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Subtract the first vertex from the remaining three to obtain edge vectors $a=(1,2,-2)$, $b=(0,1,3)$, $c=(-1,-1,0)$. Their scalar triple product isA tetrahedron occupies one sixth of the corresponding parallelepiped, so its volume isThe absolute value removes the orientation sign of the determinant.

The characteristic polynomial is $(5-\lambda )(3-\lambda )-3=(\lambda -6)(\lambda -2)$. Corresponding normalized eigenvectors areThese form an orthonormal basis. Write $x=u{e}_6+v{e}_2$; the quadratic form becomesThe locus is an ellipse centred at the origin. Its minor semiaxis has length $\sqrt{2/3}$ along the direction $30^{\circ }$ from the positive horizontal axis; its major semiaxis has length $\sqrt{2}$ along $120^{\circ }$. The right panel of the figure in part 1 shows these principal directions.

The addition formulas for the hyperbolic functions implySince $M=B(y)/2$, its fourth matrix power isAlternatively, the orthonormal eigenvectors $(1,\pm 1)/\sqrt{2}$ have eigenvalues $e^{\pm y}/2$. Raising these eigenvalues to the fourth power gives the same result by spectral decomposition.

An integrating factor for the first-order linear differential equation is $\mu (x)=e^{\sin x}$. Multiplication givesSubstitute $u=\sin x$ in the antiderivative: $\int 2u{e}^{u}du=2(u-1)e^u$. ThereforeDifferentiating verifies the equation everywhere on the real line; no division by $\cos x$ restricts the solution.

All complex logarithms of $i$ are $i(\pi /2+2\pi k)$ for integer $k$. By complex exponentiation,Every value is positive and real. If the power is defined using the principal logarithm, the single value is $\boxed{e^{-\pi /2}}$. Keeping the logarithm branch explicit distinguishes this value from the full multivalued set.