Final answer:
Yes, there exists a line L such that f(L)=L. It is a line parallel to the x-axis with a parametric representation L(t)=(x0, c+dt, c+dt-17), where x0 is any real number and d is a constant.
Step-by-step explanation:
To determine if an isometry f:E3→E3 given by f(x,y,z)=(x+30,z+17,y+17) has a line L such that f(L)=L, we need to look for a line that remains unchanged when f is applied to it. This would mean that every point (x,y,z) on L maps to a point on L after using the isometry f.
Let's assume L has a parametric representation of L(t)=(a+bt,c+dt,e+ft) for some constants a, b, c, d, e, and f where t is a parameter. Applying the given isometry f to this line, we get the line:
f(L(t)) = f(a+bt,c+dt,e+ft) = ((a+bt)+30,(e+ft)+17,(c+dt)+17).
For f(L(t)) to be the same as L(t), their components must be identical, i.e., f(L(t))=(a+bt,c+dt,e+ft). Hence:
- a+bt+30 = a+bt
- e+ft+17 = c+dt
- c+dt+17 = e+ft
Comparing both sides of these equations, we get that b=0 and f=d for the equalities to hold. As for the x-component to satisfy the equality, it should be independent of t; thus, we cannot have a line that satisfies the given isometry except for the specific case where the line is parallel to the x-axis.
The only possible L would then have a parametric representation L(t)=(x0,c+dt,c+dt-17) where x0 is any real number, and d is a constant describing the slope in the yz-plane. Therefore, such a line L does exist.