Solution

Put

c = y \cdot z, p = x \cdot y, q = x \cdot z .

Since

x, y, z

are unit vectors,

S = 2 - p^{2} - q^{2} + p + q = \frac{5}{2} - (p - \frac{1}{2})^{2} - (q - \frac{1}{2})^{2} .

\frac{p ^{2} - 2 c pq + q ^{2}}{1 - c ^{2}} \leq 1.

The desired pair

p = q = 1/2

is feasible exactly when

c \geq - 1/2

. In that case every maximizer has these inner products. If

c < - 1/2

, symmetry and strict convexity force

p = q

, and the nearest feasible diagonal point is

p = q = \frac{1 + c}{2} .

x \times (y \times z) = y (x \cdot z) - z (x \cdot y) = q y - p z .

Hence

F (y, z) = ⎩ ⎨ ⎧ \frac{1}{2} (y - z), \frac{1 + y \cdot z}{2} (y - z), y \cdot z \geq - \frac{1}{2}, y \cdot z < - \frac{1}{2} .

Rotations preserve dot products and cross-product lengths, while

z^{'} = 2 z

. Thus

S^{'} = 5 - p^{2} - 4 q^{2} + p + 2 q = \frac{11}{2} - (p - \frac{1}{2})^{2} - 4 (q - \frac{1}{4})^{2} .

Let

t = (1/2 1/4), K = (1 c c 1), W = (1004) .

The target pair is feasible precisely when

t^{T} K^{- 1} t = \frac{5 - 4 c}{16 ( 1 - c ^{2} )} \leq 1,

\frac{1 - 3 5}{8} \leq c \leq \frac{1 + 3 5}{8} .

On this interval,

G (y, z) = \frac{1}{4} y - \frac{1}{2} z .

Outside this interval, let

λ > 0

be the unique number for which

r = (W + λ K^{- 1})^{- 1} W t, r^{T} K^{- 1} r = 1,

and write

r = (p, q)^{T}

. These equations give the unique weighted projection of

t

onto the feasible ellipse, and therefore determine the maximizing inner products solely from

c = y \cdot z

. In the remaining cases,

G (y, z) = q y - p z .

Neither answer depends on

θ_{1}

θ_{2}

Solved by gpt-5.6-sol high.

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