Question #274148

Find the radius and centre of the circular section of sphere |r|= 4, cut off by the plane


r.(2i-j+4k)= 3.


1
Expert's answer
2021-12-02T16:10:43-0500


Let SS be the sphere in R3R^3 with center O(x0,y0,z0)O(x_0,y_0,z_0) and radius r,r, and let PP be the plane with equation Ax+By+Cz+D=0,Ax+By+Cz+D=0, so that n=(A,B,C)\vec n=(A,B,C)is a normal vector of P.P.

If P0P_0 is an arbitrary point on P,P, the signed distance from the center of the sphere OO to the plane PP is



ρ=(OP0)nn=Ax0+By0+Cz0DA2+B2+C2.\rho=\dfrac{(O-P_0)\vec n}{\|\vec n\|}=\dfrac{Ax_0+By_0+Cz_0-D}{\sqrt{A^2+B^2+C^2}}.


The intersection SPS\cap P is a circle if and only if r<ρ<r,r<\rho<r, and in that case, the circle has radius rc=r2ρ2r_c=\sqrt{r^2-\rho^2} and center



C=O+ρnnC=O+\rho \cdot\dfrac{\vec n}{\|\vec n\|}=(x0,y0,z0)+ρ(A,B,C)A2+B2+C2=(x_0, y_0, z_0)+\rho\cdot\dfrac{(A,B,C)}{\sqrt{A^2+B^2+C^2}}


In our case:

O(0,0,0),r=4,2xy+4z=3O(0,0,0), |r|=4, 2x-y+4z=3


S={(x,y,z):x2+y2+z2=16},S=\{(x,y,z):x^2+y^2+z^2=16\},


P={(x,y,z):2xy+4z=3}}.P=\{(x,y,z):2x-y+4z=3\}\}.




ρ=Ax0+By0+Cz0DA2+B2+C2\rho=\dfrac{Ax_0+By_0+Cz_0-D}{\sqrt{A^2+B^2+C^2}}=2010+403(2)2+(1)2+(4)2=321=\dfrac{2\cdot0-1\cdot0+4\cdot0-3}{\sqrt{(2)^2+(-1)^2+(4)^2}}=-\dfrac{3}{\sqrt{21}}rc=r2ρ2=(4)2(321)2=1097r_c=\sqrt{r^2-\rho^2}=\sqrt{(4)^2-(-\dfrac{3}{\sqrt{21}})^2}=\sqrt{\dfrac{109}{7}}3.946\approx3.946C=O+ρnnC=O+\rho \cdot\dfrac{\vec n}{\|\vec n\|}=(0,0,0)+(321)(2,1,4)(2)2+(1)2+(4)2=(0, 0, 0)+\big(-\dfrac{3}{\sqrt{21}}\big)\cdot\dfrac{(2,-1,4)}{\sqrt{(2)^2+(-1)^2+(4)^2}}=(27,17,47)=(-\dfrac{2}{7}, \dfrac{1}{7}, -\dfrac{4}{7})rc=10973.946,C=(27,17,47)r_c=\sqrt{\dfrac{109}{7}}\approx3.946, C=(-\dfrac{2}{7}, \dfrac{1}{7}, -\dfrac{4}{7})

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