Q.By computing the shortest distance determine whether following lines intersect each other. $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(2\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = (2\hat{i} + 2\hat{j} - 3\hat{k}) + \mu(\hat{i} + \hat{j} - 2\hat{k})$

Two lines intersect each other if and only if the shortest distance between them is zero. Applying Theorem 6.6's formula, lines $\vec r=\vec a_1+\lambda_1\vec b_1$ and $\vec r=\vec a_2+\lambda_2\vec b_2$ intersect if and only if

i.e. the scalar triple product of $\vec a_2-\vec a_1$, $\vec b_1$, $\vec b_2$ vanishes -- geometrically, this says the four points/directions all lie in one plane (the lines are coplanar). Written out in Cartesian form for lines $\dfrac{x-x_1}{a_1}=\dfrac{y-y_1}{b_1}=\dfrac{z-z_1}{c_1}$ and $\dfrac{x-x_2}{a_2}=\dfrac{y-y_2}{b_2}=\dfrac{z-z_2}{c_2}$, this same condition becomes the determinant equation

$\begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix}=0.$ …