JEE Advanced 2023 — Paper 1 — MATH-III solution

JEE Advanced2023Paper 1MATH-III

Let $P$ be the plane $3 x + 2 y + 3 z = 16$ and let $S = {α \hat{i} + β \hat{j} + γ \hat{k} : α^{2} + β^{2} + γ^{2} = 1 and the distance of (α, β, γ) from the plane P is \frac{7}{2}}$ . Let $u$ , $v$ and $w$ be three distinct vectors in $S$ such that $∣ u - v ∣ = ∣ v - w ∣ = ∣ w - u ∣$ . Let $V$ be the volume of the parallelepiped determined by vectors $u$ , $v$ and $w$ . Then the value of $\frac{80}{3} V$ is

Your answer

Solution

The normal to plane $P$ has magnitude $3 + 4 + 9 = 4$ , so the perpendicular distance from origin to $P$ is $\frac{16}{4} = 4$ .

Points on $S$ lie on the unit sphere and at perpendicular distance $\frac{7}{2}$ from $P$ , hence on a plane parallel to $P$ at distance $4 - \frac{7}{2} = \frac{1}{2}$ from origin. The intersection of the unit sphere with that plane is a circle of radius $1 - (\frac{1}{2})^{2} = \frac{3}{2}$ .

Let $C$ be the centre of this circle. Vectors $u$ , $v$ , $w$ are three points on the circle with equal pairwise chord lengths, so they form an equilateral triangle inscribed in a circle of radius $R = \frac{3}{2}$ , with side $s = R 3 = \frac{3}{2}$ .

Triangle area: $\frac{3}{4} s^{2} = \frac{3}{4} \cdot \frac{9}{4} = \frac{9 3}{16}$ .

Volume of the tetrahedron $O u v w$ (with apex $O$ and base the triangle) is $\frac{1}{3} \cdot \frac{9 3}{16} \cdot \frac{1}{2} = \frac{3 3}{32}$ .

Volume of parallelepiped on the three position vectors equals $6$ times the tetrahedron volume: $V = 6 \cdot \frac{3 3}{32} = \frac{9 3}{16}$ .

Therefore $\frac{80}{3} V = \frac{80}{3} \cdot \frac{9 3}{16} = \frac{720}{16} = 45$ .

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