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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 $b\ne 0$.

For case (i),Each component has the sign of its coordinate, so the gradient points outward from the minimum. Near the origin it is $(x,y{)}^T+O(x^2+y^2)$.

For case (ii),Its $x$ component has the sign of $x$, while its $y$ component has the opposite sign to $y$. Thus arrows point outward on the $x$-axis and inward on the $y$-axis, with local form $(x,-y{)}^T+O(x^2+y^2)$. These directions are included in the contour figure; in both cases the gradient is perpendicular to the level curves and points toward increasing $h$.

The Hessian matrix is diagonal:Its eigenvalues are $c$ and $d$. Hence the second-derivative test gives a strict local minimum if $c,d>0$, a strict local maximum if $c,d<0$, and a saddle point of a scalar function if $cd<0$. No degenerate case occurs because both are nonzero.

The local minimum is the unique global minimum, since both separated summands are strictly convex functions and tend to $+\infty$ at either end of the real line. Similarly, when both signs are negative there is a unique global maximum.

The stationary point equations are $a{e}^x=c$ and $b{e}^y=d$. The sign assumptions make both ratios positive, so each equation has exactly one real solution. ThereforeSubstitution gives the stationary value

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.

For the specified standard basis vectors, $\mathbf{a}\times \mathbf{b}=\hat{\mathbf{k}}$, soOn the other hand, $\mathbf{b}\times \mathbf{c}=\hat{\mathbf{i}}-\hat{\mathbf{j}}$ and henceThese results are unequal, proving that the cross product does not satisfy the associative property.

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.

All three linear equations are multiples of the first, leaving the single constraint $x+y+z=0$. Choose two real numbers $s,t$ freely:A basis for this two-dimensional linear subspace is $(1,0,-1{)}^T,(0,1,-1{)}^T$.

The original PDF places complement bars on the second $A$ in each parenthesis and on the final $A$. Thus the first parenthesis is the empty set, whereas the second is the whole sample space. Their union with any further subset is still the whole space:The bars are missing from the local OCR, so using its unbarred expression would give the wrong result.

Since $A$ is a subset of the sample space $\Omega$, taking its intersection with $\Omega$ leaves it unchanged: