Find the equation of the cylinder whose base is the circle #x^2+y^2+z^2=4#, #x+y+2z=3#?

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Answer 1 · 2017-06-21T11:03:53

See below.

Given

#p = (x,y,z)#
#vec n = (1,1,2)#
#p_1 = (0,0,3/2)#

#S->norm (p)^2-2^2 = x^2+y^2+z^2-4=0#
#Pi-> << vec n , p - p_1 >> =x+y+2z-3=0#

the line

#L->p_0+lambda vec n#

is potentially a cylinder generatrix, providing that #p_0 in S nn Pi#

then

#norm(p_0+lambda vec v)^2 = norm(p_0)^2+2lambda << p_0, vec n >> + norm(vec n)^2 = 4#

Solving for #lambda#

#lambda = (- << p_0, vec n >> pm sqrt(<< p_0, vec n >>^2+4 norm(vec n)^2 - norm(vec n)^2 norm(p_0)^2))/norm(vec n)^2#

The tangency condition requires that

#<< p_0, vec n >>^2+4 norm(vec n)^2 - norm(vec n)^2 norm(p_0)^2=0# or

#24 - 5 x_0^2 + 2 x_0 y_0 - 5 y_0^2 + 4 x_0 z_0 + 4 y_0 z_0 - 2 z_0^2 =0#

which gives the sough cylinder surface.

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