Planes, lines and points

Assume you have a plane defined by P1, P2, P3
P1( x1=2.1  yy1=5.5     zz1= 4.3  )
P2( x2=1.2  yy2=1.53    zz2= 2.54 )
P3( x3=3.9  yy3=3.5     zz3= 9.2  )
The command is det(a)
Plane xyz = a*x+b*y+c*z+d=0

       | yy1  zz1 1 |                  | x1 zz1 1 |       
a= det | yy2  zz2 1 |      b=det  (-1)*| x2 zz2 1 |   
       | yy3  zz3 1 |                  | x3 zz3 1 |   
   
        | x1 yy1 1 |
c= det  | x2 yy2 1 |
        | x3 yy3 1 |

d= -(a*x1+b*yy1+c*zz1) 
 
a = -22.973    b = 1.242   c = 8.946     d =  2.9445

[ (x1-x3), (yy1-yy3), (zz1-zz3)] =| s1 |  =[-1.8 , 2.0, -4.9 ]

[ (x2-x3), (yy2-yy3), (zz2-zz3)] =| s2 |  = [-2.7, -1.97,-6.66 ]


| s1 | dot | s2 | = 33.554    dotP(s1,s2) -> k


| s1 | dot | s2 |/(abs(s1)*abs(s2)) = t7 = 0.8055066

dotP( s1,s2 )/( sqrt( dotP( s1,s1  ) ) * sqrt( dotP( s2,s2 ) ) ) =t7

cos(t8)=t7   cos-1(t7) -> t8

t8=36.341 degrees

========

assume x4=5

s1 dot s3 = 0 =g1   s2 dot s3 = 0 = g2  solve for yy3 and zz3

yy4=3.4405   zz4= 8.772

Find distance P4 to plane

abs(a*x4+b*yy4+c*zz4+d)/sqrt(a^2+b^2+c^2) = g = 1.182

 

=========
TI-89 program

sqrt is square root symbol
|> is right triangle
:cos-1( ) is inverse cos

:zz()
:Prgm
:ClrIO
:setMode("Angle","Degree")
:setMode("Display Digits","Float 5")
:setMode("Exact/Approx","Approximate")
:DelVar yy4,zz4
:2.1->x1
:5.5->yy1
:4.3->zz1
:1.2->x2
:1.53->yy2
:2.54->zz2
:3.9->x3
:3.5->yy3
:9.2->zz3
:5->x4
:[[yy1,zz1,1][yy2,zz2,1][yy3,zz3,1]]->t1
:[[x1, zz1,1][x2,  zz2,1][x3, zz3,1]]->t2
:[[x1, yy1,1][x2, yy2,1][x3, yy3,1]]->t3
:det(t1)->a
:-det(t2)->b
:det(t3)->c
:-(a*x1+b*yy1+c*zz1)->d
:Disp "plane"
:Disp "a,b",{a,b}
:Disp "c,d",{c,d}
:Pause
:ClrIO
:
:5->x4

:[[x1-x3,yy1-yy3,zz1-zz3]]->s1
:[[x2-x3,yy2-yy3,zz2-zz3]]->s2
:[[x4-x3,yy4-yy3,zz4-zz3]]->s3	
:
:dotP(s1,s3)->g1
:dotP(s2,s3)->g2
:solvw(g1=0 and g2=0,{yy4,zz4})->t8
:part(t8,1)->n1
:part(n1,2)->yy4
:Disp "yy4",yy4
:part(t8,2)->zz4
:Disp "zz4",zz4
:
:sqrt(x3-x4)^2+(yy3-yy4)^2+(zz3-zz4)^2)->dd
:disp "dd",dd

:EndPrgm