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The cross product is $\mathbf{a}\times \mathbf{b}=(-2,0,2)$, which is perpendicular to both vectors. Any nonzero scalar multiple works, so a simple answer isIts dot product with each given vector is zero.

For positive semiaxes $a,b$, the cross product of the two parametric tangent vectors gives the outward oriented surface element of an ellipsoid of revolution:Substituting the parametrization into the vector field and taking its dot product with this element givesThe middle term has zero integral over $0\le \varphi \le 2\pi$. In the other terms, ${\int }_0^{2\pi }{\cos }^2\varphi d\varphi =\pi$, ${\int }_0^{\pi }{\sin }^3\theta {\cos }^2\theta d\theta =4/15$, and ${\int }_0^{\pi }\sin \theta {\cos }^4\theta d\theta =2/5$. The directly evaluated flux is therefore

Write $\mathbf{\Delta }=\mathbf{d}-\mathbf{b}$. At an intersection, $\mathbf{\Delta }=\lambda \mathbf{a}-\mu \mathbf{c}$, so its scalar triple product with $\mathbf{a},\mathbf{c}$ vanishes. The expression in the question is exactlyUnder a shift of origin by $\mathbf{h}$, the direction vectors $\mathbf{a},\mathbf{c}$ stay fixed, while $\mathbf{b},\mathbf{d}$ both change by $-\mathbf{h}$. Thus $\mathbf{\Delta }$ and the condition are invariant. For nonparallel lines this is also sufficient for intersection; for parallel lines it only asserts coplanarity and need not imply intersection.

Put $\delta \lambda ={\lambda }_2-{\lambda }_1$ and $\delta \mu ={\mu }_2-{\mu }_1$. Then ${\mathbf{x}}_2-{\mathbf{x}}_1=\delta \lambda \mathbf{a}$ and ${\mathbf{x}}_4-{\mathbf{x}}_1=({\mathbf{x}}_3-{\mathbf{x}}_1)+\delta \mu \mathbf{c}$. Bilinearity of the cross product givesThe tetrahedron volume is one sixth of the absolute value of this triple product. Sliding either vertex pair along its line preserves $|\delta \lambda |$ or $|\delta \mu |$, because its separation is fixed and the direction vectors are fixed. Thus the volume is unchanged, including the zero-volume coplanar case. This is tetrahedron volume under sliding opposite edges.

The identity $\nabla \cdot (\mathbf{a}\times \mathbf{v})=\mathbf{v}\cdot (\nabla \times \mathbf{a})-\mathbf{a}\cdot (\nabla \times \mathbf{v})$ reduces to $-\mathbf{a}\cdot (\nabla \times \mathbf{v})$ when $\mathbf{a}$ is constant. In components, this follows by differentiating ${\epsilon }_{ijk}{a}_j{v}_k$ and using the antisymmetry of the Levi-Civita symbol. The divergence theorem and the cyclic scalar triple product identity now giveSince $\mathbf{a}$ is arbitrary,The order of the cross product determines the minus sign, as in the vector curl form of the divergence theorem.

The determinant defining $[a,a,c]$ has two equal columns. Interchanging them both leaves it unchanged and reverses its sign, so it is zero. Equivalently, the cross product $a\times c$ is orthogonal to $a$, giving $a\cdot (a\times c)=0$.

The plane normals are $n_1=(2,-1,-1)$ and $n_2=(3,-1,-3)$. Their cross product is $(2,3,1)$, perpendicular to both normals and hence parallel to the intersection line. Setting $z=0$ in both equations gives the common point $(1,-1,0)$. ThereforeThe normals are not parallel, so their intersection is a unique line.

Put $d=b\cdot x$ and $B=|b{|}^2$. Taking the dot product of the original equation with $b$ eliminates $b\times x$ and gives $d=a\cdot b-Bd$. ThusNext take its cross product with $b$. The term $db$ disappears, and the vector triple product identity givesInsert this expression back into $x=a-db-b\times x$. The two longitudinal terms cancel, leaving $(1+B)x=a-b\times a$. ConsequentlyThe denominator is strictly positive for every real $b$, including $b=0$. Substitution using the same triple-product identity verifies the solution and establishes uniqueness. This is the cross-product operator with a longitudinal rank-one term.

The vector triple product identity givesIt is the requested linear combination of $b$ and $c$. The notation $\wedge$ in this question denotes the three-dimensional cross product.

Expand the same double cross product in two ways. First expanding with $c\times d$ as the inner cross product givesAlternatively, use $(a\times b)\times w=(a\cdot w)b-(b\cdot w)a$ with $w=c\times d$ to getEquating these expressions and moving every term to one side provesThis four-vector scalar triple product identity is valid without assuming that any three of the four vectors are independent.

The vector triple product identity isThus the expression lies in the linear span of $b$ and $c$, with respective scalar coefficients $a\cdot c$ and $-a\cdot b$. The order and the minus sign matter because the cross product is not an associative operation.

For the triangular face, take the cross product of two edges and divide by two:The origin lies on the opposite side from this normal, since $(1,1,2)\cdot \overrightarrow{PO}=-3$. Thus this is already the outward vector area, withThe vector areas of a closed polyhedron sum to zero, so the sum for the remaining three faces is

Their individual outward vector areas are $(0,0,-3/2)$, $(1/2,-1,1)$, and $(-1,1/2,-1/2)$, which directly check that sum.

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.

Expanding the cross product gives $\mathbf{u}\times \mathbf{v}=(bc,-ac,a)$. Comparing its three components in order from the last to the first gives $a=2$, $ac=6$, and $bc=3$. ThereforeThe nonzero value of $a$ permits division to find $c$, and the resulting nonzero $c$ determines $b$.

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:

For $h>0$, parametrize the curved surface by $(r\cos \theta ,r\sin \theta ,r^2/a)$, with $0\le r\le \sqrt{ah}$ and $0\le \theta \le 2\pi$. The magnitude of the cross product of its coordinate tangents gives the area element of a surface:Integrating givesThis is the curved surface area, not the area of a lid across the horizontal opening. The expression tends to zero at $h=0$ and has dimensions of area. The volume and centroid of a parabolic cup use the same geometry.

The cross product is $\mathbf{a}\times \mathbf{b}=(1,-7,-3)$. Taking its dot product with $\mathbf{c}$ gives the scalar triple product

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.

The $z$ component of the cross product is $2a-2$. For the result to lie in the $x$-$y$ plane, it must vanish, giving $a=1$. The dot product condition is then $2a+2-3b=0$, so