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By Thales's theorem, the two sides meeting on the circular arc are orthogonal, because the opposite side is a diameter. Both vectors point towards that meeting point; reversing both does not change their dot product. Consequently

This is a generalized eigenvalue problem. A nonzero vector lies in the kernel of $A-\lambda B$ precisely when its determinant vanishes:HenceThe corresponding eigenvectors can be taken as $(1,-1{)}^T$ and $(1,1{)}^T$, respectively, which also checks that both values give nontrivial solutions.

The two directed segments have vectors $\mathbf{u}=(1,0,-1)$ and $\mathbf{v}=(0,1,1)$. Their dot product is $-1$ and both norms equal $\sqrt{2}$. HenceThe common starting point fixes the obtuse angle between these segments.

Both segments start at $(1,1,1)$, so their direction vectors are $\mathbf{a}=(-1,-1,-1)$ and $\mathbf{b}=(1,1,0)$. Their dot product is $-2$, and their norms are $\sqrt{3}$ and $\sqrt{2}$. ThusThe angle between the rays lies in $[0,\pi ]$, so its sine is nonnegative:The negative cosine correctly identifies an obtuse angle. One also obtains the sine from $|\mathbf{a}\times \mathbf{b}|/(|\mathbf{a}||\mathbf{b}|)=\sqrt{2}/\sqrt{6}$, using the cross product.

The evaluated position vectors are ${\mathbf{r}}_1(1)=(2,1,0)$ and ${\mathbf{r}}_2(-1)=(-1,-1,0)$. Their difference is $(3,2,0)$, so the Euclidean distance is

The cross product of the direction vectors is ${\mathbf{n}}_1\times {\mathbf{n}}_2=(1,-1,0)$. It is nonzero, so a common perpendicular direction exists. Projecting ${\mathbf{a}}_2-{\mathbf{a}}_1=(-1,0,1)$ onto its unit vector gives the distance between skew lines:ThusThe numerator is a scalar triple product; movement along either line does not change this perpendicular projection.

The square $d(t{)}^2=(t+1{)}^2+1/2$ is minimized when its first term vanishes. ThereforeThis agrees with the distance between skew lines found in part (b). The corresponding parameter on the first line is its orthogonal projection valueThe closest pair is $(-1/2,-3/2,0)$ and $(-1,-1,0)$; their connecting vector $(-1/2,1/2,0)$ is perpendicular to both direction vectors.

The upward unit vector extracts the vertical curl component:It is constant over the unit disc, soThis agrees with part (d) by Stokes theorem: the boundary circle has counterclockwise orientation when viewed from the upward normal, including for the parametrizations $2\pi t^n$.

For $a=0$, differentiate the proposed scalar potential directly:ThusFor $a\ne 0$, the curl is the nonzero vector $(0,0,2a)$. A gradient of a twice differentiable scalar field has zero curl, so no such scalar field exists. Equivalently the closed line integral in part (d) is nonzero, contradicting the path independence of a conservative vector field. The field is smooth on all of ${\mathbb{R}}^3$, so no excluded singular set complicates this conclusion.

In cylindrical coordinates, $x^2+y^2=r^2$ and $r>0$, soThe potential vector has components $(A_r,A_{\theta },A_z)=(0,0,a\ln r)$. Expanding the prescribed cylindrical curl formula givesThereforeSince ${\hat{\mathbf{e}}}_{\theta }=(-\sin \theta ,\cos \theta ,0)$, this agrees with the Cartesian result and its clockwise orientation.

The logarithm is defined away from the $z$ axis, where $x^2+y^2>0$. For the vector potential $\psi {\hat{\mathbf{e}}}_z=(0,0,\psi )$, its curl is $({\psi }_y,-{\psi }_x,0)$. Sincewe obtainFor $a>0$ this is clockwise circulation about the axis, the velocity structure of a point vortex in each transverse plane.

The two orthogonal directions $(1,1{)}^T$ and $(1,-1{)}^T$ are preserved by this symmetric matrix. Direct multiplication, using the hyperbolic function identities $\cosh u\pm \sinh u=e^{\pm u}$, gives the eigenvalue–eigenvector pairsEach eigenvector can be multiplied by any nonzero scalar. For $u\ne 0$ these give the two one-dimensional eigenspaces. When $u=0$, the matrix is the identity: the only eigenvalue is $1$, with multiplicity two, and every nonzero vector is an eigenvector.

The dot product isBoth vectors have squared Euclidean norm $a^2+(a+1{)}^2=2a^2+2a+1$. Under the given condition $a^2+a=1/2$, their dot product is $1$ and each squared Euclidean norm is $2$. Hence

The vector $(1,-1,0{)}^T$ is an eigenvector with eigenvalue $4$. The complementary subspace of vectors with first two components equal is preserved by $A$: a vector $(s,s,t{)}^T$ maps to $(4s+2t,4s+2t,4s+2t{)}^T$. There, the eigenvalue equations reduce to the $2\times 2$ matrix $\left(\begin{array}{cc}4& 2\\ 4& 2\end{array}\right)$, whose characteristic polynomial is $\lambda (\lambda -6)$. For $\lambda =6$, $t=s$; for $\lambda =0$, $t=-2s$.

Normalising by the Euclidean norm, a complete orthonormal basis of eigenvectors isSign changes of any of these eigenvectors are equally valid. The eigenvalues also check part (a): their sum is $10$ and the sum of their squares is $52$.

Three vectors form a basis of three-dimensional space exactly when their scalar triple product is nonzero:It is the determinant of the matrix with these columns, so this condition is equivalent to linear independence and spanning the whole space.

Take the dot product of the expansion of $\mathbf{y}$ with $\mathbf{v}\times \mathbf{w}$. The terms containing $\mathbf{v}$ and $\mathbf{w}$ vanish, leaving $\mathbf{y}\cdot (\mathbf{v}\times \mathbf{w})=\alpha \Delta$. Cycling the same argument yieldsThe three cross-product numerators use the reciprocal basis.

At $A=(1/2,1/4)$, the gradient is $(-3/64,-9/32{)}^T$. The position vector has length $\sqrt{5}/4$, so its normalized direction is $(2,1{)}^T/\sqrt{5}$. Hence the directional derivative is

Expanding the cross product componentwise givesIts dot product with each original vector is zero, as required.

If the three vectors lie in one plane through the origin and $\mathbf{u},\mathbf{v}$ are independent, their cross product is a normal vector to that plane, so its dot product with $\mathbf{w}$ is zero. If $\mathbf{u},\mathbf{v}$ are dependent, their cross product is already zero. Thus coplanarity always forces the scalar triple product to vanish.

Taking the contrapositive proves that a nonzero scalar triple product rules out coplanarity. Geometrically its absolute value is the nonzero volume of the spanned parallelepiped.

Use displacements from $P_1$, since coplanarity of points concerns an affine plane that need not contain the origin. The three displacement vectors areTheir cross product is $\mathbf{u}\times \mathbf{v}=(-1,1,-1{)}^T$ and its dot product with $\mathbf{w}$ is $-1$. Since the scalar triple product is nonzero,

The vector triple product identity isConsequently a choice of the requested coefficients isThese are the standard identity's coefficients, even if the particular three vectors are linearly dependent and a representation need not be unique.