Question

find the smallest sphere which touches the lines (x-2)/1=(y-1)/-2=(z-6)/1 and (x+3)/7=(y+3)/-6=(z+3)/1 

Answer 1

$$Solution$$

We will first find the distance between the two lines. Then that is the diameter of smallest sphere touching the two st. lines. Dividing the diameter by 2 we will get the radius of the sphere.

The two lines $L_1$ and $L_2$ are

$$<t_1-2,\ 2-2t_1,\ t_1-6>\ and\ <7t_2+21,\ -6t_2-18,\ t_2+3>$$

The vector $\vec{P}$ between the two touchpoints $P _1$ and $P _2$ must be perpendicular to both direction vectors of the lines $L_1\ and\ L_2$

$\vec{P}=P_1-P_2= \begin{bmatrix} (t_1-2)-(7t_2+21)\\ (-2t_1+2)-(-6t_2-18)\\ (t_1-6)-(t_2+3) \end{bmatrix} \implies \begin{bmatrix} t_1-7t_2-23\\ -2t_1+6t_2+20\\ t_1-t_2-9 \end{bmatrix}\\ \vec{P}\begin{bmatrix}1\\ -2\\ 1 \end{bmatrix}=(t_1-7t_2-23)-2(-2t_1+6t_2+20)+(t_1-t_2-9)=0\\ \implies 6t_1-20t_2-72=0\\ \vec{P}\begin{bmatrix}7\\ -6\\ 1 \end{bmatrix}=-7(t_1-7t_2-23)+6(-2t_1+6t_2+20)+(t_1-t_2-9)=0\\ \implies -20t_1+86t_2+290=0\\$

Solving simultaneously, we get

$t_1=\frac{98}{29}\\ t_2=-\frac{75}{29}\\ P_1=\{\frac{40}{29}\, -\frac{138}{29}, \ -\frac{76}{29} \}\\ P_2=\{\frac{84}{29}\, -\frac{72}{29}, \ \frac{12}{29} \}\\$

The center of the sphere is the middle between the touchpoints:

$M=\{\frac{62}{29},\ -\frac{105}{29},\ -\frac{32}{29},\ \}\\ r^2=\frac{|\vec{P}|}{4}=\frac{(t_1-7t_2-23)^2+(-2t_1+6t_2+20)^2+(t_1-t_2-9)^2}{4}=\frac{121}{29}$

Thus the smallest sphere is the sphere with the equation

$$(x-\frac{62}{29})^2+(y+\frac{105}{29})^2+(z+\frac{32}{29})^2=\frac{121}{29}--->answer$$