Question

find the equation of right circular cylinder of radius 2 and whose axis in the line x-1/2=y-2/1=z-3/2​

Answer 1

Answer:

45x^2+/40y^2+14z^2+12xy+36yz-24zx-42x-280y-126z+294=0

Answer 2

Answer:

Step-by-step explanation:

The required equation of the cylinder is $(x-a)^2+(y-b)^2+(z-c)^2- (\frac{l(x-a)+m(y-b)+n(z-c)}{\sqrt{l^2+m^2+n^2} }) ^{2} =r^2$

Here,

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

so,

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

l=2, m=5, n=3

r=2

So,

Now

$(x-1)^2+(y-3)^2+(z+1)^2-(\frac{2(x-1)+5(y-3)+3(z+1)}{\sqrt{2^2+5^2+3^2} } )^2=2^2\\\\x^2+1-2x+y^2+9-6y+z^2+1+2z-(\frac{2x-2+5y-15+3z+3}{\sqrt{38} } )^2=4\\\\x^2+11-2x+y^2-6y+z^2+2z-(\frac{2x-2+5y-15+3z+3)^2}{{38} } )=4\\\\\\x^2+11-2x+y^2-6y+z^2+2z-(\frac{2x-2+5y-15+3z+3)^2}{{38} } )=4\\\\\\x^2+11-2x+y^2-6y+z^2+2z-(\frac{2x+5y-14+3z)^2}{{38} } )=4\\\45x^2+40y^2+14z^2+12xy+36yz-24zx-42x-280y-126z+294=0$

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