(a) Let the set P = {(x, y, z)ᵀ : x − y − z = 0 and 2x − y + z = 0} be the collection of vectors of a vector space R³(R). Then…

(a)(i) Proving P is a subspace

Define T: ℝ³ → ℝ² by T(x, y, z) = (x − y − z, 2x − y + z). Then T is linear, and P = {v ∈ ℝ³ : T(v) = (0, 0)}. Take u = (x₁, y₁, z₁) ∈ P and v = (x₂, y₂, z₂) ∈ P, and let α, β ∈ ℝ. Since u, v ∈ P, x₁ − y₁ − z₁ = 0, 2x₁ − y₁ + z₁ = 0, x₂ − y₂ − z₂ = 0, 2x₂ − y₂ + z₂ = 0. Then for αu + βv = (αx₁+βx₂, αy₁+βy₂, αz₁+βz₂), (αx₁+βx₂) − (αy₁+βy₂) − (αz₁+βz₂) = α(x₁−y₁−z₁) + β(x₂−y₂−z₂) = 0. Similarly, 2(αx₁+βx₂) − (αy₁+βy₂) + (αz₁+βz₂) = α(2x₁−y₁+z₁) + β(2x₂−y₂+z₂) = 0. Also (0, 0, 0) ∈ P. Hence P is non-empty and closed under addition and scalar multiplication. Therefore P is a subspace of ℝ³(ℝ).

(a)(ii) Basis and dimension of P

Solve the two equations: x − y − z = 0, so y = x − z. Substitute into 2x − y + z = 0: 2x − (x − z) + z = x + 2z = 0, so x = −2z. Then y = x − z = −2z − z = −3z. Thus every vector of P is (x, y, z) = (−2z, −3z, z) = z(−2, −3, 1). So a basis is {(−2, −3, 1)}. Hence dim P = 1.

(b) Common area of circle and parabola

Find intersections of x² + y² = 4 and y² = 3x. Substituting y² = 3x in the circle: x² + 3x = 4, so x² + 3x − 4 = 0 = (x + 4)(x − 1). Since y² = 3x requires x ≥ 0, x = 1, and then y² = 3, so y = ±√3.

The common region lies inside both curves. For 0 ≤ x ≤ 1, the smaller vertical bound is from the parabola, |y| ≤ √(3x). For 1 ≤ x ≤ 2, the smaller vertical bound is from the circle, |y| ≤ √(4 − x²). By symmetry about the x-axis, A = 2∫ from 0 to 1 √(3x) dx + 2∫ from 1 to 2 √(4 − x²) dx.

First, 2∫ from 0 to 1 √(3x) dx = 2√3 · (2/3) = 4√3/3. Next, 2∫ from 1 to 2 √(4 − x²) dx = [x√(4 − x²) + 4 arcsin(x/2)] from 1 to 2 = (0 + 4·π/2) − (√3 + 4·π/6) = 2π − √3 − 2π/3 = 4π/3 − √3. Therefore A = 4√3/3 + 4π/3 − √3 = (4π + √3)/3. Hence the common area is (4π + √3)/3 square units.

(c) Smallest sphere touching two skew lines

Let L₁: (x−3)/3 = (y−8)/(−1) = (z−3)/1 = t, L₂: (x+3)/(−3) = (y+7)/2 = (z−6)/4 = s. Then points on the lines are A(t) = (3+3t, 8−t, 3+t), B(s) = (−3−3s, −7+2s, 6+4s). Their direction vectors are u = (3, −1, 1), v = (−3, 2, 4).

For the common perpendicular, w = B(s) − A(t) must satisfy w·u = 0 and w·v = 0. Now w = (−6−3s−3t, −15+2s+t, 3+4s−t). Thus w·u = −7s − 11t = 0, w·v = 29s + 7t = 0. Solving, 7s + 11t = 0 and 29s + 7t = 0 gives t = 0 and s = 0.

So the closest points are A(0) = (3, 8, 3), B(0) = (−3, −7, 6). The shortest distance between the lines is d = |B(0) − A(0)| = |(−6, −15, 3)| = √(36 + 225 + 9) = √270 = 3√30.

For any sphere touching both lines, if its radius is r, the distance between the two points of contact is at most 2r. Since the shortest distance between the lines is d, we need d ≤ 2r, so r ≥ d/2. Equality is achieved when the centre is the midpoint of the common perpendicular segment. Hence the smallest radius is r = d/2 = 3√30/2.

The centre is the midpoint of A(0) and B(0): M = ((3−3)/2, (8−7)/2, (3+6)/2) = (0, 1/2, 9/2). Therefore the required sphere is x² + (y − 1/2)² + (z − 9/2)² = (3√30/2)² = 135/2. So the equation is x² + (y − 1/2)² + (z − 9/2)² = 135/2.