Question:
S$_1$ = \{x = (x$_1$, x$_2$, x$_3$)$^T$ $\in$ R$^3$ $|$ x$^T$x $\le$ 16\}.
Let S$_2$ be subspace of R$^3$ with dimension 2 then Area of S$_1$ $\cap$ S$_2$ ?
S$_1$ = \{x = (x$_1$, x$_2$, x$_3$)$^T$ $\in$ R$^3$ $|$ x$^T$x $\le$ 16\}.
Let S$_2$ be subspace of R$^3$ with dimension 2 then Area of S$_1$ $\cap$ S$_2$ ?
Show Hint
Remember the geometric meaning of algebraic expressions. $x^T x = ||x||^2$ represents the squared distance from the origin. A subspace always contains the origin. The intersection of a sphere centered at the origin with a plane through the origin will always be a great circle (or the disk it bounds).
Updated On: Feb 23, 2026
- 16 $\pi$
- 16$\pi^2$
- 4$\pi^2$
- 4$\pi$
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Solution and Explanation
Step 1: Understanding the Question:
We need to find the area of the intersection of two geometric objects in 3D space. - $S_1$ is a set of points in $R^3$ defined by an inequality. - $S_2$ is a 2-dimensional subspace of $R^3$.
Step 2: Geometric Interpretation:
- Interpreting S$_1$: The condition is $x^T x \le 16$. Let $x = (x_1, x_2, x_3)$. Then $x^T x = x_1^2 + x_2^2 + x_3^2$. So the inequality is $x_1^2 + x_2^2 + x_3^2 \le 4^2$. This is the equation of a solid sphere (a ball) centered at the origin (0,0,0) with a radius of $r=4$.
- Interpreting S$_2$: A 2-dimensional subspace of $R^3$ is, by definition, a plane that passes through the origin.
Step 3: Finding the Intersection:
We are looking for the area of the region formed by the intersection of a solid sphere (centered at the origin) and a plane that passes through the origin.
When a plane cuts through a sphere, the intersection is a circle. Since the plane passes through the center of the sphere, the intersection will be a "great circle" of the sphere. The resulting shape within the solid sphere is a circular disk.
The radius of this intersection disk is the largest possible radius, which is equal to the radius of the sphere itself.
- Radius of the sphere = 4.
- Radius of the intersection disk = 4.
Now, we calculate the area of this circular disk.
\[ \text{Area} = \pi \times (\text{radius})^2 \]
\[ \text{Area} = \pi \times (4)^2 = 16\pi \]
Step 4: Final Answer:
The area of the intersection $S_1 \cap S_2$ is $16\pi$.
We need to find the area of the intersection of two geometric objects in 3D space. - $S_1$ is a set of points in $R^3$ defined by an inequality. - $S_2$ is a 2-dimensional subspace of $R^3$.
Step 2: Geometric Interpretation:
- Interpreting S$_1$: The condition is $x^T x \le 16$. Let $x = (x_1, x_2, x_3)$. Then $x^T x = x_1^2 + x_2^2 + x_3^2$. So the inequality is $x_1^2 + x_2^2 + x_3^2 \le 4^2$. This is the equation of a solid sphere (a ball) centered at the origin (0,0,0) with a radius of $r=4$.
- Interpreting S$_2$: A 2-dimensional subspace of $R^3$ is, by definition, a plane that passes through the origin.
Step 3: Finding the Intersection:
We are looking for the area of the region formed by the intersection of a solid sphere (centered at the origin) and a plane that passes through the origin.
When a plane cuts through a sphere, the intersection is a circle. Since the plane passes through the center of the sphere, the intersection will be a "great circle" of the sphere. The resulting shape within the solid sphere is a circular disk.
The radius of this intersection disk is the largest possible radius, which is equal to the radius of the sphere itself.
- Radius of the sphere = 4.
- Radius of the intersection disk = 4.
Now, we calculate the area of this circular disk.
\[ \text{Area} = \pi \times (\text{radius})^2 \]
\[ \text{Area} = \pi \times (4)^2 = 16\pi \]
Step 4: Final Answer:
The area of the intersection $S_1 \cap S_2$ is $16\pi$.
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