Glenn Hopkins

2022-07-23

The sphere ${x}^{2}+{y}^{2}+{z}^{2}-2x+6y+14z+3=0$ meets the line joining $A(2,-1,4),B(5,5,5)$ in the points $C$ and $D$. Prove that $AC:CB=-AD:DB=1:2$

Sandra Randall

Beginner2022-07-24Added 17 answers

Hint : The equation of the line through $A$ and $B$ is $(x,y,z)=(2+3t,-1+6t,4+t)$. Substitute this into the equation for the circle, solve the quadratic to find the points $C$ and $D$ ... etc ...

Livia Cardenas

Beginner2022-07-25Added 5 answers

The line through $A(2,-1,4),B(5,5,5)$ has a direction given by $\overrightarrow{u}=B-A=\{3,6,1\}$

so the line $AB$ has equation $A+s\overrightarrow{v}=\{2+3s,-1+6s,4+s\}$

plug in the equation of the sphere

$(s+4{)}^{2}+(3s+2{)}^{2}+(6s-1{)}^{2}-2(3s+2)+6(6s-1)+14(s+4)+3=0$

$2(23{s}^{2}+26s+35)=0$

Which has no real solutions

so the line $AB$ has equation $A+s\overrightarrow{v}=\{2+3s,-1+6s,4+s\}$

plug in the equation of the sphere

$(s+4{)}^{2}+(3s+2{)}^{2}+(6s-1{)}^{2}-2(3s+2)+6(6s-1)+14(s+4)+3=0$

$2(23{s}^{2}+26s+35)=0$

Which has no real solutions

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