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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:

The three eigenvectors form an orthonormal basis, so an arbitrary vector has expansion $\mathbf{r}=a{\mathbf{v}}_{-}+b{\mathbf{v}}_0+c{\mathbf{v}}_{+}$. Multiplication by $A$ removes the zero-eigenvalue component:Its dot product with ${\mathbf{v}}_0$ is zero for every $a,b,c$. Thus one convenient answer is the nonzero vectorEquivalently, $A^{\mathsf{T}}\mathbf{u}=A\mathbf{u}=0$, so $(A\mathbf{r})\cdot \mathbf{u}=\mathbf{r}\cdot (A^{\mathsf{T}}\mathbf{u})=0$. Any nonzero scalar multiple of this vector in the null space also works.