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Choose two distinct space-diagonal directions $\mathbf{u}=(1,1,1)$ and $\mathbf{v}=(-1,1,1)$. Their dot product is one and their norms are both $\sqrt{3}$. Thus the acute angle between their unoriented lines isAny two distinct space diagonals have an absolute dot product of one when represented with components $\pm 1$, so they give the same acute angle.

Let $\Delta =[\mathbf{a},\mathbf{b},\mathbf{c}]\ne 0$. These three vectors are linearly independent, so write $\mathbf{d}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}$. Taking the dot product with $\mathbf{b}\times \mathbf{c}$ gives $\alpha =[\mathbf{b},\mathbf{c},\mathbf{d}]/\Delta$. Likewise, the dot products with $\mathbf{c}\times \mathbf{a}$ and $\mathbf{a}\times \mathbf{b}$ giveHere the cyclic invariance and sign change under an interchange of the scalar triple product fix every sign. Substituting the three coefficients provesThis is precisely coordinate extraction using the reciprocal basis $(\mathbf{b}\times \mathbf{c},\mathbf{c}\times \mathbf{a},\mathbf{a}\times \mathbf{b})/\Delta$.

Write $s=\hat{\mathbf{n}}\cdot \mathbf{x}$. Taking the dot product of the equation with the unit vector $\hat{\mathbf{n}}$ gives $2s+s^2=-1$, so $(s+1{)}^2=0$ and $s=-1$.

Now take the cross product with $\hat{\mathbf{n}}$. The vector triple product gives $\hat{\mathbf{n}}\times (\hat{\mathbf{n}}\times \mathbf{x})=s\hat{\mathbf{n}}-\mathbf{x}=-\hat{\mathbf{n}}-\mathbf{x}$, soThe original equation, now with $s^2=1$, also gives $\hat{\mathbf{n}}\times \mathbf{x}=\mathbf{b}-\hat{\mathbf{n}}-2\mathbf{x}$. Substitution eliminates the remaining cross product of the unknown and yieldsIts dot product with $\hat{\mathbf{n}}$ is $-1$, and direct substitution using the vector triple product verifies the original equation. The elimination gives a unique solution without choosing components. It is also an application of the inverse of a scalar plus a cross-product operator after the scalar component has been fixed.

Parametrize the circle by $\mathbf{r}(t)=(\cos t,\sin t)$, $0\le t\le 2\pi$. Its tangent is ${\mathbf{r}}^{\prime }=(-\sin t,\cos t)=\mathbf{F}(\mathbf{r}(t))$, so the dot product $\mathbf{F}\cdot {\mathbf{r}}^{\prime }$ is one. For the other field,Integrating around the full circle gives

The vector field is radial, while the differential displacement along a centred circle is tangential. These directions are orthogonal, so the dot product vanishes at every point. Hence

By Thales's theorem, the two sides meeting on the circular arc are orthogonal, because the opposite side is a diameter. Both vectors point towards that meeting point; reversing both does not change their dot product. Consequently

For the $C$ obtained above, the characteristic polynomial isThus the eigenvalues are $5$ and $-5$. For $\mu =5$, $(C-5I)\mathbf{v}=0$ gives $v_2=-3v_1$; for $\mu =-5$, it gives $v_1=3v_2$. Normalizing the eigenvectors yieldsBoth are unit vectors, their dot product is zero, and either overall sign is allowed for each eigenvector.

Compute the flux directly using the outward vector area element, as required. Restrict the vector field to the surface parametric equation and take the dot product:The terms quadratic in $r$ cancel. Therefore the surface integral isThis is a direct evaluation of the flux of the position vector through a cone.

The new cone can be parametrized byThe outward vector area element is $r(\frac{b}{a}\cos \theta ,1,\frac{b}{a}\sin \theta )drd\theta$. This is ${\mathbf{R}}_{\theta }^{\prime }\times {\mathbf{R}}_r^{\prime }drd\theta$; reversing the order is needed to retain the outward orientation after the exchange of coordinates. Its dot product with $\mathbf{F}({\mathbf{R}}^{\prime })$ is again $brdrd\theta$. HenceThis equality also follows because a rigid rotation takes the first cone into the second and leaves the radial vector field $\mathbf{F}(\mathbf{x})=\mathbf{x}$ invariant.

Parametrize the segment in its stated direction by $\mathbf{x}(t)=(1-t,t,t)$, $0\le t\le 1$. Then $d\mathbf{x}=(-1,1,1)dt$, and substitution into the two vector fields givesTaking each dot product with $d\mathbf{x}$ evaluates the line integrals:

The two directed segments have vectors $\mathbf{u}=(1,0,-1)$ and $\mathbf{v}=(0,1,1)$. Their dot product is $-1$ and both norms equal $\sqrt{2}$. HenceThe common starting point fixes the obtuse angle between these segments.

Both segments start at $(1,1,1)$, so their direction vectors are $\mathbf{a}=(-1,-1,-1)$ and $\mathbf{b}=(1,1,0)$. Their dot product is $-2$, and their norms are $\sqrt{3}$ and $\sqrt{2}$. ThusThe angle between the rays lies in $[0,\pi ]$, so its sine is nonnegative:The negative cosine correctly identifies an obtuse angle. One also obtains the sine from $|\mathbf{a}\times \mathbf{b}|/(|\mathbf{a}||\mathbf{b}|)=\sqrt{2}/\sqrt{6}$, using the cross product.

Use only the dot product to minimize the squared Euclidean distance from the origin:Completing the square givesThe square is minimized at $t_0=-({\mathbf{a}}_2\cdot {\mathbf{n}}_2)/\Vert {\mathbf{n}}_2{\Vert }^2$, yieldingGeometrically this subtracts the squared length of the orthogonal projection parallel to the line. At the minimizing point, $({\mathbf{a}}_2+t_0{\mathbf{n}}_2)\cdot {\mathbf{n}}_2=0$, as required for perpendicular distance.

The dot product isBoth vectors have squared Euclidean norm $a^2+(a+1{)}^2=2a^2+2a+1$. Under the given condition $a^2+a=1/2$, their dot product is $1$ and each squared Euclidean norm is $2$. Hence

Three vectors form a basis of three-dimensional space exactly when their scalar triple product is nonzero:It is the determinant of the matrix with these columns, so this condition is equivalent to linear independence and spanning the whole space.

Take the dot product of the expansion of $\mathbf{y}$ with $\mathbf{v}\times \mathbf{w}$. The terms containing $\mathbf{v}$ and $\mathbf{w}$ vanish, leaving $\mathbf{y}\cdot (\mathbf{v}\times \mathbf{w})=\alpha \Delta$. Cycling the same argument yieldsThe three cross-product numerators use the reciprocal basis.

Use $g=x^2+y^2+z^2$. The Lagrange multiplier equations areTaking the dot product with $(x,y,z{)}^T$ gives $2T=2\lambda$, so $\lambda =T$.

If $\lambda \ne 0$, the first and third equations give $x=z$. The middle equation then gives $x=\lambda y$, so $2{\lambda }^2=1$. The norm constraint gives $x^2=z^2=1/4$ and $y^2=1/2$. Thus the hottest points and their temperature areThe coldest points areFor $\lambda =0$, the equations give $y=0$, $z=-x$, and hence the other two constrained stationary points areThey are saddle points of the restriction to the sphere: with $u=(x+z)/\sqrt{2}$ and $v=y$ as local tangent coordinates there, $T=\sqrt{2}uv$ changes sign in every neighbourhood. The surface is compact and smooth, so the listed extreme temperatures are global. Here “zero gradient on the surface” means zero surface gradient; the ambient gradient at the hottest and coldest points is normal to the sphere, rather than zero.