Question #155298

Find the equation of the right circular cone with 

vertex at (1, 2, —1), axis as the x-axis and semi-

vertical angle π/3.


1
Expert's answer
2021-01-14T16:29:53-0500

Find the equation of the right circular cone with 

vertex at (1, 2, —1), axis as the x-axis and semi-

vertical angle π/3.

solution:


equation of rigth circular cone is cosΘ=l(xα)+m(yβ)+n(zΥ)l2+m2+n2x2+y2+z2cos \Theta={ l(x-\alpha)+ m(y-\beta)+ n(z-\Upsilon) \over\sqrt{l^2+m^2+n^2}\sqrt{x^2+y^2+z^2}}


the directional ratios (l,m,n)=(a,0,0) since the axis is the x-axis

θ=π3=1803=60o\theta = {\pi \over 3}={180 \over3} = 60^o


cos60o=a(x1)+0(yβ)+0(zγ)x2+y2+z2a2+02+02\because cos 60^o={a(x-1)+0(y-\beta)+0(z-\gamma) \over \sqrt{x^2 +y^2 + z^2} \sqrt{a^2 + 0^2+0^2}}


12=a(x1)ax2+y2+z2{1 \over 2} = {a(x-1) \over a\sqrt{x^2 +y^2 + z^2}}


12=(x1)x2+y2+z2{1 \over 2} = {(x-1) \over \sqrt{x^2 +y^2 + z^2}}


Square both sides


14=(x1)2x2+y2+z2{1 \over 4} = {(x-1)^2 \over x^2 +y^2 + z^2}


x2+y2+z2=4(x22x+1)x^2 +y^2 + z^2 = 4(x^2 -2x +1)


x2+y2+z2=4x28x+4x^2 +y^2 + z^2 = 4x^2 -8x +4

3x2y2z28x+4=03x^2 -y^2-z^2-8x +4=0


the equation of the right circular cone is

3x2y2z28x+4=03x^2 -y^2-z^2-8x +4=0



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