Question

Show that every normal line to the sphere\({x^2} + {y^2} + {z^2} = {r^2}\)passes through the center of the sphere.

Short answer

The normal to the sphere\({x^2} + {y^2} + {z^2} = {r^2}\)passes through the center.

Step-by-step solution

Definition

The normal line to a curve is the line that is perpendicular to the tangent of the curve at a particular point.

Find the derivative of the function w.r.t\(x,y\,\& \,z\)

The equation of the sphere is\({x^2} + {y^2} + {z^2} = {r^2}\).

Let\(f(x,y,z) = {x^2} + {y^2} + {z^2}\)

Then \({f_x}(x,y,z) = 2x\)

\(\begin{aligned}{l}{f_y}(x,y,z) = 2y\\{f_z}(x,y,z) = 2z\end{aligned}\)

Equation of normal to sphere

The equation of the normal line to be sphere at \(\left( {{x_0},{y_0},{z_0}} \right)\)is given by

\(\frac{{x - {x_0}}}{{{f_x}\left( {{x_0},{y_0},{z_0}} \right)}} = \frac{{y - {y_0}}}{{{f_y}\left( {{x_0},{y_0},{z_0}} \right)}} = \frac{{z - {z_0}}}{{{f_z}\left( {{x_0},{y_0},{z_0}} \right)}}\)

i.e. \(\frac{{x - {x_0}}}{{2{x_0}}} = \frac{{y - {y_0}}}{{2{y_0}}} = \frac{{z - {z_0}}}{{2{z_0}}}\)

Substituting value

Now the centre of the sphere is\((0,0,0)\).

For the normal line to pass through\((0,0,0)\)

\(\begin{aligned}{l}\frac{{0 - {x_0}}}{{2{x_0}}} = \frac{{0 - {y_0}}}{{2{y_0}}} = \frac{{0 - {z_0}}}{{2{z_0}}}\\\frac{{ - 1}}{2} = \frac{{ - 1}}{2} = \frac{{ - 1}}{2}\end{aligned}\)

Which is true

Hence every normal line to the sphere passes through the centre of the sphere