S1 = { x = (x1, x2, x3)^T  ∈ R3 | x^T  x ≤16 }. Let S₂ be subspace of R³ with dimension 2 then Area of S₁ ∩ S₁ ?

The correct option is 16 π.

Here is the step-by-step logical explanation of why this is the correct answer:

1. Understand the set $S_1$:
The set $S_1$ is defined as:

$$S_1=\{x={(x_1,x_2,x_3)}^T\in {ℝ}^3|x^{T}x\le 16\}$$

Here, $x^{T}x$ represents the dot product of the vector with itself, which is:

$$x_1^2+x_2^2+x_3^2\le 16$$

This equation represents a solid sphere in 3-dimensional space (${ℝ}^3$) centered at the origin $(0,0,0)$ with a radius $R$ calculated as:

$$R=\sqrt{16}=4$$

2. Understand the subspace $S_2$:
We are given that $S_2$ is a subspace of ${ℝ}^3$ with a dimension of 2.
By definition, any 2-dimensional subspace of ${ℝ}^3$ is a flat plane that passes through the origin $(0,0,0)$.

3. Find the intersection $S_1\cap S_2$:
The intersection of a solid 3D sphere of radius $R=4$ centered at the origin and a plane passing through the origin is a flat 2D disk (circle and its interior) lying on that plane.
Since the plane passes exactly through the center of the sphere (the origin), this intersection forms a great disk of the sphere.

The radius of this intersecting disk is equal to the radius of the sphere itself:

$$r=R=4$$

4. Calculate the Area of the intersection:
The area of a flat circular disk of radius $r$ is given by the standard formula:

$$\text{Area}=\pi r^2$$

Substituting $r=4$ into the formula:

$$\text{Area}=\pi \times 4^2=16\pi$$

Thus, the area of the intersection is indeed 16 π.