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Expanding the cross product componentwise givesIts dot product with each original vector is zero, as required.

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 three displacements from $Q_1$ areHere $\mathbf{u}\times \mathbf{v}=(-1,-2,3{)}^T$ and its dot product with $\mathbf{w}$ is zero. Because the cross product itself is nonzero, the first three points determine a plane and the fourth lies on it. Explicitly all four satisfyThus the four points are coplanar.

Use displacements from $P_1$, since coplanarity of points concerns an affine plane that need not contain the origin. The three displacement vectors areTheir cross product is $\mathbf{u}\times \mathbf{v}=(-1,1,-1{)}^T$ and its dot product with $\mathbf{w}$ is $-1$. Since the scalar triple product is nonzero,

On the first segment, parameterize ${\mathbf{r}}_1(t)=(t,t,0{)}^T$, $0\le t\le 1$. The dot products with ${\mathbf{r}}_1^{\prime }=(1,1,0{)}^T$ areTheir line integrals are one and $\sin 1$, respectively.

On the second segment take ${\mathbf{r}}_2(t)=(1,1,t{)}^T$, with tangent vector $(0,0,1{)}^T$. The corresponding integrands are $a-2+2t$ and $c$, whose integrals are $a-1$ and $c$. Adding the segments yields

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 $b\ne 0$.

The two prescribed directions have dot product zero. Normalize them toTheir cross product is the perpendicular unit vector ${\mathbf{q}}_3=(2,-1,-1{)}^T/\sqrt{6}$. Using these as columns givesThe columns form an orthonormal basis, so $Q^{T}Q=I$ and $Q$ is an orthogonal matrix.

Normalize the direction to the unit vector $\hat{\mathbf{u}}=(1,1,0{)}^T/\sqrt{2}$. The directional derivative is the dot product of this unit vector with the gradient:

Contracting the Kronecker delta identifies $i=k$. The remaining two independent sums are dot products, so the Einstein summation convention givesThe index expression here is verified against the PDF because the OCR corrupts it completely.

Write $z_j=x_j+i{y}_j$. Multiplication givesTherefore the real part is the dot product of the corresponding vectors:For nonzero vectors, this is $|z_1||z_2|\cos (\arg z_1-\arg z_2)$, measuring their alignment.

The modulus is the Euclidean norm, obtained by taking the square root of the dot product of the vector with itself:

The dot product is $1\cdot 0+2(-1)+(-1)\cdot 1=-3$, while the Euclidean norms are $\sqrt{6}$ and $\sqrt{2}$. Therefore the angle $\theta \in [0,\pi ]$ satisfies

The scalar $\mathbf{r}\cdot \hat{\mathbf{u}}$ is the signed scalar projection along the unit vector $\hat{\mathbf{u}}$. Multiplying by that unit vector makes $P_{\hat{\mathbf{u}}}(\mathbf{r})$ the orthogonal projection of $\mathbf{r}$ onto its span.

For the other expression, the vector triple product identity givesThis is the orthogonal projection onto a plane perpendicular to $\hat{\mathbf{u}}$ through the origin. Its dot product with $\hat{\mathbf{u}}$ is zero, while the difference from $\mathbf{r}$ is parallel to $\hat{\mathbf{u}}$.

The curl of a gradient is zero. Consequently the dot product in the exponent vanishes and

The dot product with the given unit vector is $\mathbf{r}\cdot \hat{\mathbf{n}}=(x+z)/\sqrt{2}$. Subtracting the corresponding orthogonal projection givesTaking its squared Euclidean norm yields the equivalent Cartesian equation

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 reversed product is the dot product, a scalar represented by a $1\times 1$ matrix:

Since $\mathbf{v}=-\mathbf{u}$ and $\mathbf{u}\ne 0$, the vectors point in opposite directions. More explicitly, their dot product givesHence their angle is $\boxed{\theta =\pi \text{ radians}=180^{\circ }}$.

Put $\mathbf{W}=\nabla \times \mathbf{F}$ and $\mathbf{v}=\dot{\mathbf{r}}$. All vector fields and their spatial partial derivatives below are evaluated at $\mathbf{r}(t)$. The product rule for a dot product givesBecause the vector field has no explicit time dependence, the multivariable chain rule gives $d\mathbf{W}/dt=J_{\mathbf{W}}\dot{\mathbf{r}}$, where the Jacobian matrix has entries $(J_{\mathbf{W}}{)}_{ij}={\partial }_j{W}_i$. ConsequentlyThe first term measures the change of the curl sampled as the point moves; the second measures the change in the velocity vector used in the dot product.

To give the result explicitly in $P,Q,R$, define the directional derivative operator $L=\dot{x}{\partial }_x+\dot{y}{\partial }_y+\dot{z}{\partial }_z$. Sincethe same expression isFor example $L(R_y-Q_z)=\dot{x}(R_{yx}-Q_{zx})+\dot{y}(R_{yy}-Q_{zy})+\dot{z}(R_{yz}-Q_{zz})$; the other two terms expand in the same way. Thus only the first and second spatial partial derivatives of $P,Q,R$, together with the velocity and acceleration of the parametric curve, occur.