The shortest distance between lines L 1 and L 2 , where L 1 : x − 1 2 = y + 1 − 3 = z + 4 2 and L 2 is the…

Given: $L_2$ passes through $A\left(-4,4,3\right) \text{and} B\left(-1,6,3\right)$.

$\Rightarrow L_2:\frac{x+4}{3}=\frac{y-4}{2}=\frac{z-3}{0}$

Also, $L_1:\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$

We know that, shortest distance between $L_1: \frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $L_2: \frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ is given by,

$\left|\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{{\left(a_1{b}_2-a_2{b}_1\right)}^2+{\left(b_1{c}_2-b_2{c}_1\right)}^2+{\left(c_1{a}_2-c_2{a}_1\right)}^2}}\right|$.

So, shortest distance, $D=\left|\frac{\left|\begin{array}{ccc}-4-1& 1+4& 4+3\\ 2& -3& 2\\ 3& 2& 0\end{array}\right|}{\sqrt{{\left(4+9\right)}^2+{\left(0-4\right)}^2+{\left(6-0\right)}^2}}\right|$

$\Rightarrow D=\left|\frac{\left|\begin{array}{ccc}-5& 5& 7\\ 2& -3& 2\\ 3& 2& 0\end{array}\right|}{\sqrt{169+16+36}}\right|$

$\Rightarrow D=\left|\frac{31\times 3+48}{\sqrt{221}}\right|$

$\Rightarrow D=\frac{141}{\sqrt{221}}$