Q.Find the shortest distance between the lines whose vector equations are $\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k}$ and $\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}$

Concept: Shortest distance between two skew lines — we rewrite each line in standard form $\vec{r} = \vec{a} + \lambda \vec{b}$, then apply the formula

$d = \frac{|(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)|}{|\vec{b}_1 \times \vec{b}_2|}$.

Step 1: Identify direction vectors and points.

First line: $\vec{r} = \hat{i} - 2\hat{j} + 3\hat{k} + t(-\hat{i} + \hat{j} - 2\hat{k})$, so

$\vec{a}_1 = \hat{i} - 2\hat{j} + 3\hat{k}$, $\vec{b}_1 = -\hat{i} + \hat{j} - 2\hat{k}$.

Second line: $\vec{r} = \hat{i} - \hat{j} - \hat{k} + s(\hat{i} + 2\hat{j} - 2\hat{k})$, so

$\vec{a}_2 = \hat{i} - \hat{j} - \hat{k}$, $\vec{b}_2 = \hat{i} + 2\hat{j} - 2\hat{k}$.

Step 2: Compute $\vec{b}_1 \times \vec{b}_2$.

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} = ( -2 + 4)\hat{i} - (2 + 2)\hat{j} + (-2 - 1)\hat{k} = 2\hat{i} - 4\hat{j} - 3\hat{k}$.

Magnitude: $|\vec{b}_1 \times \vec{b}_2| = \sqrt{4 + 16 + 9} = \sqrt{29}$. …

Rewrite each line in point–direction form:

Cross product of the direction vectors:

Vector joining a point on each line:

$\vec a_2 - \vec a_1 = 0\,\hat i + \hat j - 4\hat k.$ …

Method: Shortest distance when lines are given in collapsed parametric form

Sometimes a line is written as a single position vector whose components each contain the parameter, e.g. $\vec r=(1-t)\hat i+(t-2)\hat j+(3-2t)\hat k$. Before any distance work, split it into the standard $\vec a+\lambda\vec b$ shape.

Steps

Step 1: Separate constants from the parameter. Group the parameter-free terms into the point $\vec a$ and the coefficients of the parameter into the direction $\vec b$:

Do this for both lines to obtain $\vec a_1,\vec b_1$ and $\vec a_2,\vec b_2$.

Step 2: Cross the directions — $\vec b_1\times\vec b_2$ and its magnitude. …

Common Mistakes

Mistake 1: Working directly with the collapsed form without splitting into point and direction.

Why it's wrong: $\vec r=(1-t)\hat i+(t-2)\hat j+(3-2t)\hat k$ must first become $(\hat i-2\hat j+3\hat k)+t(-\hat i+\hat j-2\hat k)$; reading a "vector" straight from the raw components mixes point and direction. Correct approach: group the parameter-free part as $\vec a$ and the $t$-coefficients as $\vec b$.

Mistake 2: Mis-signing the second line's $\hat k$ coefficient. …