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In the parametric description, $a$ is a point of the straight line and the nonzero vector $b$ gives its direction. The real parameter runs through every point of that line. For the cross product description, take another cross product with $c$ and use the vector triple product identity:Hence every solution has the representationHere $a\cdot b=0$ and $|b|=|c|$, as required. Conversely,Thus the perpendicularity assumption is sufficient, and the solutions form a straight line parallel to $c$. When $d\ne 0$, the vector $d$ is perpendicular both to $c$ and to every position vector on the line; it is the normal to the plane through the origin containing that line. The point $a=c\times d/|c{|}^2$ is the perpendicular foot from the origin. Its distance from the origin is $|d|/|c|$, since $c\cdot d=0$.

For a general parametric straight line, minimize the squared Euclidean distanceThis strictly convex quadratic function has its unique minimum at $\lambda =b\cdot (y-a)/|b{|}^2$, giving the orthogonal projectionFor the cross product description, $a\cdot c=0$, so this becomes

For the two planes, put $s=m\cdot n$ and $c=m\times n$. Nonparallel unit normals give $|s|<1$ and $|c{|}^2=1-s^2$. Seek the perpendicular foot in their span, $a=um+vn$. Its two dot products impose $u+sv=\mu$ and $su+v=\nu$. Solving gives both requested descriptions of the intersection:andIndeed the vector triple product identity gives $x\times (m\times n)=m(x\cdot n)-n(x\cdot m)$. Conversely, since $m,n$ are linearly independent, equality to $\nu m-\mu n$ forces both plane equations. Thus neither description introduces extra points.

Use change of basis matrices whose columns are the respective basis vectors in standard coordinates:The coordinate relation is $[\Phi (x){]}_C=A[x{]}_B$. Therefore the standard-coordinate matrix of a linear map is $T=S_{C}A{S}_B^{-1}$, and the new coordinate matrix is $A^{\prime }=S_{C^{\prime }}^{-1}T{S}_{B^{\prime }}$. Multiplying givesOne can also read off the columns: the two new domain basis vectors map to twice the first and four times the second new codomain basis vector. All four change of basis matrices are invertible. In particular the printed codomain vectors are three-dimensional; retaining their third coordinates is essential.

The fundamental theorem of calculus connects the Riemann integral with differentiation. Its first part says that for continuous $h:[a,b]\to \mathbb{R}$, the accumulation function $H(x)={\int }_a^{x}h(t)dt$ is continuous on $[a,b]$ and differentiable inside it, withTo prove this, boundedness of $h$ gives $|H(y)-H(x)|\le \Vert h{\Vert }_{\infty }|y-x|$, proving continuity. For an interior point and a sufficiently small nonzero increment $u$,Its absolute value is at most ${\sup }_{|t-x|\le |u|}|h(t)-h(x)|$, which tends to zero by continuity of $h$ at $x$. This proves the derivative formula. The same estimate gives the appropriate one-sided endpoint derivatives.

The second part says that if $Q$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $Q^{\prime }=h$ with $h$ continuous, thenIndeed $(Q-H{)}^{\prime }=0$, so the mean value theorem makes $Q-H$ constant, and evaluating at the endpoints proves the formula. A useful stronger version only requires $Q^{\prime }$ to be Riemann integrable. For a partition $a=x_0<\cdots <x_N=b$, the mean value theorem on each subinterval provides ${\xi }_j\in (x_{j-1},x_j)$ such thatAs the mesh tends to zero, these tagged Riemann sums converge to ${\int }_a^b{Q}^{\prime }$, proving that version too. Its integrability hypothesis must not be omitted.

An integrable integrand need not give an everywhere differentiable accumulation function. For a counterexample, let $f(t)=0$ for $t<1/2$ and $f(t)=1$ for $t\ge 1/2$. This bounded step function is Riemann integrable, butIts left and right derivatives at $1/2$ are zero and one. Thus $F$ need not be differentiable at every interior point. The first part of the fundamental theorem of calculus does apply at every point where the integrand is continuous.

A derivative need not have a Riemann integral. Define on all of $\mathbb{R}$At zero $f(h)/h=h\sin (1/h^2)\to 0$, so $f^{\prime }(0)=0$; elsewhere the product rule and chain rule giveAt $x_n=(2\pi n{)}^{-1/2}$, this equals $-2/x_n\to -\infty$. Hence $f^{\prime }$ is unbounded on $[0,1]$, whereas a properly Riemann integrable function on a compact interval must be bounded. The requested Riemann integral therefore need not exist, even though $f$ is differentiable everywhere. This example is an unbounded derivative obstruction to Riemann integrability; it does not contradict the theorem's version that explicitly assumes integrability of the derivative.