1,1)\mathrm{P}(-2,-1,1)P(−2,−1,1) and Q(5617,4317,11117)\mathrm{Q}\left(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\right)Q(1756​,1743​,17111​) be the vertices of the rhombus PRQS. If the direction ratios of the diagonal RS are α,−1,β\alpha,-1, \betaα,−1,β, where both α\alphaα and β\betaβ are integers of minimum absolute values, then α2+β2\alpha^{2}+\beta^{2}α2+β2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 450

  1. Use the property of a rhombus

In rhombus PRQSPRQSPRQS, the diagonals are PQPQPQ and RSRSRS. A rhombus is a parallelogram with equal sides, so its diagonals:

  • bisect each other, and
  • are perpendicular.

Hence, the line RSRSRS must:

  1. pass through the midpoint of PQPQPQ,
  2. be perpendicular to PQPQPQ.

Since only direction ratios of RSRSRS are asked, we only need a vector perpendicular to PQ→\overrightarrow{PQ}PQ​.

  1. Find the direction vector of diagonal PQPQPQ

Given P(−2,−1,1),Q(5617,4317,11117)P(-2,-1,1), \quad Q\left(\frac{56}{17},\frac{43}{17},\frac{111}{17}\right)P(−2,−1,1),Q(1756​,1743​,17111​)

So, PQ→=(5617+2,4317+1,11117−1)\overrightarrow{PQ}=\left(\frac{56}{17}+2,\frac{43}{17}+1,\frac{111}{17}-1\right)PQ​=(1756​+2,1743​+1,17111​−1)

=(56+3417,43+1717,111−1717)=\left(\frac{56+34}{17},\frac{43+17}{17},\frac{111-17}{17}\right)=(1756+34​,1743+17​,17111−17​)

=(9017,6017,9417)=\left(\frac{90}{17},\frac{60}{17},\frac{94}{17}\right)=(1790​,1760​,1794​)

Thus a simpler direction ratio of PQPQPQ is 90:60:94=45:30:4790:60:94 = 45:30:4790:60:94=45:30:47

So we take direction vector of PQPQPQ as d⃗PQ=(45,30,47).\vec d_{PQ}=(45,30,47).dPQ​=(45,30,47).

  1. Let the direction ratios of RSRSRS be (α,−1,β)(\alpha,-1,\beta)(α,−1,β)

Since diagonals of a rhombus are perpendicular, (α,−1,β)⋅(45,30,47)=0(\alpha,-1,\beta) \cdot (45,30,47)=0(α,−1,β)⋅(45,30,47)=0

Therefore, 45α−30+47β=045\alpha -30 +47\beta =045α−30+47β=0

or 45α+47β=30.45\alpha +47\beta =30. 45α+47β=30.

We need integer values of α,β\alpha,\betaα,β with minimum absolute values.

  1. Solve the Diophantine equation

We solve 45α+47β=30.45\alpha +47\beta =30.45α+47β=30.

Reducing modulo 454545: 47β≡30(mod45)47\beta \equiv 30 \pmod{45}47β≡30(mod45) 2β≡30(mod45)2\beta \equiv 30 \pmod{45}2β≡30(mod45)

Since inverse of 222 modulo 454545 is 232323 (because 2⋅23=46≡1(mod45)2\cdot 23=46\equiv 1 \pmod{45}2⋅23=46≡1(mod45)), β≡30⋅23=690≡15(mod45).\beta \equiv 30\cdot 23 = 690 \equiv 15 \pmod{45}.β≡30⋅23=690≡15(mod45).

So, β=15+45t\beta = 15 + 45tβ=15+45t for some integer ttt.

Substitute into the equation: 45α+47(15+45t)=3045\alpha +47(15+45t)=3045α+47(15+45t)=30 45α+705+2115t=3045\alpha +705 +2115t =3045α+705+2115t=30 45α=−675−2115t45\alpha = -675 -2115t45α=−675−2115t α=−15−47t.\alpha = -15 -47t.α=−15−47t.

Hence the general integer solution is α=−15−47t,β=15+45t.\alpha = -15-47t, \qquad \beta = 15+45t.α=−15−47t,β=15+45t.

  1. Choose integers with minimum absolute values

We test nearby integer values of ttt:

  • For t=0t=0t=0: α=−15,β=15\alpha=-15,\quad \beta=15α=−15,β=15
  • For t=−1t=-1t=−1: α=32,β=−30\alpha=32,\quad \beta=-30α=32,β=−30
  • For t=1t=1t=1: α=−62,β=60\alpha=-62,\quad \beta=60α=−62,β=60

Clearly, the minimum absolute values occur at α=−15,β=15.\alpha=-15,\quad \beta=15.α=−15,β=15.

Thus direction ratios of RSRSRS can be taken as (−15,−1,15).(-15,-1,15).(−15,−1,15).

  1. Compute α2+β2\alpha^2+\beta^2α2+β2

α2+β2=(−15)2+(15)2=225+225=450.\alpha^2+\beta^2 = (-15)^2 + (15)^2 = 225+225=450.α2+β2=(−15)2+(15)2=225+225=450.

  1. Compare with stored answer

Our derived answer is 450,450,450, which matches the stored correct answer.

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