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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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Formula 9 or formula 10 from here would be useful... – J. M. Jul 8 '12 at 13:10
As $2^2+3^2+6^2=7^2$ the correct direction cosines are $(\frac 27, \frac {-3}7, \frac 67)$ – Ross Millikan Nov 18 '12 at 5:11

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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DCs are proportional to them, not exactly equal. – Hyperbola Jul 9 '12 at 3:08

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