Let P be the plane √3x + 2y + 3z = 16 and let S = {α i + β j + γ k : α² + β² + γ² = 1 and the distan

Q: Wait, why is the intersection of a plane and a sphere always a circle?

Think of it like slicing an orange. If you cut a sphere with a flat plane, the cross-section is always a circle. Mathematically, if you have a sphere $x^2 + y^2 + z^2 = r^2$ and a plane $ax + by + cz = d$, the points satisfying both equations form a circle because the distance from any point on the intersection to the center of the sphere is constant ($r$), and the distance to the plane is also constant. This geometric constraint forces the intersection to be a circle.

Q: Why is the volume of the parallelepiped six times the volume of the tetrahedron?

This is a beautiful property of vector geometry. The volume of a parallelepiped defined by vectors $\vec{u}, \vec{v}, \vec{w}$ is given by the absolute value of the scalar triple product $|\vec{u} \cdot (\vec{v} \times \vec{w})|$. The volume of a tetrahedron with vertices at the origin and the tips of these vectors is $\frac{1}{6} |\vec{u} \cdot (\vec{v} \times \vec{w})|$. Thus, the parallelepiped is exactly six times larger than the tetrahedron.

Q: Can we just use the coordinates of the vectors to solve this?

You could, but it would be an algebraic nightmare! The beauty of this problem lies in its symmetry. By recognizing the geometric shapes—the sphere, the circle, the equilateral triangle, and the tetrahedron—we bypass complex coordinate geometry and solve it using elegant, high-level spatial reasoning. Always look for the geometry first!

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Sphere and the Plane

We are given a unit sphere centered at the origin $O(0,0,0)$ and a plane $P$ defined by the equation $\sqrt{3}x + 2y + 3z = 16$. To determine the relationship between the sphere and the plane, we calculate the perpendicular distance $D$ from the origin to the plane using the formula:

$$D = \frac{|d|}{\sqrt{a^2 + b^2 + c^2}}$$

Substituting the given coefficients, we find:

$$D = \frac{16}{\sqrt{(\sqrt{3})^2 + 2^2 + 3^2}} = \frac{16}{\sqrt{3 + 4 + 9}} = \frac{16}{\sqrt{16}} = 4$$

Thus, the plane $P$ is exactly $4$ units away from the origin.

### The Parallel Slice

The set $S$ consists of points on the sphere at a distance of $3.5$ (or $\frac{7}{2}$) from the plane $P$. These points must lie on a plane $P'$ parallel to $P$. Since the plane $P$ is $4$ units from the origin, the plane $P'$ is located at a distance $h$ from the origin:

$$h = 4 - 3.5 = 0.5 = \frac{1}{2}$$

The intersection of the plane $P'$ and the unit sphere (radius $r=1$) forms a circle. The radius $R$ of this circle is determined by the Pythagorean theorem:

$$R^2 + h^2 = r^2 \Rightarrow R^2 + \left(\frac{1}{2}ight)^2 = 1^2$$

$$R^2 = 1 - \frac{1}{4} = \frac{3}{4} \Rightarrow R = \frac{\sqrt{3}}{2}$$

### The Equilateral Triangle and the Tetrahedron

Three vectors $\vec{u}, \vec{v}, \vec{w}$ in $S$ form an equilateral triangle inscribed in the circle of radius $R$. The side length $a$ of this equilateral triangle is given by $a = R\sqrt{3}$:

$$a = \frac{\sqrt{3}}{2} 	imes \sqrt{3} = \frac{3}{2}$$

The area of this equilateral triangle is calculated as:

$$	ext{Area} = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \left(\frac{3}{2}ight)^2 = \frac{9\sqrt{3}}{16}$$

By connecting the origin $O$ to the vertices of this triangle, we form a tetrahedron with base area $\frac{9\sqrt{3}}{16}$ and height $h = 0.5$. The volume $V_T$ of this tetrahedron is:

$$V_T = \frac{1}{3} 	imes 	ext{Area} 	imes h = \frac{1}{3} 	imes \frac{9\sqrt{3}}{16} 	imes \frac{1}{2} = \frac{3\sqrt{3}}{32}$$

### Final Calculation

The volume $V$ of the parallelepiped determined by the vectors $\vec{u}, \vec{v}, \vec{w}$ is $6$ times the volume of the tetrahedron:

$$V = 6 	imes V_T = 6 	imes \frac{3\sqrt{3}}{32} = \frac{18\sqrt{3}}{32} = \frac{9\sqrt{3}}{16}$$

We are asked to evaluate the expression $\frac{80}{\sqrt{3}} V$:

$$\frac{80}{\sqrt{3}} 	imes \frac{9\sqrt{3}}{16} = 5 	imes 9 = 45$$

The final result is **45**.

Enter Numerical Value:

Visualized Solution

The Unit Sphere and Plane $P$

Distance from Origin to Plane $P$

Setting up Distance $O Q$

Evaluating Distance $O Q$

Locating the Set $S$

Distance of Plane $P^{'}$ from Origin

The Circle of Intersection

Evaluating Radius $R$

The Equilateral Triangle

Area of the Triangle

The Tetrahedron

Volume of the Tetrahedron

Volume of the Parallelepiped

Final Evaluation

The Sigma Insight: Equation of a Plane

Analyzing the Geometry of the Sphere and the Plane

The Parallel Slice

The Equilateral Triangle and the Tetrahedron

Final Calculation

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