Answer to Question #104627 in Analytic Geometry for Deepak

Question #104627
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.
1
Expert's answer
2020-03-10T10:37:24-0400

Let P(x1,y1,z1)P(x_1,y_1,z_1) be any point on the cylinder. Draw PMPM perpendicular to the axis of the cylinder.

Then PM=rPM=r .



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


xal=ybm=zcn\frac{x-a}{l}=\frac{y-b}{m}=\frac{z-c}{n}

Now


AP2=(x1a)2+(y1b)2+(z1c)2AP^2=(x_1-a)^2+(y_1-b)^2+(z_1-c)^2

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

MA=l(xa)+m(yb)+n(zc)(l2+m2+n2)MA=\frac{l(x-a)+m(y-b)+n(z-c)}{\sqrt(l^2+m^2+n^2)}

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



AP2MA2=MP2AP^2-MA^2=MP^2

(x1a)2+(y1b)2+(z1c)2(l(x1a)+m(y1b)+m(z1c)(l2+m2+n2))2=r2(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


(xa)2+(yb)2+(zc)2(l(xa)+m(yb)+m(zc)(l2+m2+n2))2=r2(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)P(1,-1,4), the line


x12=y35=z+13\frac{x-1}{2}=\frac{y-3}{5}=\frac{z+1}{3}

x1=1,y1=1,z1=4,x_1=1,y_1=-1,z_1=4,

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

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

r2=(11)2+(12)2+(4(1))2(3(11)+5(13)+3(4(1))(22+52+32))2r^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


r2=153338r^2=\frac{1533}{38}

Answer:

The equation of the right circular cylinder


(x1)2+(y3)2+(z+1)2(2(x1)+5(y3)+3(z1))238=153338(x-1)^2+(y-3)^2+(z+1)^2-\frac{(2(x-1)+5(y-3)+3(z-1))^2}{38}=\frac{1533}{38}


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