300 Reputation

12 years, 141 days

Thank you...

First, many thanks to Kitonum. Because i want to have a beautiful answer, therefore, i write the equation circle passing three points A, B, C. And then, write the equation tangent of the circle (ABC) at the point C. Thus, the line is this equation of this tangent. Now, i hope Kitonum understand my idea. Thank you again.

Thank you...

Thank you very much.

Thank you...

Thank you very much.

My code...

restart:

with(geom3d):

A:=<-1,3,3>:  B:=<3,-5,3>:  C:=<2,1,-1>:M:=<x,y,z>: o:=<0,0,0>:

sol:=solve([seq(o[i] = (A-M - 2*(B-M) - 3*(C-M))[i], i=1..3)]):

point(T,op(2,op(1,sol)), op(2,op(2,sol)), op(2,op(3,sol))):

plane(P,2*x - y +z  - 1 = 0,[x,y,z]):

coordinates(projection(H,T,P));

Another way...

restart;

with(geom3d):

point(o,0,0,0):

point(H,2,1,1):

n:=dsegment(oH,o,H):

plane(P,[H,n]):

sort(Equation(P,[x,y,z]));

Please comment to me. Thank you.

My code...

This is my code.

with(geom3d):

point(A,-2,3,4): point(B,1,2,-4):

line(l,[-t+1,2*t,3*t+1],t):

a:=ParallelVector(l):

line(AB,[A,B],t):

v:=ParallelVector(AB):

with(linalg):

n:=crossprod(a,v):

plane(P,[A,a],[x,y,z]):

projection(H,B,P):

Equation(line(AH,[A,H],t));# minimum

Equation(line(Delta,[A,n],t));# maximum

Please comment to me. Thank you very much.

Explain...

Let I be midpoint of the segment AB. Denote v(AB) is vector AB. We have

v(MA).v(MB) = (v(MI) + v(IA)).(v(MI) + v(IB)) =  (v(MI) + v(IA)).(v(MI) - v(IA)) = MI^2 - IA^2.

Because IA is a constant, therefore v(MA).v(MB) minimum when and only when MI^2 minimum, thus M is projection of I on (P).

My code...

`with(geom3d):`
`point(A,3,-1,2):point(B,1,-5,0):plane(P,2*x -y +2*z + 9 = 0,[x,y,z]):midpoint(T,A,B):coordinates(projection(H,T,P));`
`Please comment for me. Thank you very much.`

This is my code...

This is my idea:

1) Let a be direction vector of d; n be normal vector of plane (P). Then direction vector of Delta is  v=crossprod(a,n).

2) Write the equation of the plane (Q) passing the line Delta and parallel to the line d.

3) Distance between (d) and (Delta) is also  between the point A and the plane Q.

4) Delta is intersection of (P) and (Q).

and this is my code.

restart;with(geom3d):

line(d,[2*t+1,t-2,t],t);

plane(P,x + 2*y -z - 3 = 0,[x,y,z]);

a:=ParallelVector(d);

n:=NormalVector(P);

with(linalg):

v:=crossprod(a,n);

delta:=crossprod(a,v);

plane(Q,y - z + m = 0,[x,y,z]);

d:=distance(point(A,1,-2,0),Q);

eq:=solve(d = sqrt(2),m);

plane(Q1, y - z + 4=0,[x,y,z]);

intersection(Delta_1,P,Q1);

Equation(Delta_1,t);

plane(Q2, y - z =0,[x,y,z]);

intersection(Delta_2,P,Q2);

Equation(Delta_2,t);

I'm really not agree with this code. Please comment for me. Thank you.

Thank you...

Yes, i will to try. Thank you very much.

Thank you...

Yes, i will to try. Thank you very much.

Note that,  the system of...

Note that,  the system of equations

solve({a^2+b^2+c^2=1, a*(0-2)+b*(1-(-1))+c*(-2-0)=0, abs(a+2*b+c)/sqrt(1+2^2+1)=1/2});

has four solutions. But we only have two planes.

Note that,  the system of...

Note that,  the system of equations

solve({a^2+b^2+c^2=1, a*(0-2)+b*(1-(-1))+c*(-2-0)=0, abs(a+2*b+c)/sqrt(1+2^2+1)=1/2});

has four solutions. But we only have two planes.

Thank you...

Thank you. Can you write a procedure?

Thank you...

Thank you. Can you write a procedure?

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