Answer
$(-7,-2,-7); (\dfrac{19}{3}, \dfrac{14}{3},\dfrac{19}{3})$
Work Step by Step
Our aim is to determine the normal line equation.
The general form is:
$\dfrac{(x_2-x_1)}{f_x(x_1,y_1,z_1)}=\dfrac{(y_2-y_1)}{f_y(x_1,y_1,z_1)}=\dfrac{(z_2-z_1)}{f_x(x_1,y_1,z_1)}$ ...(1)
From the given data, we have $f(x,y,z)=(1,2,1)$
Equation (1), becomes:
Thus,
$\dfrac{(x-1)}{8}=\dfrac{(y-2)}{4}=\dfrac{(z-1)}{8}$
Let us consider
$\dfrac{(x-1)}{8}=\dfrac{(y-2)}{4}=\dfrac{(z-1)}{8}=s$
Therefore, $x=1+8s;y=2+4s,z=1+8s$
After simplifications we get $s=-1,\dfrac{2}{3}$
$x=-7,y=-2,z=-7;$ or, $x=\dfrac{19}{3}, y=\dfrac{14}{3},z=\dfrac{19}{3}$
Therefore, the normal line through the point (1,2,1) on the ellipsoid intersect the sphere $x^2+y^2+z^2=102$ are: $(-7,-2,-7); (\dfrac{19}{3}, \dfrac{14}{3},\dfrac{19}{3})$
