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 .7/2 \ . Let overrightarrow u , overrightarrow v and overrightarrow w be three distinct vectors in S such that overrightarrow u -overrightarrow v = overrightarrow v -overrightarrow w = overrightarrow w -overrightarrow u . Let V be the volume of the parallelepiped determined by vectors, overrightarrow u , overrightarrow v and overrightarrow w . Then the value of frac 80 √(3) V is

Mathematics · Vector Algebra

JEE Advanced 2023 — Paper 1 — Question 11

Let $P$ be the plane $\sqrt{3} x+2 y+3 z=16$ and let $S=\left\{\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}: \alpha^{2}+\beta^{2}+\gamma^{2}=1\right.$ 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

Answer: 45

Numerical answer — enter this value.

Step-by-step solution

$\text{Plane:\; } P: \sqrt{3}x + 2y + 3z = 16\; , \text{ point\; } (\alpha,\beta,\gamma) \in\; S \text{ has\; distance\; } \frac{7}{2}$

$|\sqrt{3}\alpha + 2\beta + 3\gamma - 16| / \sqrt{3^2+2^2+3^2} = 7/2 \implies \sqrt{3}\alpha + 2\beta + 3\gamma = 2 \text{ (intersection\; with\; unit\; sphere)}$

$\text{Distance\; from\; origin\; to\; plane\; } d_0 = 2/4 = 1/2,\; \text{ circle\; radius\; } r = \sqrt{1 - d_0^2} = \sqrt{3}/2$

$\text{Three \; vectors\; form\; equilateral\; triangle\; on\; this\; circle,\; side\; length } s = r \sqrt{3} = 3/2$

$\text{Volume\; of\; parallelepiped\; } V = \frac{\sqrt{3}}{16} \cdot 9 = 9\sqrt{3}/16$

$\frac{80}{\sqrt{3}} V = \frac{80}{\sqrt{3}} \cdot \frac{9\sqrt{3}}{16} = 45$

$\boxed{45}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2023
Paper: Paper 1
Subject: Mathematics
Chapter: Vector Algebra
Topic: Volume of parallelopiped, tetrahedron.