Equation of right circular cylinder with radius of the base as 2 units.

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

• Formula 9 or formula 10 from here would be useful... – J. M. isn't a mathematician 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
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.