Question

Consider a set \(S_1 = \{x = (x_1, x_2, x_3)^T \in \mathbb{R}^3 \mid x^Tx \leq 16\}\). Let \(S_2\) be another set which is a subspace of \(\mathbb{R}^3\) with dimension two.
Which of the following gives the area of \(S_1 \cap S_2\)?

• A. \(16\pi\)
• B. \(4\pi\)
• C. \(4\pi^2\)
• D. \(16\pi^2\)

Hint:

\(S_1\) is a solid ball of radius 4 centered at the origin, \(S_2\) is a plane through the origin. Their intersection is a disk of radius 4, so use area \(= \pi r^2\).

Answer 1

The Correct Option is A

Step 1: Set up coordinates inside the plane.
Since $S_2$ is a two-dimensional subspace of $\mathbb{R}^3$, we can pick an orthonormal basis $u_1, u_2$ for $S_2$ (two mutually perpendicular unit vectors that span the plane). Any point $x$ in $S_2$ can then be written as:
\[ x = a\,u_1 + b\,u_2, \quad a, b \in \mathbb{R} \]

Step 2: Rewrite the ball condition in these coordinates.
Because $u_1, u_2$ are orthonormal ($u_1 \cdot u_1 = 1$, $u_2 \cdot u_2 = 1$, $u_1 \cdot u_2 = 0$), the squared length of $x$ works out cleanly:
\[ x^Tx = (a u_1 + b u_2)^T (a u_1 + b u_2) = a^2(u_1^Tu_1) + 2ab(u_1^Tu_2) + b^2(u_2^Tu_2) = a^2 + b^2 \]
So the condition $x^Tx \leq 16$ becomes, purely in terms of the plane's own coordinates $(a,b)$:
\[ a^2 + b^2 \leq 16 \]

Step 3: Recognize the shape in the $(a,b)$ plane.
This is precisely the equation of a filled disk of radius $4$ centered at the origin of the $(a,b)$ coordinate system. Since $(a,b)$ are just orthonormal coordinates within $S_2$, distances measured in $(a,b)$ match distances measured in $\mathbb{R}^3$, so this disk is a true, undistorted disk of radius $4$.

Step 4: Compute the area.
\[ \text{Area} = \pi r^2 = \pi (4)^2 = 16\pi \]

Step 5: Conclusion.
This coordinate-based derivation confirms, without relying on intuition alone, that intersecting a solid ball of radius 4 through its center with any plane through that center always produces a disk of the same radius 4. \[ \boxed{16\pi} \]