Q.Find the shortest distance between the lines $\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = 2\hat{i} - \hat{j} - \hat{k} + \mu(2\hat{i} + \hat{j} + 2\hat{k})$

Concept: Shortest Distance Between Skew Lines

The lines are skew (not parallel, not intersecting). The shortest distance is the length of the common perpendicular.

Step 1 – Identify vectors

First line: point $\vec{a}_1 = \hat{i} + 2\hat{j} + \hat{k}$, direction $\vec{b}_1 = \hat{i} - \hat{j} + \hat{k}$.

Second line: point $\vec{a}_2 = 2\hat{i} - \hat{j} - \hat{k}$, direction $\vec{b}_2 = 2\hat{i} + \hat{j} + 2\hat{k}$.

Step 2 – Compute cross product

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} = (-2 - 1)\hat{i} - (2 - 2)\hat{j} + (1 + 2)\hat{k} = -3\hat{i} + 0\hat{j} + 3\hat{k}$.

Magnitude: $|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$.

Step 3 – Compute scalar triple product

$\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (-1-2)\hat{j} + (-1-1)\hat{k} = \hat{i} - 3\hat{j} - 2\hat{k}$. …

Concept First: Why This Formula Works

Two lines in 3D 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 line segment that is perpendicular to both lines simultaneously.

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

Where $\vec{a}_1$ and $\vec{a}_2$ are position vectors of points on the two lines, and $\vec{b}_1$, $\vec{b}_2$ are their direction vectors.

Step-by-Step Solution

1. Identify the vectors from the given equations

First line: $\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$

So $\vec{a}_1 = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b}_1 = \hat{i} - \hat{j} + \hat{k}$

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

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

2. Find the vector connecting a point on each line

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

Expanding:

• $\hat{i}$ component: $(-1)(2) - (1)(1) = -2 - 1 = -3$
• $\hat{j}$ component: $-(1)(2) - (1)(2) = -(2 - 2) = 0$
• $\hat{k}$ component: $(1)(1) - (-1)(2) = 1 + 2 = 3$

So $\vec{b}_1 \times \vec{b}_2 = -3\hat{i} + 0\hat{j} + 3\hat{k} = -3\hat{i} + 3\hat{k}$ …

Method: Shortest distance between two skew lines

Two lines that are neither parallel nor intersecting (skew) have a unique shortest distance, measured along the common perpendicular to both. One formula delivers it.

Steps

Step 1: Extract points and directions. Write each line as $\vec r=\vec a+\lambda\vec b$ and read $\vec a_1,\vec b_1$ and $\vec a_2,\vec b_2$.

Step 2: Find the common-perpendicular direction via the cross product $\vec b_1\times\vec b_2$ (perpendicular to both lines), and its magnitude $|\vec b_1\times\vec b_2|$.

Step 3: Form the join vector $\vec a_2-\vec a_1$ between a point on each line.

Step 4: Apply the formula. …

Common Mistakes

Mistake 1: Omitting the absolute value in the numerator.

Why it's wrong: the scalar triple product $(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=-9$ is negative here, but a distance can't be negative. Correct approach: take $|-9|=9$ before dividing.

Mistake 2: Dividing by the cross-product vector instead of its magnitude.

Why it's wrong: the denominator is the length $|\vec b_1\times\vec b_2|=3\sqrt2$, a scalar. Correct approach: compute $\sqrt{(-3)^2+0^2+3^2}$ and divide by that. …