## 15564 Reputation

12 years, 144 days

## Slip...

Sorry! I made a slip of the pen. Instead  2  must be  d .

After  v:=crossprod(w,a)  insert two lines into your code:

d:=igcd(v[1],v[2],v[3]):

v:=map(x->x/2,v):

The lower boundary your region is a common segment  [-1, 1] . So look for  maximum and minimum of the function of one variable  f(x,0)=x^2+x  on this segment by the commands  maximize  and  minimize . In this decision, the derivatives are not needed.

## Explanation...

Computed plane passes through the line AB so that AB and its projection onto given plane  are perpendicular to the line of intersection of computed plane with given plane. Draw a picture and everything will be clear!

## Code...

restart:

with(LinearAlgebra):

A:=<1, -2, 4>:  B:=<3, 5, -1>:  n1:=<1, 1, 1>:  M:=<x, y, z>:

n2:=CrossProduct(n1, B - A):  n:=CrossProduct(n2, B - A):  d:=igcd(seq(n[i],i=1..3)):

sort((n/d).(M - A)) = 0;   # Equation of the required plane

## Examples...

nops(x^3+x+1);

3

nops(expand((x+1)^100));

101

## Solution...

If this statement, i.e.  (abcabcabc...abc‾)  Ξ (abc‾)(mod91) , is true for any number of repeats of abc, then the number abc itself  must be divisible by 91. This follows from the fact that  abcabc - abc=abc*1000, but 1000 and 91 are coprimes.

The example:

15 mod(4);

3

## Possible solution...

None sequence is  uniquely determined by several members without additional conditions! If we look for a formula in a class of polynomials of degree <=5, then the decision will be

a[n]=CurveFitting[PolynomialInterpolation]([1, 2, 3, 4, 5, 6],[1, 3, 6, 12, 33, 51],n);

## All right!...

I got the same answer. In my decision  and  t  are the parameter values ​​for those points on the lines, for which is realized a distance equal to 1.

The code of my decision:

restart:

A:=<1, -1, 3>:   u:=<a, b, c>:   v:=<2, -5, 6>:   M:=A+s*u:   N:=t*v:

solve([a^2+b^2+c^2=1, DotProduct(<1, 1, 5>,u,conjugate = false)=0, DotProduct(M-N,u,conjugate = false)=0, DotProduct(M-N,v,conjugate = false)=0, Norm(M-N,2)=1],[a,b,c,s,t]);

## Parametric equation...

If you subtract the first equation of the second equation, it is easy to find  z = 3/2 , ie all points of the curve lie in the horizontal plane  z = 3/2 . Substituting  z = 3/2  in any equation, we obtain the equation of a circle. The parametric equation of this circle will be  x=3*sqrt(3)/2*cos(t), x=3*sqrt(3)/2*sin(t), z=3/2 .

Code for the construction of the curve and surfaces:

A:=plots[spacecurve]([3*sqrt(3)/2*cos(t),3*sqrt(3)/2*sin(t), 3/2],t=0..2*Pi,color=red,axes=normal,thickness=3):

B:=plots[implicitplot3d]({x^2+y^2+z^2 = 9, x^2+y^2+(z-3)^2 = 9}, x = -5 .. 5, y = -5 .. 5, z = -3.5 .. 6.5, numpoints = 27000, style = surface):

plots[display](A,B);

## Error...

There is a serious mistake in your decision! Why do you define the line l in such a strange way, taking on it arbitrarily point C? This line is already defined by its equation! And if you specify another point  C, you would get a different line.

I would write your code like this:

restart: with(geometry):

point(A,-1,5):

point(B,-2,-2):

point(T,a,b):

line(l,3*x-4*y-27=0,[x,y]):

sys:=solve([distance(T,A) = distance(T,B), distance(T,A)=distance(T,l)],[a,b]):

point(o, rhs(sys[1,1]), rhs(sys[1,2])):

point(K, rhs(sys[2,1]), rhs(sys[2,2])):

sort(Equation(circle(c1,[o,distance(o,A)],[x,y]))); # First solution

sort(Equation(circle(c2,[K,distance(K,A)],[x,y]))); # Second solution

## Possible way...

Unfortunately, I haven't found a way for keeping such a format. So everything had to be done by hand. If you need this often, you can write the procedure.

See an example:

## Impossible...

Since none of the numbers 5, 10, 20, 50 does not have a factor 3, then i think it's  impossible to express the log [9] (40) as rational function  of c and d.

## Two ways...

f(alpha):=1+2*cos(alpha): # as assigning an expression

plot(f(alpha), alpha=-2*Pi/3..2*Pi/3, coords=polar);  # or

f:=alpha->1+2*cos(alpha): # as assigning a function

plot(f, -2*Pi/3..2*Pi/3, coords=polar);  # or

plot(f(t), t=-2*Pi/3..2*Pi/3, coords=polar);

Second way more universal! We see that the argument of a function can be no specify or denoted by other names.

 First 237 238 239 240 241 242 Page 239 of 242
﻿