Answer to Question #104279 in Analytic Geometry for Gayatri Yadav

Question

Show that x = y = z+1 is a secant line of the sphere x^2 +y^2 +z^2 −x−y+z−1 = 0.
Also find the intercept made by the sphere on the line.

Answer 1

Suppose that $x=y=z+1=t$ .

$x=t, \ y=t,\ z=t-1$

Substituting this values in the equation of sphere, we have:

$t^2+t^2+(t-1)^2-t-t+(t-1)-1=0$

$3t^2-3t-1=0$

$t_{1,2}=\frac{3\pm \sqrt{3^2-4\times 3\times (-1)}}{2\times 3}=\frac{3\pm \sqrt{21} }{6}$

$t_1=\frac{3+\sqrt{21}}{6}, \ t_2=\frac{3-\sqrt{21}}{6}$

So, we have two points $A(\frac{3+\sqrt{21}}{6}, \frac{3+\sqrt{21}}{6}, \frac{-3+\sqrt{21}}{6}), \ B(\frac{3-\sqrt{21}}{6}, \frac{3-\sqrt{21}}{6}, \frac{-3-\sqrt{21}}{6})$ .

The line intersects the sphere at two points A and B. Therefore, this line is a secant line.

Length of the intercept made by the sphere on the line is

$\sqrt {(x_a-x_b)^2+(y_a-y_b)^2+(z_a-z_b)^2}=\sqrt{ (\frac{2\sqrt{21}}{6})^2 + (\frac{2\sqrt{21}}{6})^2+ (\frac{2\sqrt{21}}{6})^2}=$

$= \sqrt{\frac{21}{3}}=\sqrt{7}$