Chapter 1: Vectors, Lines and Planes
Section 1.7: Planes
Example 1.7.11
If plane P, and lines L1 and L2, are given respectively by
x−y+z=6
x=2 t+1,y=t−1,z=2−2 t
x=2−m,y=3+m,z=4+m
find an equation for any line L that is parallel to P, and intersects the lines in such a way as to form, between them, a segment of length 36.
Solution
Mathematical Solution
A vector V from L2 to L1 and N, the normal to P, are
V=2⁢t−1+st−5−s−2−2 t−s N=1−11
Line L is parallel to P if V·N=0, and the condition on the length of the segment between L1 and L2 is V=36, or equivalently, V2=54.
These two conditions give two equations in t and s, thus determining any line L satisfying these conditions. Two such lines exist, one (red, in Figure 1.7.11(a)) actually lying in plane P, and the other (green) above the plane.
use plots in module() local P,L1,L2,V,N,Ans1,Ans2,p1,p2,p3,p4,p5,p6; P:=x-y+z=6; L1:=<2*t+1,t-2,2-2*t>; L2:=<2-m,3+m,4+m>; V:=L1-L2; N:=<1,-1,1>; Ans1:=<-1-2*s,-3-s,4+s>; Ans2:=<5+s,-s,-2-2*s>; p1:=implicitplot3d(P,x=-5..10,y=-5..4,z=-5..8,style=surface,color=pink,transparency=.5); p2:=spacecurve(Ans1,s=-5..2,color=red,thickness=3,numpoints=2); p3:=spacecurve(Ans2,s=-4..1,color=green,thickness=3,numpoints=2); p4:=spacecurve(L1,t=-3..3,color=black,thickness=3,numpoints=2); p5:=spacecurve(L2,m=-6..2,color=blue,thickness=3,numpoints=2); p6:=display([p1,p2,p3,p4,p5],scaling=constrained,axes=none,orientation=[-35,-45,25]); print(p6); end module: end use:
Figure 1.7.11(a) Lines L1 (black), L2 (blue), plane P, and two solutions for L, in red and green.
The conditions V·N=0 and V2=54 lead to the equations
−t+2+s=0,2⁢t−1+s2+t−5−s2+−2−2⁢t−s2=54
whose solutions are s,t=−3,−1 and s,t=0,2. Hence there are two possible lines, namely,
xyz=−1−6 p−3−3 p4+3 p and xyz=5+3 p−3 p−2−6 p
Maple Solution - Interactive
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
For k=1,2, define lines Lk as the position vectors Lk Define N, the normal for plane P
Context Panel: Assign to a Name≻L[1]
2 t+1,t−2,2−2 t→assign to a nameL1
Context Panel: Assign to a Name≻L[2]
2−s,3+s,4+s→assign to a nameL2
Context Panel: Assign to a Name≻N
1,−1,1→assign to a nameN
Obtain V, a vector from L2 to L1
Write V=L1−L2 Context Panel: Assign Name
V=L1−L2→assign
Impose the conditions V·N=0 and V2=54 and solve the resulting equations
Write the sequence V·N=0,V2=54 Press the Enter key.
Context Panel: Solve≻Solve
Context Panel: Assign to a Name≻S
V·N=0,V2=54
→solve
s=−3,t=−1,s=0,t=2
→assign to a name
S
In terms of the parameter p, obtain line L as a position vector
Evaluate L1+p V at S1
Context Panel: Evaluate and Display Inline
L1+p Vx=a|f(x)S1 =
Evaluate L1+p V at S2
L1+p Vx=a|f(x)S2 =
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
For k=1,2, define lines Lk as the position vectors Lk
Define N, the normal for plane P and define V=L1−L2
L1,L2,N,V≔2 t+1,t−2,2−2 t,2−s,3+s,4+s,1,−1,1,L1−L2:
Via the Norm, DotProduct and solve commands:
Impose and solve the equations arising from the conditions V·N=0 and V2=54
S≔solveDotProductV,N=0,NormV,Euclidean2=54,s,t
Use the eval command to obtain line L
x,y,z=evalL1+p V,S1
x,y,z=evalL1+p V,S2
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