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Obtain the equation of right circular cylinder with radius of the base as 2 units. Its axis passes through $(1, 2, 3)$ and direction cosines are given as $(2, -3, 6)$

I got $45x^2+40y^2+13z^2+12xy-36yz-24zx-42x-280y-126z+294 = 0$

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  • $\begingroup$ Formula 9 or formula 10 from here would be useful... $\endgroup$ – J. M. isn't a mathematician Jul 8 '12 at 13:10
  • $\begingroup$ As $2^2+3^2+6^2=7^2$ the correct direction cosines are $(\frac 27, \frac {-3}7, \frac 67)$ $\endgroup$ – Ross Millikan Nov 18 '12 at 5:11
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I am not sure but are you talking about direction cosines because they should lie between $+1 $ and $-1$, which your values (2,3-6) are not. Or is it the end point of the axis, since the axis end points have to be specified for the equation of a cylinder

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  • $\begingroup$ DCs are proportional to them, not exactly equal. $\endgroup$ – Hyperbola Jul 9 '12 at 3:08
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Put $$ \begin{gathered} A = \left( {1,2,3} \right) \hfill \\ X = \left( {x,y,z} \right) \hfill \\ \mathbf{n} = \frac{1} {7}\left( {2, - 3,6} \right) \hfill \\ \end{gathered} $$ Then the required cylinder is the lieu of points such that: $$ 2 = \left| {\mathop {AX}\limits^ \to \times \mathbf{n}} \right| $$ that is: $$ 14 = \left| {\left( {6\left( {y - 2} \right) + 3\left( {z - 3} \right),\; - 6\left( {x - 1} \right) + 2\left( {z - 3} \right),\; - 3\left( {x - 1} \right) - 2\left( {y - 2} \right)} \right)} \right| $$ i.e. $$ \left( {6\left( {y - 2} \right) + 3\left( {z - 3} \right)} \right)^2 + \left( { - 6\left( {x - 1} \right) + 2\left( {z - 3} \right)} \right)^2 + \left( { - 3\left( {x - 1} \right) - 2\left( {y - 2} \right)} \right)^2 - 14^2 = 0 $$ which confirm your equation, except for the sign of $36yz$ which shall be positive, not negative.

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