Find the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$.

Solution

1. Identify the points ${\vec{a}}_1=(1,2,3)$ and ${\vec{a}}_2=(2,4,5)$, and the direction vectors ${\vec{b}}_1=(2,3,4)$ and ${\vec{b}}_2=(3,4,5)$.
2. Calculate the cross product ${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|=-\mathbf{i}+2\mathbf{j}-\mathbf{k}$, with magnitude $$\begin{array}{rl}|{\vec{b}}_1\times {\vec{b}}_2|& =\sqrt{(-1{)}^2+2^2+(-1{)}^2}\\ & =\sqrt{6}\end{array}$$.
3. Compute the vector ${\vec{a}}_2-{\vec{a}}_1=(1,2,2)$ and find the scalar triple product $$\begin{array}{rl}|({\vec{a}}_2-{\vec{a}}_1)\cdot ({\vec{b}}_1\times {\vec{b}}_2)|& =|(1)(-1)+(2)(2)+(2)(-1)|\\ & =|-1+4-2|\\ & =1\end{array}$$.
4. Divide the scalar triple product by the magnitude of the cross product to get the distance $d=\frac{1}{\sqrt{6}}$.
5. Hence the answer is (A).