Answer to Question #104265 in Analytic Geometry for Sourav Mondal

Question

Show that x = y = z+1 is a secant line of the sphere x² +y² +z²−x−y+z−1 = 0. Also find the intercept made by the sphere on the line.

Answer 1

The equation of the sphere

$x^2+y^2+z^2-x-y+z-1=0\\ (x-\frac{1}{2})^2+(y-\frac{1}{2})^2+(z+\frac{1}{2})^2=1+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\\ (x-\frac{1}{2})^2+(y-\frac{1}{2})^2+(z+\frac{1}{2})^2=\frac{7}{4}$

center $O(\frac{1}{2},\frac{1}{2},-\frac{1}{2})$ , $R=\sqrt\frac{7}{4}=\frac{\sqrt7}{2}$

Let

$x=y=z+1=t$

now putting the value of $x,y$ and $z$ in the above equation of the sphere we get

$(t-\frac{1}{2})^2+(t-\frac{1}{2})^2+(t-1+\frac{1}{2})^2=\frac{7}{4}\\ 3(t-\frac{1}{2})^2=\frac{7}{4}\\ (t-\frac{1}{2})^2=\frac{7}{12}\\ t-\frac{1}{2}=\sqrt\frac{7}{12} \\and \\t-\frac{1}{2}=-\sqrt\frac{7}{12}\\ t=\frac{1}{2}+\sqrt\frac{7}{12}\\ and\\ t=\frac{1}{2}-\sqrt\frac{7}{12}$

so we get two points where the line cuts the sphere :

$A(\frac{1}{2}+\sqrt\frac{7}{12},\frac{1}{2}+\sqrt\frac{7}{12},-\frac{1}{2}+\sqrt\frac{7}{12})\\ and\\ B(\frac{1}{2}-\sqrt\frac{7}{12},\frac{1}{2}-\sqrt\frac{7}{12},-\frac{1}{2}-\sqrt\frac{7}{12})$

distance between these points is

$AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2}$

computable

$(x_B-x_A)^2=(\frac{1}{2}-\sqrt\frac{7}{12}-\frac{1}{2}-\sqrt\frac{7}{12})^2=\\ =(2\sqrt\frac{7}{12})^2=\frac{7}{3}\\ (y_B-y_A)^2=(\frac{1}{2}-\sqrt\frac{7}{12}-\frac{1}{2}-\sqrt\frac{7}{12})^2=\frac{7}{3}\\ (z_B-z_A)^2=(-\frac{1}{2}-\sqrt\frac{7}{12}+\frac{1}{2}-\sqrt\frac{7}{12})^2=\frac{7}{3}$

$AB=\sqrt{\frac{7}{3}+\frac{7}{3}+\frac{7}{3}}=\sqrt{7}=2R$

distance between the two points is equal the diameter of the sphere

hence the given line is a secant the sphere and the intercept made by the line to the sphere is$\sqrt{7}$