Answer to Question #114934 in Analytic Geometry for ANJU JAYACHANDRAN

Question

A right circular cylinder passes through the point (1,−1,4) and has the axis along
the line (x−1)/2 =(y−3)/5 =(z+1)/3. Is this information sufficient to determine the equation of the cylinder? If it is, determine the equation of the cylinder. Otherwise, state another condition so that the equation can be determined uniquely, and also find the equation.

Answer 1

$\frac{x-1}{2}=\frac{y-3}{5}=\frac{z+1}{3};M_1(1;-1;4);\\x=1+2\times t;\\y=3+5\times t;\\z=-1+3\times t;\\p(M_1;l)=\frac{(\overline{M_0M_1};\overline{p})}{|\overline{p}|};\overline{M_0M_1}=\{0;-4;5\};\\(\overline{MM_1};\overline{p})=\begin{vmatrix}-4 & 5 \\ 5 &3 \end{vmatrix};\begin{vmatrix} 0& 5 \\ 2 & 3 \end{vmatrix};\begin{vmatrix} 0 & -4 \\ 2 & 5 \end{vmatrix};\\(\overline{MM_1};\overline{p})=\begin{Bmatrix} -37; & -10; & -8\\ \end{Bmatrix};\\p(M_0;l)=\frac{\sqrt{37^2+10^2+8^2}}{\sqrt{2^2+5^2+3^2}}=\sqrt{\frac{1533}{38}};\\p(M;l)=\frac{|\overline{M_0M};\overline{p|}}{|p|}=\sqrt{\frac{1533}{38}}\times (\overline{M_0M})=\\=\begin{Bmatrix} x; & y+4;&z-5 \\ \end{Bmatrix};p=\begin{Bmatrix} 2; & 5; &3\\ \end{Bmatrix};\\(\overline{M_0M}; \overline p) =\begin{vmatrix} y+4 & z-5 \\ 5 & 3 \end{vmatrix};\begin{vmatrix} x & z-5 \\ 2 & 3 \end{vmatrix};\\\begin{vmatrix} x & y+4 \\ 2 & 5 \end{vmatrix}=\\=3\times y-5\times z+37;3\times x-2\times z+10;\\5\times x-2\times y-8;\\|\overline{M_0M}; \overline p|=\sqrt{(3\times y-5\times z+37)^2};\\\sqrt{(3\times x-2\times z+10)^2;(5\times x-2\times y-8)^2}=\\=\sqrt{34\times x^2+9\times y^2-45\times y\times z-18\times x\times z-}\\\sqrt{-30\times x\times y+222\times y+140\times x+1533}\\34\times x^2+9\times y^2-45\times y\times z-18\times x\times z-\\-30\times x\times y+222\times y+140\times x=0$The last expression is the equation of the desired cylinder