Question

Find the equation to the right circular cylinder whose guiding curve is x²+y²+z²=9, x-y+z=3​

Answer 1

Answer:

This is the correct answer

Answer 2

Given:

The guiding curve is $x^{2}+y^{2}+z^{2}=9, x-y+z=3$.

To find:

The equation to the right circular cylinder.

Step-by-step explanation:

The axis of the cylinder is perpendicular to the plane $x-y+z=3$ ,its direction cosines are proportional to 1,-1,1.

$\frac{x-x_{1} }{1} =\frac{y-y_{1} }{-1} =\frac{z-z_{1} }{1} =r(say)$

$(x_{1}+r,y_{1}-r,z_1+r} )$

if this point lies on the guiding curve, then

$(x_{1} +r)^2+(y_{1} -r)^2+(z_{1} +r)^2=9$

$x_{1}^{2}+y_{1}^{2}+z_{1}^{2}+2r(x_{1}-y_{1}+z_{1})+3r^2=9$ and $x_{1}-y_{1}+z_{1}+3r=3$

We get,

$x_{1}^{2}+y_{1}^{2}+z_{1}^{2}-\frac{2(x_{1}-y_{1}+z_{1})(x_{1}-y_{1}+z_{1})}{3}+\frac{3(x_{1}-y_{1}+z_{1}-3)^{2}}{9}=9$

$x_{1}^{2}+y_{1}^{2}+z_{1}^{2}+y_{1}z_{1}-z_{1}x_{1}+x_{1}y_{1}-9=0$

The locus of $(x_{1},y_{1},z_{1})$ is,

$x^{2}+y^{2}+z^{2}+yz-zx+xy-9=0$

Answer:

Therefore, The equation to the right circular cylinder $x^{2}+y^{2}+z^{2}+yz-zx+xy-9=0$.