Answer to Question #104627 in Analytic Geometry for Deepak

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

Let $P(x_1,y_1,z_1)$ be any point on the cylinder. Draw $PM$ perpendicular to the axis of the cylinder.

Then $PM=r$ .

Let $A(a,b,c)$ be the point which lies on the axis

$$\frac{x-a}{l}=\frac{y-b}{m}=\frac{z-c}{n}$$

Now

$$AP^2=(x_1-a)^2+(y_1-b)^2+(z_1-c)^2$$

$$MA=projection\space of\space AP\space on\space the\space axis$$

$$MA=\frac{l(x-a)+m(y-b)+n(z-c)}{\sqrt(l^2+m^2+n^2)}$$

Now, from the right angled $\bigtriangleup AMP$, we get

$$AP^2-MA^2=MP^2$$

$(x_1-a)^2+(y_1-b)^2+(z_1-c)^2-(\frac{l(x_1-a)+m(y_1-b)+m(z_1-c)}{\sqrt(l^2+m^2+n^2)})^2=r^2$

Then the required equation of the cylinder is

$$(x-a)^2+(y-b)^2+(z-c)^2-(\frac{l(x-a)+m(y-b)+m(z-c)}{\sqrt(l^2+m^2+n^2)})^2=r^2$$

Given $P(1,-1,4)$, the line

$$\frac{x-1}{2}=\frac{y-3}{5}=\frac{z+1}{3}$$

$x_1=1,y_1=-1,z_1=4,$

$a=1,b=3,c=-1,$

$l=2,m=5,n=3$

$r^2=(1-1)^2+(-1-2)^2+(4-(-1))^2-(\frac{3(1-1)+5(-1-3)+3(4-(-1))}{\sqrt(2^2+5^2+3^2)})^2$

$$r^2=\frac{1533}{38}$$

Answer:

The equation of the right circular cylinder

$$(x-1)^2+(y-3)^2+(z+1)^2-\frac{(2(x-1)+5(y-3)+3(z-1))^2}{38}=\frac{1533}{38}$$