Question

find the equation of right circular cylinder whose axis is x-2=z,y=0 and which passes through the point A(3,0,0)

Answer 1

Topic - The Cylinder

Formula to find the distance of a point from a straight line:

Let $P$ be the point $(x_{1},\:y_{1},\:z_{1})$ and a straight line be given by

$\quad \frac{x-\alpha}{l}=\frac{y-\beta}{m}=\frac{z-\gamma}{n}\quad ...(i)$

where $l,\:m,\:n$ are direction cosines of the straight line.

$\therefore$ the distance of the point $P$ from the straight line $(i)$ is given by

$\bigg({\left|\begin{array}{cc}x_{1}-\alpha&y_{1}-\beta\\l&m\end{array}\right|}^{2}+{\left|\begin{array}{cc}y_{1}-\beta&z_{1}-\gamma\\m&n\end{array}\right|}^{2}+{\left|\begin{array}{cc}z_{1}-\gamma&x_{1}-\alpha\\n&l\end{array}\right|}^{2}\bigg)^{\frac{1}{2}}$

Finding equation of a right circular cylinder:

Let the cylinder be of radius $r$ and let the axis be the straight line

$\quad \frac{x-\alpha}{l}=\frac{y-\beta}{m}=\frac{z-\gamma}{n}$

Let $(x',\:y',\:z')$ be any point on the cylinder.

We have to use this point to determine $r$.

Then the equation of the right circular cylinder be

$(x-\alpha)^{2}+(y-\beta)^{2}+(z-\gamma)^{2}-\bigg[\frac{l(x-\alpha)+m(y-\beta)+n(z-\gamma)}{\sqrt{l^{2}+m^{2}+n^{2}}}\bigg]^{2}=r^{2}$

Solution:

The axis of the right circular cylinder is given by

$\quad\quad x-2=z,\:y=0$

This can be put in symmetrical form as follows,

$\quad \frac{x-2}{\frac{1}{\sqrt{2}}}=\frac{y-0}{0}=\frac{z-0}{\frac{1}{\sqrt{2}}}\quad ...(i)$

The point on the cylinder is $A\:(3,\:0,\:0)$.

Then the distance of the point $A$ from the straight line $(i)$ is

$=\bigg({\left|\begin{array}{cc}3-2&0-0\\ \frac{1}{\sqrt{2}}&0\end{array}\right|}^{2}+{\left|\begin{array}{cc}0-0&0-0\\0&\frac{1}{\sqrt{2}}\end{array}\right|}^{2}+{\left|\begin{array}{cc}0-0&3-2\\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{array}\right|}^{2}\bigg)^{\frac{1}{2}}$

$=\big(0+0+\frac{1}{2}\big)^{\frac{1}{2}}$

$=\frac{1}{\sqrt{2}}$

This distance is radius of the right circular cylinder. That is $r=\frac{1}{\sqrt{2}}$.

Therefore the equation of the right circular cylinder is

$\quad (x-2)^{2}+(y-0)^{2}+(z-0)^{2}-\bigg[\frac{\frac{1}{\sqrt{2}}(x-2)+0(y-0)+\frac{1}{\sqrt{2}}(z-0)}{\sqrt{(\frac{1}{\sqrt{2}})^{2}+0^{2}+(\frac{1}{\sqrt{2}})^{2}}}\bigg]^{2}=(\frac{1}{\sqrt{2}})^{2}$

$\Rightarrow (x-2)^{2}+y^{2}+z^{2}-\frac{1}{2}(x+z-2)^{2}=\frac{1}{2}$

$\Rightarrow 2(x^{2}-4x+4+y^{2}+z^{2})-(x^{2}+z^{2}+4+2zx-4x-4z)=1$

$\Rightarrow x^{2}+2y^{2}+z^{2}-2zx-4x+4z+4=1$

$\Rightarrow \boxed{x^{2}+2y^{2}+z^{2}-2zx-4x+4z+3=0}$

Answer: Therefore the equation of the required right circular cylinder is

$x^{2}+2y^{2}+z^{2}-2zx-4x+4z+3=0.$

Answer 2

Step-by-step explanation:

mark me as a branlist !!!!!!!