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Choose two perpendicular unit directions in the $-2$ eigenspace:Their pairwise dot products vanish and their Euclidean norms are one. They form an orthonormal basis of eigenvectors, with eigenvalues $4,-2,-2$, as guaranteed for a real symmetric matrix by the spectral theorem for real symmetric matrices.

The normal from (ii) gives the plane equation $x+y+2z=3$. The distance from a point to a plane is the absolute value of the normalized plane equation at that point. At the origin this givesFor a geometric check, the perpendicular foot is $H=(1/2,1/2,1)$, parallel to the normal and satisfying the plane equation. Its Euclidean norm is the same distance.

Use outward normal vectors to the two faces. The cross productspoint away from the opposite vertices, since ${\mathbf{n}}_1\cdot \overrightarrow{OQ}=-3$ and ${\mathbf{n}}_2\cdot \overrightarrow{OP}=-3$. Their Euclidean norms are $3$ and $\sqrt{6}$, and their dot product is $-6$. The interior dihedral angle of a convex polyhedron is supplementary to the outward-normal angle, soThis also equals the cosine of the acute angle between the unoriented face planes. The outward-normal angle itself has cosine $-\sqrt{2/3}$; keeping that distinction avoids a sign ambiguity.

A real square orthogonal matrix $Q$ satisfiesequivalently $Q^{-1}=Q^T$. Its columns, and also its rows, form an orthonormal basis. Consequently it preserves dot products and Euclidean norms.

The cross product of the two in-plane directions is $\mathbf{u}\times \mathbf{v}=(-5,1,2)$, with Euclidean norm $\sqrt{30}$. Either sign gives a unit normal vector:The plane equation is $\hat{\mathbf{n}}\cdot (\mathbf{r}-\mathbf{A})=0$, so the distance from a point to a plane gives

The common ratio in the two plane equations can be denoted by $u$. Their intersection then has $x=1+4u$, $y=2+3u$, and $z=-1+2u$. The direction vector $(4,3,2)$ has Euclidean norm $\sqrt{29}$, so the requested unit vector form isSubstituting this expression into both equations gives the same ratio $\lambda /\sqrt{29}$, verifying the entire line.

The Euclidean norm of each position vector gives

A row's squared Euclidean norm is $[(1+\sqrt{3}{)}^2+1+(1-\sqrt{3}{)}^2]/6=3/2$, and the dot product of the two rows is $[2(1-3)+1]/6=-1/2$. Thus the row Gram matrix isFor the column Gram matrix, the first and third squared column norms are $4/3$, the middle one is $1/3$, and the respective cross-products are $1/3,-2/3,1/3$. HenceThe two products are square matrices of different sizes; their unequal dimensions are consistent with $A$ being rectangular.

For a matrix with equal diagonal entries, the symmetric and antisymmetric directions $(1,1)$ and $(1,-1)$ are eigenvectors. Direct multiplication of $A{A}^{\mathsf{T}}$ gives eigenvalues $3/2-1/2=1$ and $3/2+1/2=2$. Dividing each direction by its Euclidean norm givesThe two normalized eigenvectors are orthogonal; either may independently be replaced by its negative.

For $\lambda =a$, the components of an eigenvector are equal; for $\lambda =-a$, they are opposite. Dividing each vector by its Euclidean norm givesThey are mutually orthogonal and have unit norm. The labels specify their eigenvalues even when $a<0$.

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 $52$.

The displayed matrix is not orthogonal: its first column has squared Euclidean norm three, whereas every column of an orthogonal matrix must have squared length one.

For the general assertion, put $C=AB$. In the Einstein summation convention, orthogonality gives $A_{ki}{A}_{kj}={\delta }_{ij}$ and $B_{ki}{B}_{kj}={\delta }_{ij}$. The entries of $C$ satisfyThis is exactly $C^{T}C=I$, proving that the matrix product $AB$ is orthogonal. The Kronecker delta contracts the two middle indices without changing the multiplication order of the matrices.

Parameterize the surface by $\mathbf{r}(x,z)=(x,1+z,z{)}^T$. Its two tangent vectors are $(1,0,0{)}^T$ and $(0,1,1{)}^T$, whose cross product is $(0,-1,1{)}^T$ of Euclidean norm $\sqrt{2}$. Thus the scalar area element of a surface isEquivalently, projection onto the $xz$-plane reduces the surface area by $|\mathbf{n}\cdot {\mathbf{e}}_y|=1/\sqrt{2}$.

The traversal fixes the orientation of the vector area as half the cross product of the edges $AB$ and $AC$:Its Euclidean norm $1/\sqrt{2}$ is the area of the triangle, and its direction follows the right-hand rule.

The first coordinate direction is already an eigenvector of $A$ with eigenvalue $4$. On the last two coordinates, the matrix is $\left(\begin{array}{cc}2& 4\\ 4& 2\end{array}\right)$, with characteristic polynomialThus the three eigenvalues are $4,6,-2$. For $\lambda =6$, the last two components are equal; for $\lambda =-2$, they are opposite. Normalising by the Euclidean norm gives the complete listEach sign-reversed vector is an equally valid choice. The three displayed eigenvectors are mutually orthogonal, as expected for a real symmetric matrix with distinct eigenvalues.

The modulus is the Euclidean norm, obtained by taking the square root of the dot product of the vector with itself:

The dot product is $1\cdot 0+2(-1)+(-1)\cdot 1=-3$, while the Euclidean norms are $\sqrt{6}$ and $\sqrt{2}$. Therefore the angle $\theta \in [0,\pi ]$ satisfies

The removed vector is the orthogonal projection of $\mathbf{r}$ onto the line through the origin in direction $\hat{\mathbf{n}}$. The remaining vector measures the perpendicular displacement from that line. Fixing its Euclidean norm at $2$ therefore describes an infinite circular cylinder of radius $2$, with axis $x=z$, $y=0$.

The dot product with the given unit vector is $\mathbf{r}\cdot \hat{\mathbf{n}}=(x+z)/\sqrt{2}$. Subtracting the corresponding orthogonal projection givesTaking its squared Euclidean norm yields the equivalent Cartesian equation