Question #152945
Find the equation of the cylinder whose
axis is x = y = - z and radius is 2.
1
Expert's answer
2020-12-28T19:11:19-0500

Let O(0,0,0),A(1,1,1)O(0,0,0), A(1,1,-1) be two points on the cylinder axis, X(x,y,z)X(x,y,z) be any point on the cylinder surface, P(px,py,pz)P(p_x, p_y, p_z) be the orthogonal projection of the point X onto the axis. Then the length |XP| is a distance from X to the axis.

The vector OPOP is a projection of OXOX , it is equal to OP=OA<OX,OA>/OA2OP = OA<OX,OA>/|OA|^2 .

<OX,OA>=x+yz<OX,OA> = x + y - z

OA2=12+12+(1)2=3|OA|^2 = 1^2 + 1^ 2 + (-1)^2 = 3

PX=OXOP=(xpx,ypy,zpz)PX =OX - OP = (x-p_x,y-p_y,z-p_z)

xpx=x(x+yz)/3=(2xy+z)/3x-p_x = x - (x+y-z)/3 = (2x-y+z)/3

ypy=y(x+yz)/3=(2yx+z)/3y-p_y = y - (x+y-z)/3 = (2y-x+z)/3

zpz=z+(x+yz)/3=(x+y+2z)/3z-p_z = z + (x+y-z)/3 = (x+y+2z)/3

4=PX2=((2xy+z)2+(2yx+z)2+(x+y+2z)2)/94 =|PX|^2=((2x-y+z)^2 + (2y-x+z)^2+(x+y+2z)^2)/9

36=6(x2+y2+z2xy+yz+zx)36 = 6(x^2+y^2+z^2-xy+yz+zx)

x2+y2+z2xy+yz+zx=6x^2+y^2+z^2-xy+yz+zx=6

This is the equation of the cylinder with the axis OAOA and radius 2.


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