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Put $D=[a,m,n]\ne 0$ and rewrite the intersection equation as $b=a+sm-tn$. Take scalar triple products with $a$ and $n$:Taking the triple product with $a$ and $m$ instead givesThus the intersection parameters from scalar triple products areThe assumed nonzero denominator makes these expressions valid and guarantees nonparallel directions. The resulting point is $r=a+sm=b+tn$; these formulas are conditional on the intersection assumed in the question.

At an intersection, the two position vectors agree, so $a-b=tn-sm$. The scalar triple product with $m,n$ is thereforesince a determinant with repeated columns vanishes. Geometrically the displacement of the two anchor points must lie in the same plane as the two line directions.

This is a necessary condition. If $m,n$ are not parallel, it is also sufficient, because their span is exactly the plane perpendicular to $m\times n$. For parallel directions the triple product vanishes automatically, even for distinct lines, so the condition alone cannot guarantee intersection.

Expand the same double cross product in two ways. First expanding with $c\times d$ as the inner cross product givesAlternatively, use $(a\times b)\times w=(a\cdot w)b-(b\cdot w)a$ with $w=c\times d$ to getEquating these expressions and moving every term to one side provesThis four-vector scalar triple product identity is valid without assuming that any three of the four vectors are independent.

The vector triple product identity givesIt is the requested linear combination of $b$ and $c$. The notation $\wedge$ in this question denotes the three-dimensional cross product.

Square the two-term form and integrate directly:The first two integrals are each $\pi$ before their prefactors, while the cross term is zero because its integrand is $\frac{1}{2}\sin 2x$. Thus the result is again $4\pi$, agreeing with Parseval's identity.

In the convention with constant term $a_0/2$, Parseval's identity isOnly the first harmonic contributes here, so

The curve starts at $y(0)=-\sqrt{3}$ and initially rises. It crosses zero at $x=\pi /3$ and $4\pi /3$, attains its maximum at $(5\pi /6,2)$ and minimum at $(11\pi /6,-2)$, then returns to $y(2\pi )=-\sqrt{3}$. These follow either from the shifted sine form or from $y^{\prime }=2\cos (x-\pi /3)$.

Using the sine addition formula, combine the surviving modes:It has amplitude $2$ and a phase delay $\pi /3$. The full general solution would add $C{e}^{-x/\sqrt{3}}$, but this exponential is periodic only for $C=0$, so the displayed periodic solution is unique.

The constant coefficient is $a_0=0$. For every $n\ge 1$ the two mode equations arewhere ${\delta }_{n1}$ is the Kronecker delta. The first equation gives $a_n=-\sqrt{3}n{b}_n$, so the second becomes $(1+3n^2)b_n=4{\delta }_{n1}$. HenceNo other modes survive. This is the single-frequency case of the Fourier solution of a first-order equation with periodic forcing.

Fourier orthogonality isolates a coefficient. Multiplication by $\sin (mx)$ and integration over $[0,2\pi ]$ annihilates the constant, all cosine terms and all sine terms except the one with $n=m$, whose squared integral is $\pi$. Thus the corresponding coefficients on the two sides must agree for each positive integer $m$.

Multiplication by $\cos (mx)$ works in the same way, while integration against the constant mode extracts $a_0$. This is the uniqueness of the Fourier coefficients; it does not follow from comparing a few point values of the series.

Differentiate the Fourier series term by term and collect cosine and sine terms. Substitution into the equation givesFor a smooth periodic solution, differentiation of the series is justified. Alternatively, multiply the differential equation by each trigonometric mode and integrate by parts; periodicity cancels the boundary terms and gives the same coefficient constraints without first differentiating an infinite sum.

In the real Fourier series convention for period $2\pi$,where the Fourier coefficients areThese formulas apply to integrable periodic functions. Writing equality at each point requires appropriate convergence hypotheses; the periodic solution sought below is smooth, so its series causes no such difficulty.

The time already elapsed since the noise is $32$ minutes. Conditional on a particular noise time $T=t$ and on still being in progress, memorylessness of the exponential distribution givesThis law does not depend on $t$. Averaging over the conditional distribution of $T$ therefore leaves the same remaining-time law. HenceThe elapsed $32$ minutes are not subtracted from the original mean; the exponential residual lifetime has the same mean after any elapsed duration.