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

The line is secant of the sphere if the line has two intersection points with the sphere. Let us find these points. Let us substitute $y = x = z+1$ in the equation of sphere:

$x^2+x^2+(x-1)^2 - x - x + (x-1 )-1 = 0.$ Therefore, $3x^2-3x-1=0.$

Next, we calculate the discriminant of the equation:

$D^2 = b^2-4ac =$ 21. We see that $D^2>0,$ so the equation has two roots and the line has two points of intersection with the sphere, because every root corresponds to one $y$ and $z$ value.

Now let us determine the roots:

$x_1 = \dfrac{3+\sqrt{21}}{6}, \;\; x_2 = \dfrac{3-\sqrt{21}}{6}.$

Next we calculate $y$ and $z$ coordinates:

$y_1=x_1 = \dfrac{3+\sqrt{21}}{6}, \;\; y_2= x_2 = \dfrac{3-\sqrt{21}}{6}; \\ z_1 = x_1-1 = \dfrac{-3+\sqrt{21}}{6}, \;\; z_2=x_2-1 = \dfrac{-3-\sqrt{21}}{6}.$

Therefore, points are

$\left( \dfrac{3+\sqrt{21}}{6}, \dfrac{3+\sqrt{21}}{6}, \dfrac{-3+\sqrt{21}}{6} \right) \mathrm{and} \left( \dfrac{3-\sqrt{21}}{6}, \dfrac{3-\sqrt{21}}{6}, \dfrac{-3-\sqrt{21}}{6} \right).$

We may also calculate the length of a segment between these points:

$l = \sqrt{\left(x_2-x_1 \right)^2 +\left(y_2-y_1 \right)^2 + \left(z_2-z_1 \right)^2} = \sqrt{7}.$