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For the particular eigenvectors chosen in (iii), the dot products areThus only ${\mathbf{e}}_1$ and ${\mathbf{e}}_2$ are orthogonal for this choice. The matrix is not symmetric, so distinct eigenvalues alone do not imply orthogonality. Another basis within the repeated eigenspace may change which chosen vectors are perpendicular; the dot products above specify the convention completely.

Perpendicular vectors have zero dot product. Here $\mathbf{u}\cdot \mathbf{v}=3\cdot 4-2c$, so $12-2c=0$. Therefore

Use orthogonality to the image of a matrix. For $n=(1,2,1{)}^T$, the condition is $n^{T}A=0$, since $n^{T}Ar$ must vanish for every nonzero input. Direct multiplication givesEach coefficient must vanish; choosing the three coordinate basis vectors as inputs already proves necessity. Solve $2b+c=-3$, $b+2c=1$ to obtainThese values make every output have zero dot product with $n$, including any zero output. Here orthogonality is understood in this algebraic sense; no direction is assigned to the zero vector.

The nonzero vectors $b,c$ are normal vectors to the two planes. If $b\times c=0$, then $c=qb$ for some nonzero real $q$. Their equations are therefore $r\cdot b=\lambda$ and $q(r\cdot b)=\mu$.

When $\mu =q\lambda$, the planes coincide and their intersection is an entire plane. Otherwise they are distinct parallel planes and have no intersection. Neither case has exactly a line as its intersection. Hence $b\times c\ne 0$ is necessary for intersection in a line. It means that the two normal vectors are not parallel; in that case it is also sufficient.

A directional derivative measures change per unit distance, so first normalize the direction: $\hat{v}=(4,3)/5$. The gradient isTaking the dot product with the unit vector,Using the unnormalized vector would instead give $-2$, the change per unit parameter along $(1,0)+tv$; it is not the rate per unit distance requested here.

Solving $(A-4I)\mathbf{v}=0$ gives $v_2=0$ and $v_1=v_3$. A convenient normalized eigenvector isThe displayed nonzero vector proves that four is an eigenvalue. Every nonzero multiple represents the same eigendirection.

A vector perpendicular to ${\mathbf{e}}_1$ has the form $\mathbf{v}=(s,t,-s)$. Direct matrix multiplication givesThus every nonzero such vector is an eigenvector with $\boxed{\lambda =-2}$. The zero vector must be excluded from the literal claim about “any” perpendicular vector, since an eigenvector is nonzero. The eigenspace is two-dimensional. Equivalently, $A=-2I+6{\mathbf{e}}_1{\mathbf{e}}_1^{\mathsf{T}}$ exhibits the spectrum of a scalar matrix plus a rank-one projector.

The volume of a parallelepiped is the absolute scalar triple product of its three edge vectors. Since $\overrightarrow{OP}\times \overrightarrow{OQ}=(0,0,3)$,The factor $1/6$ appropriate to a tetrahedron must not be applied to this parallelepiped.

If either vector is zero, both sides vanish. Otherwise let $\theta$ be the angle between them. The magnitudes of the cross product and dot product are described bySquaring and using ${\sin }^2\theta =1-{\cos }^2\theta$ proves Lagrange's identity:

Use $(0,1,1)$ as an anchor. Two directions in the required plane areTheir cross product is $(-2,8,2)$, so a normal vector is $\mathbf{n}=(-1,4,1)$. The Cartesian equation of a plane isDirect substitution verifies all three given points. The independent direction vectors ensure the plane is uniquely determined.

The axis intercepts of the plane areUse two edge vectors from $P$:The triangle area is half this cross product magnitude:

The two derivatives are $2x$ and $2(x-1)$. At the intersection their slopes are $1$ and $-1$, so their direction vectors $(1,1)$ and $(1,-1)$ have zero dot product. Thus the tangent lines are perpendicular and their angle is

Project the radius vector onto the $x$-$y$ plane. The polar angle $\theta =\pi /6$ lies between the positive $z$ axis and $OP$. The azimuth $\varphi =\pi /3$ lies in the $x$-$y$ plane between the positive $x$ axis and the projection of $OP$, measured toward positive $y$.

The symmetry between the first and third coordinates makes $(1,0,-1{)}^{\mathsf{T}}$ an eigenvector, with eigenvalue $-6$. In the remaining invariant plane, write a vector as $(s,t,s{)}^{\mathsf{T}}$. Its image is $(8s-4t,-8s+4t,8s-4t{)}^{\mathsf{T}}$, so the reduced eigenvalue equation has determinantThe corresponding directions have $t=2s$ for $\lambda =0$ and $t=-s$ for $\lambda =12$. Normalizing givesTheir squared unnormalized norms are $2,6,3$. The pairwise unnormalized dot products are $1-1=0$, $1-1=0$, and $1-2+1=0$, so these unit eigenvectors are mutually orthogonal. The eigenvalues sum to $6$ and have product zero, agreeing with part (a).

Expand an arbitrary vector in the orthonormal eigenbasis:Its image has no component along ${\mathbf{e}}_0$, soAlternatively, symmetry gives $(A\mathbf{r})\cdot {\mathbf{e}}_0=\mathbf{r}\cdot A{\mathbf{e}}_0=0$. This is image-kernel orthogonality for an adjoint, with $A^{\mathsf{T}}=A$.

The nonparallel direction unit vectors have nonzero cross product. LetFor any joining vector $\mathbf{s}=\mathbf{c}-\mathbf{a}+\mu \hat{\mathbf{d}}-\lambda \hat{\mathbf{b}}$, its normal component is fixed:Thus its length is at least the absolute value of that component. Since the two line directions span ${\mathbf{n}}^{\perp }$, suitable $\lambda ,\mu$ cancel the entire component in that plane, attaining the bound. Therefore the distance between skew lines, including the zero-distance intersecting case, isThe numerator is a scalar triple product. The nonparallel hypothesis is essential to the denominator and to the spanning argument.

Choose any nonzero auxiliary vector $\mathbf{q}$ not collinear with $\mathbf{p}$. Then $\mathbf{p},\mathbf{q},\mathbf{p}\times \mathbf{q}$ form a basis, so writeTaking the dot product with $\mathbf{p}$ gives $\alpha |\mathbf{p}{|}^2+\beta \mathbf{p}\cdot \mathbf{q}=k$. This separates the two cases requested in the hint.

If $\mathbf{p}\cdot \mathbf{q}\ne 0$, choose $\alpha ,\gamma$ freely and solve for $\beta$:If $\mathbf{p}\cdot \mathbf{q}=0$, the coefficient of $\mathbf{p}$ is fixed and the other two are free:These formulas describe the whole plane with normal $\mathbf{p}$, including $k=0$. Completeness follows from the basis expansion, not only from checking that the displayed vectors satisfy the equation. The parametrization of a plane with an auxiliary vector also gives a single formula valid in both cases.

Collect the coefficients of the unknowns into the matrix and the constants into a column. The requested linear system isThe unknown coordinate $y$ here is a scalar component; the right-hand column is the vector denoted by the question's matrix-form notation.

The vectors $(1,1)$ and $(1,-1)$ are eigenvectors, with respective eigenvalues $\cosh \varphi +\sinh \varphi$ and $\cosh \varphi -\sinh \varphi$. The exponential definitions reduce these toAt $\varphi =0$ they coincide and the matrix is the identity.

Let ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$ be the Cartesian basis vectors, with the given vector $\mathbf{x}={\mathbf{e}}_1$. The permutation matrix cycles them forward:Thus three applications return the original vector.