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#<h4>Problem:</h4>
Find the angle between the planes x-y-z = 7 and x-2y+z=1.<br>
Find the symmetric equations for the line of intersection L of these planes.
<hr>
<p> The two planes are:

> P1 := x-y-z=7;  P2 := x-2*y+z=1;

                              P1 := x - y - z = 7

                             P2 := x - 2 y + z = 1
#The normal vectors to these planes are:
>n1 := vector([1,-1,-1]); n2 := vector([1,-2,1]);

                              n1 := [ 1, -1, -1 ]

                               n2 := [ 1, -2, 1 ]

#The angle "theta" between the planes is given by:
> theta := angle(n1,n2);

                                              1/2  1/2
                         theta := arccos(1/9 3    6   )
#which in degrees is approximately
>evalf(convert(theta,degrees),3);

                                  61.9 degrees
#The line of intersection of the two planes is: 
>L := solve({P1,P2},{x,y,z});

                    L := {y = 2 z + 6, x = 3 z + 13, z = z}
#from here we can sort out things to get the symmetric equations for L
<center> (x-13)/3 = (y-6)/2 = (z-0)/1 </center>
 Another way to find the line of intersection is to find a direction vector
"v" and a position vector "a" for this line. We have,

> v := crossprod(n1,n2); a := subs(z=0,L);

                              v := [ -3, -2, -1 ]

                          a := {y = 6, x = 13, 0 = 0}
# re-writing "a" in vector form as,

> a := vector([13,6,0]);

                               a := [ 13, 6, 0 ]
# we can write the vector form for the Line as,
>Line := a + t*v;

                                Line := a + t v
#or in component form,
>evalm(Line);

                           [ 13 - 3 t, 6 - 2 t, - t ]

>