Sketch the surface x2 z2 4 and find the points of intersec

Sketch the surface x^2 + z^2 = 4 and find the point(s) of intersection, if any, to the line L : <1 + t, 2 t, 3 + 2t>.

Solution

x^2 + z^2 = 4

is a circle of radius 2 on xz plane

in 3 D it is a cylinder

points of intersection with the line (1+t, 2-t ,3+2t)

(1+t)^2 + (3+2t)^2 = 4

t^2 + 2t + 1+4t^2 + 12t + 9 = 4

5t^2 + 12t + 6 = 0

t1 = (-12 + 2\\sqr6)/10 and t2 = (-12 - 2\\sqr6)/10 are two real solutions

two points of intersection are

(1+t1, 2-t1, 3+2t1) (1+t2, 2-t2 ,3+2t2)