Let P be the plane $\sqrt{3} \mathrm{x}+2 \mathrm{y}+3 \mathrm{z}=16$ and let $\mathrm{S}=\left\{\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}}: \alpha^{2}+\beta^{2}+\gamma^{2}=1\right.$ and the distance of $(\alpha, \beta, \gamma)$ from the plane P is $\left.\frac{7}{2}\right\}$. Let $\overrightarrow{\mathrm{u}}, \overrightarrow{\mathrm{v}}$ and $\overrightarrow{\mathrm{w}}$ be three distinct vectors in S such that $|\overrightarrow{\mathrm{u}}-\overrightarrow{\mathrm{v}}|=|\overrightarrow{\mathrm{v}}-\overrightarrow{\mathrm{w}}|=|\overrightarrow{\mathrm{w}}-\overrightarrow{\mathrm{u}}|$. Let V be the volume of the parallelepiped determined by vectors $\overrightarrow{\mathrm{u}}, \overrightarrow{\mathrm{v}}$ and $\overrightarrow{\mathrm{w}}$. Then the value of $\frac{80}{\sqrt{3}} \mathrm{~V}$ is

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Vector Algebra

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

Answer: 45

The plane is

3 x + 2 y + 3 z = 16

with normal

n = (3, 2, 3)

and

∣ n ∣ = 4

.

Distance from origin to plane:

O Q = \frac{16}{4} = 4

.

Points in S are at distance

\frac{7}{2}

from the plane and lie on the unit sphere, so they form a circle.

The center P of this circle is at distance

O P = O Q - P Q = 4 - \frac{7}{2} = \frac{1}{2}

from origin.

Radius of circle:

R = 1 - (\frac{1}{2})^{2} = \frac{3}{2}

.

Since

∣ u - v ∣ = ∣ v - w ∣ = ∣ w - u ∣

, triangle UVW is equilateral with side length:

U V = 2 R cos (30°) = 2 \cdot \frac{3}{2} \cdot \frac{3}{2} = \frac{3}{2}

Area of triangle UVW:

\frac{3}{4} (\frac{3}{2})^{2} = \frac{9 3}{16}

Volume of tetrahedron:

\frac{1}{3} \times \frac{9 3}{16} \times \frac{1}{2} = \frac{3 3}{32}

Volume of parallelepiped:

V = 6 \times \frac{3 3}{32} = \frac{9 3}{16}

Therefore:

\frac{80}{3} V = \frac{80}{3} \times \frac{9 3}{16} = 45

Answer: 45

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Integer answer