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MathematicsInteger+4Question 12 of 51

JEE Advanced 2023 · Paper 1, Mathematics question 12

Let PP be the plane 3x+2y+3z=16\sqrt3x + 2y + 3z = 16 and let

S={αi^+βj^+γk^:α2+β2+γ2=1 and the distance of (α,β,γ) from the plane P is 72}.S = \left\{\alpha\hat i + \beta\hat j + \gamma\hat k : \alpha^2 + \beta^2 + \gamma^2 = 1 \text{ and the distance of } (\alpha, \beta, \gamma) \text{ from the plane } P \text{ is } \frac72\right\}.

Let u⃗\vec u, v⃗\vec v and w⃗\vec w be three distinct vectors in SS such that ∣u⃗−v⃗∣=∣v⃗−w⃗∣=∣w⃗−u⃗∣|\vec u - \vec v| = |\vec v - \vec w| = |\vec w - \vec u|. Let VV be the volume of the parallelepiped determined by vectors u⃗\vec u, v⃗\vec v and w⃗\vec w. Then the value of 803V\dfrac{80}{\sqrt3}V is _____.

Answer and worked solution

Official answer: 45

The set SS. The normal (3,2,3)(\sqrt3,2,3) has length 4, so the distance condition is
∣3α+2β+3γ−16∣4=72  ⟹  3α+2β+3γ=2  (30 is impossible on the unit sphere).\frac{|\sqrt3\alpha+2\beta+3\gamma-16|}{4}=\frac72\implies\sqrt3\alpha+2\beta+3\gamma=2\ \ (\text{30 is impossible on the unit sphere}).
So SS is the circle where the unit sphere meets a plane at distance 12\tfrac12 from OO. Its radius is 32\frac{\sqrt3}{2}.

The triangle. u⃗,v⃗,w⃗\vec u,\vec v,\vec w form an equilateral triangle inscribed in this circle:

  • side =3⋅32=32=\sqrt3\cdot\frac{\sqrt3}2=\frac32;
  • area =34⋅94=9316=\frac{\sqrt3}{4}\cdot\frac94=\frac{9\sqrt3}{16}.

Volume. The parallelepiped is 6 times the tetrahedron O uvwO\,uvw, whose height is 12\tfrac12:
V=6⋅13⋅9316⋅12=9316.V=6\cdot\frac13\cdot\frac{9\sqrt3}{16}\cdot\frac12=\frac{9\sqrt3}{16}.

803V=80⋅916=45.\frac{80}{\sqrt3}V=\frac{80\cdot9}{16}=45.

Topics:Scalar and vector triple productsEquation of a planeAngles and distances
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