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For $\mathbf{F}$, all three curl components vanish for every $a$. More constructively,Hence $\mathbf{F}$ is conservative for every real $a$. Its endpoint vector-field potential difference is $a$, checking both line integrals.

For the other vector field, direct partial differentiation givesIt therefore has a vector-field potential exactly when $b=0$, with no restriction on $c$:The value of $a$ is irrelevant to $\mathbf{G}$. Its rotational part for general $b$ is $b(1,1,1{)}^T\times \mathbf{r}$. Both selected paths lie in $x=y$, and this rotational part has zero dot product with every tangent vector to that plane. That explains why these paths cannot detect its nonzero curl when $b\ne 0$.

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 $|\mathbf{n}\cdot {\mathbf{e}}_y|=1/\sqrt{2}$.

For spherical coordinates, the distance from the origin is $\sqrt{26}$ and the polar angle satisfies $\cos \theta =z/r=-1/\sqrt{26}$. The azimuth is the same as in part (a). ThusThe polar angle is in $(\pi /2,\pi )$ because the point lies below the $xy$-plane.

For each possibility in part (i), squaring the diagonal entries gives $M^2=I$. Equivalently, the linear transformation fixes a vector space basis after being applied twice. ThereforeGeometrically, reflecting twice in the same plane returns every vector to itself.

The three coordinate basis vectors are eigenvectors, so $M$ is diagonal in those coordinates. Exactly one coordinate is multiplied by $-1$, while the other two are fixed. The three possibilities areThey are respectively reflections in the coordinate planes $x=0$, $y=0$ and $z=0$. Each is a Householder reflection with unit normal equal to the coordinate eigenvector carrying the eigenvalue $-1$.

The vector $\mathbf{c}=\mathbf{p}\times \hat{\mathbf{a}}$ is a normal vector to the span of $\mathbf{p}$ and $\hat{\mathbf{a}}$. Since $\mathbf{c}\ne 0$, these vectors determine a plane. Thus the missing words are

the line and the origin.

Indeed this plane consists of all linear combinations of $\mathbf{p}$ and the line's direction vector.

Call the three planes $P_1,P_2,P_3$ in their given order.

For $\alpha \ne 3,-1/2$, the nonzero determinant makes their normal vectors linearly independent. The three planes meet at the single point found in part (c)(i), for every $\beta$.

For $\alpha =3$, $P_2$ and $P_3$ coincide. Their common plane and $P_1$ have nonparallel normal vectors, since their first two coefficients are $(1,-1)$ and $(1,6)$. Thus they intersect in the straight line from part (c)(ii), for every $\beta$.

For $\alpha =-1/2$, write the three plane equations asThe last two meet on $z=0$, $x-y=-1$, which never lies on $P_1$. If $\beta =3$, $P_1$ and $P_2$ are distinct parallel planes, while $P_3$ cuts both. If $\beta =-1/2$, $P_1$ and $P_3$ are distinct parallel planes, while $P_2$ cuts both. For every other $\beta$, each pair of planes intersects in a straight line; these three pairwise intersection lines are distinct and parallel to $(1,1,0{)}^T$, with no common point. These exhaust the geometric cases.

In spherical coordinates, the radial distance is $\sqrt{x^2+y^2+z^2}=\sqrt{3}$ and $\cos \theta =z/r$. Since the point lies below the equatorial plane, the polar angle exceeds $\pi /2$:The angle is approximately $2.1863$ radians, or $125.26^{\circ }$.

The chain rule gives $d(e^{xy})=e^{xy}(ydx+xdy)$, so the given total differential is exact. Thus $f=e^{xy}+C$; two such functions differ by a constant because their gradient difference vanishes on the connected plane. The assigned value at the origin forces $C=-1$, giving

No. Every vector in the plane can be written asThe two linearly independent vectors in part (ii) therefore form a vector space basis for this vector subspace. Its dimension is $2$, so any three vectors in it have linear dependence.

This is a plane through the origin in three-dimensional Euclidean space, with nonzero normal vector $(1,2,-3{)}^T$:

The plane equation $x+z=0$ has normal vector $(1,0,1{)}^T=\sqrt{2}\hat{\mathbf{n}}$. Thus its normal vector is parallel to the cylinder's axis and the plane is perpendicular to that axis. It passes through the origin, so the intersection is a circle of radius $2$ centred at the origin in that plane.

Indeed, with the orthonormal basis $\mathbf{e}=(1,0,-1{)}^T/\sqrt{2}$ and $\mathbf{f}=(0,1,0{)}^T$ of the plane, write $\mathbf{r}=s\mathbf{e}+t\mathbf{f}$. The cylinder equation becomes $s^2+t^2=4$. Equivalently, the intersection has the parametric equation

Use the distance from a point to a plane formula with the unit normal $\hat{\mathbf{n}}$. Since the plane passes through the origin, the distance isFor a direct check, the closest point on a plane is $\mathbf{p}-(\mathbf{p}\cdot \hat{\mathbf{n}})\hat{\mathbf{n}}=(1,\sqrt{2},-1{)}^T$, whose displacement from $\mathbf{p}$ has Euclidean norm $\sqrt{2}$.

For the stipulated linearly independent vectors ${\mathbf{v}}_1,{\mathbf{v}}_2$, the denominator in part (i) is nonzero. ThusEquivalently, ${\mathbf{a}}_2-{\mathbf{a}}_1$ belongs to the linear span of ${\mathbf{v}}_1,{\mathbf{v}}_2$. Geometrically the lines lie in one plane, and, since their direction vectors are not parallel, they intersect at a unique point. The linear combination expressing ${\mathbf{a}}_2-{\mathbf{a}}_1$ gives the two parameters of that intersection.

Let $\mathbf{d}={\mathbf{a}}_2-{\mathbf{a}}_1$ and $\mathbf{w}={\mathbf{v}}_1\times {\mathbf{v}}_2$. The assumed linear independence gives $\mathbf{w}\ne 0$. Every displacement joining one line to the other isIts dot product with $\mathbf{w}$ is always $\mathbf{d}\cdot \mathbf{w}$. The component in that direction is therefore fixed, so orthogonal decomposition gives $\Vert \mathbf{s}\Vert \ge |\mathbf{d}\cdot \mathbf{w}|/\Vert \mathbf{w}\Vert$.

The two linearly independent vectors ${\mathbf{v}}_1,{\mathbf{v}}_2$ span the plane perpendicular to $\mathbf{w}$. Suitable ${\lambda }_1,{\lambda }_2$ cancel the component of $\mathbf{d}$ in that plane, attaining the lower bound. Thus the distance between skew lines formula, also valid when the lines intersect, gives

The matrix issoHere $\Vert \mathbf{p}{\Vert }^2=2$, so $P=I-2\hat{\mathbf{p}}{\hat{\mathbf{p}}}^T$ with $\hat{\mathbf{p}}=\mathbf{p}/\sqrt{2}$. This is a Householder transformation: reflection in the plane $x_1+x_2=0$. The component parallel to the normal vector $\mathbf{p}$ changes sign, while the perpendicular component is fixed.

The second matrix acts byApplying $P$ from part (e) then givesThe Householder transformation $Q$ reflects in the plane $x_1-x_2=0$. Its normal vector $\mathbf{q}$ is perpendicular to $\mathbf{p}$, so the two reflecting planes meet at a right angle along the $x_3$ axis. Their composition is a rotation through $\pi$ about the $x_3$ axis, as is also immediate from the displayed rotation matrix.

From part (f), $PQ=\mathrm{diag}(-1,-1,1)$. To obtain $PQR=-I$, choose the Householder transformation that reverses only the third coordinate:Indeed $\mathbf{r}{\mathbf{r}}^T=\mathrm{diag}(0,0,2)$, so $R_{ij}={\delta }_{ij}-r_i{r}_j$, and direct matrix multiplication gives $PQR=-I$. The alternative vector $-\mathbf{r}$ produces the same matrix. Geometrically $R$ reflects in the plane $x_3=0$, completing the reversal of all three coordinates.

The relevant mixed partial derivatives agree:The vector field is smooth on the entire simply connected plane, so this condition establishes that it is a conservative vector field. We can also construct its vector-field potential explicitly. From ${\varphi }_x=y^2-2x$, take an antiderivative in $x$:Then ${\varphi }_y=2xy+h^{\prime }(y)=2xy-y$ requires $h^{\prime }(y)=-y$. HenceThe fundamental theorem for line integrals checks the previous answer: $\varphi (1,1)-\varphi (0,0)=-1/2$.

No. Every one of the three vectors has equal first and second Cartesian coordinates, so their linear span is contained in the plane $x=y$. That linear subspace has dimension two and cannot equal ${\mathbb{R}}^3$.

More explicitly, call the three given vectors $\mathbf{u},\mathbf{v},\mathbf{w}$ in their listed order. ThenThis nontrivial linear relation proves linear dependence, whereas a basis must be linearly independent. Being unit vectors alone does not ensure that they form a basis.