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The independent vectors $\mathbf{a},\mathbf{b}$ span the plane through the origin, and their cross product is a nonzero normal vector. Thus the required nonparametric equation is

Take two edge vectors in the face, $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$. Their cross product is the face normal vectorThe face plane is $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$. Every term in $\mathbf{a}\cdot \mathbf{n}$ except $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$ vanishes, givingThe scalar triple product is nonzero, so this normal cannot vanish.

The PDF diagram specifies the prism asIts triangular horizontal cross-section has area four. Calculate the outward flux on each face using its outward normal vector.

On $x=0$, the normal is $-{\mathbf{e}}_x$ and $F_x=0$, so the flux is zero. On $y=0$, the normal is $-{\mathbf{e}}_y$, givingOn the sloping vertical face $y+2x=4$, the outward normal is proportional to $(2,1,0)$. Both $F_x$ and $F_y$ vanish there, while $F_z$ is tangent to the face, so the flux is zero.

On the bottom $z=0$, the normal is $-{\mathbf{e}}_z$ and $F_z=1$, so the flux is minus the base area, namely $-4$. On the top $z=1$, $F_z=0$, so the flux is zero. Thus all five face values and their sum are

The specified unit normal vector is ${\mathbf{e}}_z$, while $F_z=0$ everywhere. Thus the normal component of the field vanishes pointwise, giving flux

Here $y=0$, the unit normal vector is ${\mathbf{e}}_y$, and $\mathbf{F}\cdot {\mathbf{e}}_y=x^2$. The square has $-1/2\le x,z\le 1/2$, so its outward flux is

The direction vectors are $\mathbf{u}=(1,-2,1)$ and $\mathbf{v}=(0,-2,1)$. Their cross product is $(0,-1,-2)$, with norm $\sqrt{5}$. A unit normal to the plane is thereforeThe two intersecting directions span the plane, so a vector orthogonal to both is a normal vector to that plane.

The plane contains $(1,0,1)$ and has normal vector $(0,1,2)$, so its equation is $y+2z=2$. In the equation $\mathbf{n}\cdot \mathbf{r}=d$ with a unit normal, $|d|$ is the perpendicular distance from the origin. Consequently

The normal vector direction is orthogonal to the tangent line. For finite nonzero slopes, their product is $-1$. The tangent has slope one, so the normal line has

The sloping height relation on the cone is $z=b(1-r/a)$. A parametric equation of its curved surface is thereforeIts two tangent vectors are ${\mathbf{R}}_r=(\cos \theta ,\sin \theta ,-b/a)$ and ${\mathbf{R}}_{\theta }=r(-\sin \theta ,\cos \theta ,0)$. Multiplication by $dr$ and $d\theta$ produces the two infinitesimal sides requested in the paper. Their cross product is the vector area element:Its radial component is positive, so this normal vector points outwards from the solid cone. Taking its norm gives the scalar area elementThe parametric equation degenerates at the apex, but that single point contributes no area.

By implicit differentiation of the ellipse equation,The tangent line through $(1,1)$ is thereforeIts normal vector is $(9,16)$, parallel to the gradient $(18,32)$ of $9x^2+16y^2$ at that point.

At a regular constrained extremum, every tangent vector to the level set $g=A$ gives zero directional derivative of $f$. Thus $\nabla f$ is perpendicular to the tangent plane, and must be a scalar multiple of its nonzero normal vector $\nabla g$. The Lagrange multiplier equations areOne solves these four equations for $x,y,z,\lambda$, evaluates $f$ at the candidates, and classifies or compares the values as required. Points where $\nabla g=0$, and any boundary or limiting cases of the feasible domain, require separate checks because the regularity argument does not cover them.

Introduce the second constraint $q(x,y,z)=y-h(x)=0$ and a second Lagrange multiplier $\mu$. At points where the two constraint gradients are linearly independent, solveThese are five scalar equations for $x,y,z,\lambda ,\mu$. The two constraint normal vectors span the normal plane to the feasible curve. Points where those gradients become dependent must be examined separately; the regular two-multiplier condition need not cover them.

The solution set is the plane through the origin with normal vector $(1,1,1{)}^T$. The three original planes coincide, so their common intersection is this entire plane.

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.

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.

On $OAC$, $y=0$ and the outward unit normal is $-{\mathbf{e}}_y$. Its flux isOn $OBC$, $x=0$ and the outward unit normal is $-{\mathbf{e}}_x$. HenceOn $ABC$, write $z=1-x-y$ over $x\ge 0$, $y\ge 0$, $x+y\le 1$. The oriented surface element is $(1,1,1{)}^{T}dxdy$, pointing away from the origin. Here $x+y+z=1$, so $\mathbf{F}=(1,1,0{)}^T$. ThusThe unnormalized normal vector in the oriented surface element already incorporates the factor $\sqrt{3}$ from the surface area; no additional factor is needed.

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.

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

Apply matrix multiplication to obtainSince $\hat{\mathbf{n}}$ is a unit vector, the subtracted term is the orthogonal projection of $\mathbf{x}$ along $\hat{\mathbf{n}}$. Hence $M$ is the orthogonal projection onto a plane through the origin with normal vector $\hat{\mathbf{n}}$. It fixes every vector perpendicular to $\hat{\mathbf{n}}$ and sends every scalar multiple of $\hat{\mathbf{n}}$ to zero. In particular $\hat{\mathbf{n}}\cdot M\mathbf{x}=0$.