Q.Find the shortest distance between the lines $l_1$ and $l_2$ whose vector equations are $\vec{r} = \hat{i} + \hat{j} + \lambda(2\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = 2\hat{i} + \hat{j} - \hat{k} + \mu(3\hat{i} - 5\hat{j} + 2\hat{k})$.

Concept: Skew Lines — lines that are neither parallel nor intersecting; the shortest distance is the length of the common perpendicular.

Step 1: Identify vectors.

For $l_1$: $\vec{a}_1 = \hat{i} + \hat{j}$, $\vec{b}_1 = 2\hat{i} - \hat{j} + \hat{k}$.

For $l_2$: $\vec{a}_2 = 2\hat{i} + \hat{j} - \hat{k}$, $\vec{b}_2 = 3\hat{i} - 5\hat{j} + 2\hat{k}$.

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

$\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (1-1)\hat{j} + (-1-0)\hat{k} = \hat{i} - \hat{k}$.

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix} = \hat{i}(-2+5) - \hat{j}(4-3) + \hat{k}(-10+3) = 3\hat{i} - \hat{j} - 7\hat{k}$.

Step 3: Shortest distance formula.

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

Numerator: $(\hat{i} - \hat{k}) \cdot (3\hat{i} - \hat{j} - 7\hat{k}) = 3 + 7 = 10$.

Denominator: $\sqrt{3^2 + (-1)^2 + (-7)^2} = \sqrt{9+1+49} = \sqrt{59}$.

Why This Works: The Concept of Skew Lines

Two lines in space that are neither parallel nor intersecting are called skew lines. They don't lie in the same plane, so the shortest distance between them is the length of the unique line segment that is perpendicular to both lines simultaneously — the common perpendicular.

Think of it this way: if you take a vector along each line ($\vec{b}_1$ and $\vec{b}_2$), their cross product $\vec{b}_1 \times \vec{b}_2$ gives a direction perpendicular to both. The shortest distance is then the projection of the vector joining any point on one line to any point on the other line onto this common perpendicular direction.

Step-by-Step Solution

1. Identify the vectors from the given equations

From $l_1: \vec{r} = \hat{i} + \hat{j} + \lambda(2\hat{i} - \hat{j} + \hat{k})$, we have:

• $\vec{a}_1 = \hat{i} + \hat{j} + 0\hat{k}$ (a point on $l_1$)
• $\vec{b}_1 = 2\hat{i} - \hat{j} + \hat{k}$ (direction vector of $l_1$)

From $l_2: \vec{r} = 2\hat{i} + \hat{j} - \hat{k} + \mu(3\hat{i} - 5\hat{j} + 2\hat{k})$, we have:

• $\vec{a}_2 = 2\hat{i} + \hat{j} - \hat{k}$
• $\vec{b}_2 = 3\hat{i} - 5\hat{j} + 2\hat{k}$

2. Find the vector joining the two points

So $\vec{a}_2 - \vec{a}_1 = \hat{i} - \hat{k}$.

3. Compute the cross product $\vec{b}_1 \times \vec{b}_2$

Expanding:

• $\hat{i}$ component: $(-1)(2) - (1)(-5) = -2 + 5 = 3$
• $\hat{j}$ component: $-( (2)(2) - (1)(3) ) = -(4 - 3) = -1$
• $\hat{k}$ component: $(2)(-5) - (-1)(3) = -10 + 3 = -7$

Thus $\vec{b}_1 \times \vec{b}_2 = 3\hat{i} - \hat{j} - 7\hat{k}$.

4. Find the magnitude of this cross product

5. Compute the scalar triple product $(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)$

6. Apply the shortest distance formula

Method: Shortest Distance Between Two Skew Lines (Vector Form)

Use this when two lines $\vec{r}=\vec{a_1}+\lambda\vec{b_1}$ and $\vec{r}=\vec{a_2}+\mu\vec{b_2}$ are skew (non-parallel, non-intersecting) and you need the shortest distance between them.

Steps

Step 1: Extract points and directions.

Read $\vec{a_1},\vec{b_1}$ from the first line and $\vec{a_2},\vec{b_2}$ from the second, treating any absent component as $0$.

Step 2: Compute $\vec{b_1}\times\vec{b_2}$ and the join vector $\vec{a_2}-\vec{a_1}$.

The cross product points along the common perpendicular; take special care with the sign of its middle ($\hat{j}$) term, a frequent slip.

Step 3: Form the distance.

This is (triple-product volume)/(base area) = height, i.e. the perpendicular gap. Keep the modulus. Before using it, confirm the lines really are skew: if $\vec{b_1}\times\vec{b_2}=\vec{0}$ they are parallel and this formula's denominator vanishes — switch to the parallel-line method.

Common Mistakes

Mistake 1: Sign slip on the $\hat{j}$ term of the cross product.

Why it's wrong: the middle cofactor carries a leading minus, $-\big[(2)(2)-(1)(3)\big]=-1$; mishandling it changes $\vec{b_1}\times\vec{b_2}$ and the whole answer. Correct approach: expand as $\hat{i}(\cdots)-\hat{j}(\cdots)+\hat{k}(\cdots)$, giving $(3,-1,-7)$.

Mistake 2: Omitting the modulus in the numerator.

Why it's wrong: the triple product can come out negative, but distance is non-negative. Correct approach: take the absolute value before dividing, giving $d=\tfrac{10}{\sqrt{59}}$.

Mistake 3: Dropping the missing $\hat{j}$ component of a point.

Why it's wrong: $\hat{i}+\hat{j}$ has $z=0$, and $2\hat{i}+\hat{j}-\hat{k}$ must be read as $(2,1,-1)$; a mis-read join vector breaks the numerator. Correct approach: write each point as a full $(x,y,z)$ triple.