Find the shortest distance between the two skew lines: r1 = (1-t)i + (t-2)j + (3-2t)k and r2 = (s+1)i + (2s-1)j - (2s+1)k.

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Step-by-Step Solution

Step 1: Identify position vectors and direction vectors

First, we need to identify the position vectors and direction vectors for each line. For a line in the form $\vec{r} = \vec{a} + \lambda\vec{b}$, $\vec{a}$ is the position vector of a point on the line and $\vec{b}$ is the direction vector of the line. By comparing the given equations with this standard form, we extract the respective vectors.

Step 2: Calculate $(\vec{a_2} - \vec{a_1})$

Next, we calculate the vector connecting a point on the first line to a point on the second line. This is done by subtracting the position vector of the first line from the position vector of the second line.

Step 3: Calculate the cross product $(\vec{b_1} \times \vec{b_2})$

To find the shortest distance between two skew lines, we need a vector that is perpendicular to both direction vectors. This vector is given by the cross product of the direction vectors $\vec{b_1}$ and $\vec{b_2}$. We compute the determinant of the matrix formed by the unit vectors and the components of $\vec{b_1}$ and $\vec{b_2}.

Step 4: Calculate the magnitude of $(\vec{b_1} \times \vec{b_2})$

We need the magnitude of the cross product vector to normalize it later. The magnitude is calculated as the square root of the sum of the squares of its components.

Step 5: Apply the shortest distance formula

The shortest distance $d$ between two skew lines is given by the formula $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$. We substitute the calculated values into this formula and perform the dot product and division.