Let $\Delta =[\mathbf{a},\mathbf{b},\mathbf{c}]\ne 0$. These three vectors are linearly independent, so write $\mathbf{d}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}$. Taking the dot product with $\mathbf{b}\times \mathbf{c}$ gives $\alpha =[\mathbf{b},\mathbf{c},\mathbf{d}]/\Delta$. Likewise, the dot products with $\mathbf{c}\times \mathbf{a}$ and $\mathbf{a}\times \mathbf{b}$ giveHere the cyclic invariance and sign change under an interchange of the scalar triple product fix every sign. Substituting the three coefficients provesThis is precisely coordinate extraction using the reciprocal basis $(\mathbf{b}\times \mathbf{c},\mathbf{c}\times \mathbf{a},\mathbf{a}\times \mathbf{b})/\Delta$.

The expected value follows by integration by parts:The same result follows from symmetry of the probability density function about $\pi /2$.

Normalization of the probability density function requires $1=\alpha {\int }_0^{\pi }\sin xdx=2\alpha$. Thus

The plane contains $(1,0,1)$ and has normal vector $(0,1,2)$, so its equation is $y+2z=2$. In the equation $\mathbf{n}\cdot \mathbf{r}=d$ with a unit normal, $|d|$ is the perpendicular distance from the origin. Consequently

The direction vectors are $\mathbf{u}=(1,-2,1)$ and $\mathbf{v}=(0,-2,1)$. Their cross product is $(0,-1,-2)$, with norm $\sqrt{5}$. A unit normal to the plane is thereforeThe two intersecting directions span the plane, so a vector orthogonal to both is a normal vector to that plane.

In the completed-square equation, the right side is the squared radius of the circle. Taking its positive square root gives

Completing the squares gives the standard equation of a circle:Its centre is therefore

The initial term at $n=-1$ is three, and the common ratio is $1/3$. The convergent geometric series therefore sums to

The binomial theorem givesThus, to three decimal places,The first three nonconstant terms already determine this rounding; the remaining positive terms total less than $1.3\times 10^{-6}$.

Choose two distinct space-diagonal directions $\mathbf{u}=(1,1,1)$ and $\mathbf{v}=(-1,1,1)$. Their dot product is one and their norms are both $\sqrt{3}$. Thus the acute angle between their unoriented lines isAny two distinct space diagonals have an absolute dot product of one when represented with components $\pm 1$, so they give the same acute angle.

A space diagonal of the unit cube has displacement vector $(1,1,1)$. Its Euclidean norm, or two successive applications of the Pythagorean theorem, givesThe requested diagonal joins opposite vertices; $\sqrt{2}$ would be the diagonal of a face.

The two radii enclose the angle $2\pi /3$, so the triangle has area $T=\frac{1}{2}{r}^2\sin (2\pi /3)=\sqrt{3}{r}^2/4$. From the sector calculation, $r^2=3B/\pi$. Consequently

The two labelled regions in the original diagram are complementary circular sectors. Their areas add to $\pi r^2$ and satisfy $A=2B$, so $B=\pi r^2/3$. The sector with central angle $\theta$ has area $r^2\theta /2$. Hence

Since $d({\sin }^{2}x)/dx=2\sin x\cos x=\sin 2x$, integration by substitution $u={\sin }^{2}x$ immediately givesHere $C$ is an arbitrary constant of integration.

Use integration by parts, taking $u=x$ and $dv=\sin 3xdx$:

The stated derivative formula needs a domain qualification. The original PDF calls $\alpha$ real, but the integral is even in $\alpha$; its derivative therefore cannot equal the printed positive expression on both sides of zero.

First take $\alpha >0$. The improper integral converges, with an $O(x^2)$ integrand near zero and $O(\ln x/x^2)$ decay at infinity. On compact positive parameter intervals, an integrable bound also justifies differentiation under the integral sign:For $\alpha \ne 1$, the partial fraction decomposition isIntegrating the rational functions givesAt $\alpha =1$, the same result follows by continuity, or directly from $2{\int }_0^{\infty }{x}^2/(1+x^2{)}^{2}dx=\pi /2$.

Integrating gives $E(\alpha )=\pi \ln (1+\alpha )+C$ on the positive half-line. To fix $C$, take $\alpha \to 0^{+}$: for $0\le \alpha \le 1$ the integrand is bounded by $\ln (1+x^2)/(1+x^2)$, an integrable function. The dominated convergence theorem gives $E(\alpha )\to E(0)=0$, so $C=0$. Evenness then extends the logarithmic parameter integral with Cauchy weight to its full stated domain:For negative $\alpha$, the derivative is $-\pi /(1-\alpha )$. At zero the right and left derivatives are $\pi$ and $-\pi$, so $E$ is not differentiable there. The pointwise derivative of the integrand at zero is zero, but interchanging it with the infinite integral at that parameter is invalid. Finally, substituting $\alpha =1$ gives the requested special value

For $\alpha \ne 0$, integration by substitution $t=\alpha x$ givesAt $\alpha =0$ the integrand is one, so $I(0)=1$, also the continuous limit of this expression. For nonzero $\alpha$, differentiation under the integral sign is valid on the finite interval and yieldsSet $\alpha =\sqrt{3}$ and use $\arctan \sqrt{3}=\pi /3$. The requested value is

The quadratic formula gives the two polynomial roots $(-3\pm \sqrt{5})/2$. Its leading coefficient is positive, so the quadratic polynomial is strictly negative between them. Therefore

Move the constant term to the left and factor the cubic polynomial:Setting each factor to zero gives all three polynomial roots: