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Write $\mathbf{u}=\mathbf{b}-\mathbf{a}$. Since the triangle is nondegenerate, $\mathbf{u}\ne 0$. Every point on the line $AB$ has a unique position vector $\mathbf{p}=\mathbf{a}+t\mathbf{u}$. The condition that $CP$ is orthogonal to $AB$ is the dot product equationIts coefficient of $t$ is $-\Vert \mathbf{u}{\Vert }^2\ne 0$, so it has exactly one solution. The foot of a perpendicular is thereforeThis is the orthogonal projection of $C$ onto the line $AB$, expressed using the Euclidean norm.

For the parametrized surface in i,These tangent vectors are orthogonal, and their Euclidean norms are $\sqrt{(c-a{)}^2+(d-b{)}^2}$ and $\rho$, respectively. Their cross product therefore has Euclidean norm $\rho \sqrt{(c-a{)}^2+(d-b{)}^2}$. Integration givesThis is the lateral surface area of a conical frustum, with no end disks included.

Let $\mathbf{r}(u,v)$ parametrize $S$. Then $A\mathbf{r}$ parametrizes its image, with tangent vectors $A{\mathbf{r}}_u$ and $A{\mathbf{r}}_v$. An orthogonal matrix preserves dot products, so the Lagrange identity for the cross product gives, for any $\mathbf{p},\mathbf{q}$,The area element of a surface is therefore the same pointwise before and after applying $A$. Integrating provesThis includes orthogonal transformations with determinant $-1$: reversal of orientation does not affect the Euclidean norm in the surface area formula.

Use the parametric curveHere $|y(t)|\le \sqrt{5}<3$, so the ellipse stays strictly above the axis of rotation $y=-3$. Rotating in three dimensions gives a surface of revolution with parametrizationThe two parameter tangent vectors are orthogonal, with Euclidean norms $\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}$ and $y(t)+3$. Hence the area element of a surface isThe generating ellipse is simple and has positive distance from the axis of rotation; its rotation sweeps the surface once, apart from the identified parameter endpoints. AlsoWriting $w(t)=\sqrt{8{\sin }^{2}t+2{\cos }^{2}t}$, the surface area is therefore $2\pi {\int }_0^{2\pi }(3+y(t))w(t)dt$. Since $y(t+\pi )=-y(t)$ and $w(t+\pi )=w(t)$,ConsequentlyThe cancellation is an instance of the surface area of revolution of a centrally symmetric curve: the surface area is the generating curve’s arc length times the distance $6\pi$ traveled by its center.

A normal vector to the plane spanned by $\mathbf{a},\mathbf{b}$ is their cross product:Dividing by this Euclidean norm gives the two possible choices of unit normal,

The two edge vectors based at $A$ are $\mathbf{b}-\mathbf{a}=(-3,-1,3)$ and $\mathbf{c}-\mathbf{a}=(-3,2,2)$. Their cross product is $(-8,-3,-9)$, with squared Euclidean norm $64+9+81=154$. The triangle occupies half the corresponding parallelogram, so

Call the given matrix $M$. Its middle coordinate decouples, immediately giving the eigenvalue $3$ with normalized eigenvector $(0,1,0{)}^T$. The remaining characteristic polynomial comes from the $xz$ block:For $\lambda =-4$, the first row of $(M+4I)v=0$ gives $x=\sqrt{3}z$; for $\lambda =8$, the first row of $(M-8I)v=0$ gives $z=-\sqrt{3}x$. After vector normalization, one possible complete set isEach has unit Euclidean norm. The dot products with $v_2$ vanish because the other two middle components are zero, and $v_1\cdot v_3=(-\sqrt{3}+\sqrt{3})/4=0$. This explicitly verifies orthonormality, also predicted by the spectral theorem for real symmetric matrices with distinct eigenvalues. Changing any eigenvector's sign gives an equally valid normalized choice.

With normal vector $n=(-1,-2,2)$, a plane through $A=(2,-1,1)$ has equation $n\cdot (r-A)=0$. As $n\cdot A=2$, its plane equation isThe distance from a point to a plane is the absolute signed offset divided by the Euclidean norm of its normal vector. Hence the Euclidean distance from the origin is $\boxed{2/3}$, since $|n|=3$. The closest point is $2n/9=(-2,-4,4)/9$, whose Euclidean distance from the origin is $2/3$ and which satisfies the plane equation.

The displacement vectors are $B-A=(-1,2,-2)$ and $C-A=(1,0,1)$. Their cross product is $(2,-1,-2)$, with Euclidean norm three. Thus either sign ofis a unit normal vector to the second plane, whose equation is $2x-y-2z=3$. A direction vector for the intersection is the cross product of the two normal vectors:It is orthogonal to each normal vector. Since both planes contain $A$, the whole intersection line is $r=(2,-1,1)+t(6,2,5)$, $t\in \mathbb{R}$.

The characteristic polynomial is $(5-\lambda )(2-\lambda )-4={\lambda }^2-7\lambda +6=(\lambda -1)(\lambda -6)$. For $\lambda =1$, the eigenvector equation gives $y=2x$; for $\lambda =6$, it gives $x=-2y$. The normalized eigenvectors can therefore be chosen asBoth have unit Euclidean norm, and their dot product is $(-2+2)/5=0$, explicitly verifying orthogonality. This also follows from the spectral theorem for a real symmetric matrix with distinct eigenvalues.

The trace sums the diagonal entries, so $\mathrm{tr}P={\sum }_{i}1-{\sum }_i{a}_i^2-{\sum }_i{b}_i^2$. The two unit vectors have squared Euclidean norms one, giving $\boxed{\mathrm{tr}P=n-2}$.

The Euclidean norms are $\Vert v_{\pm }\Vert =2$ and $\Vert v_0\Vert =\sqrt{2}$. With normalized eigenvectors as rows,The rows form an orthonormal basis, so $R{R}^T=I$ and $R$ is an orthogonal matrix. The zero eigenvalue means $M{v}_0=0$ for a nonzero vector; thus $M$ is singular and has no matrix inverse.

The distance from a point to a plane, applied at the origin, is the plane constant divided by the Euclidean norm of its normal vector. ThereforeThe closest point on a plane lies on the normal through the origin; this is an orthogonal projection onto an intercept plane.

Expanding the determinant gives $\det (\lambda I-M)=(\lambda -1)[(\lambda -1{)}^2-25]$. Solving each eigenvalue equation yieldsTheir pairwise inner products are $16-25+9=0$, $12-12=0$ and $12-12=0$. Their squared Euclidean norms are $50,50,25$, respectively, so division by $\sqrt{50},\sqrt{50},5$ gives an orthonormal basis. This also follows from the spectral theorem for real symmetric matrices: symmetry makes eigenvectors for distinct eigenvalues orthogonal.

From the line point $r_0=(1,-1,0)$ to $P=(2,2,1)$, take $w=P-r_0=(1,3,1)$ and line direction $d=(2,3,1)$. Orthogonal projection gives the nearest parameter $t=(w\cdot d)/(d\cdot d)=12/14=6/7$. The perpendicular displacement has squared Euclidean normHence the distance from a point to a line in three dimensions isThe nearest point is $(19/7,11/7,6/7)$; subtracting it from $P$ gives a vector orthogonal to $d$, verifying the minimum.

A tangent displacement has the form $(dx,dy,z_{x}dx+z_{y}dy)$. At the specified point, $z_y=0$ and $z_x=q>0$, so the fastest decrease of height per unit surface distance has $dy=0$ and $dx<0$. The unit downhill tangent is $(-1,0,-q)/\sqrt{1+q^2}$. Therefore, for speed $u$ measured along the surface,Its Euclidean norm is $u$, and it is orthogonal to the surface normal. The general construction is steepest descent on a graph surface; using a horizontal speed equal to $u$ would overestimate the full three-dimensional speed.

The three columns of $N=\sqrt{8}A$ are $(\sqrt{3},1,2)$, $(-\sqrt{2},\sqrt{6},0)$, and $(-\sqrt{3},-1,2)$. Each has squared Euclidean norm eight. Their pairwise dot products areHence $N^{\mathsf{T}}N=8I$, so $\boxed{A^{\mathsf{T}}A=I}$. Thus $A$ is an orthogonal matrix, with $A^{-1}=A^{\mathsf{T}}$. The determinant from (i) places it in the determinant-one component.

The dot product is $\mathbf{a}\cdot \mathbf{b}=-1+2+3=4$, while the Euclidean norms are $\sqrt{3}$ and $\sqrt{14}$. Thus the angle in the stated range isThe positive cosine shows that it is an acute angle.

Hold $r$ fixed and differentiate the position vector $\mathbf{X}(\theta ,\varphi )$. Its coordinate tangent vectors areTheir dot product is zero, and their squared Euclidean norms are $r^2$ and $r^2{\sin }^2\theta$. To first order, the displacement is $\delta \mathbf{X}={\mathbf{X}}_{\theta }\delta \theta +{\mathbf{X}}_{\varphi }\delta \varphi +O((|\delta \theta |+|\delta \varphi |{)}^2)$. Squaring givesThis is the line element on a sphere. The chord distance and the short surface arc length agree to the order retained here.

A plane through the origin has equation $lx+my+nz=0$ for some nonzero normal $(l,m,n)$. Dividing that normal by its Euclidean norm makes $l^2+m^2+n^2=1$ without changing the plane. Substitute the spherical coordinates from (a) and cancel the nonzero radius:Its intersection with the sphere is the desired great circle; changing the sign of all three coefficients describes the same circle.