Q.Find the shortest distance between the lines given by $\vec{r} = (8 + 3\lambda)\hat{i} - (9 + 16\lambda)\hat{j} + (10 + 7\lambda)\hat{k}$ and $\vec{r} = 15\hat{i} + 29\hat{j} + 5\hat{k} + \mu(3\hat{i} + 8\hat{j} - 5\hat{k})$.

Concept: Shortest Distance Between Skew Lines — we use 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 vectors

First line: $\vec{a}_1 = 8\hat{i} - 9\hat{j} + 10\hat{k}$, $\vec{b}_1 = 3\hat{i} - 16\hat{j} + 7\hat{k}$.

Second line: $\vec{a}_2 = 15\hat{i} + 29\hat{j} + 5\hat{k}$, $\vec{b}_2 = 3\hat{i} + 8\hat{j} - 5\hat{k}$.

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

Step 3: Compute $(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)$ …

Concept First: Why This Formula Works

Two lines in 3D that are not parallel and do not intersect are called skew lines. The shortest distance between them is the length of the line segment that is perpendicular to both lines simultaneously — this is the common perpendicular.

Think of it geometrically:

• Each line has a direction vector ($\vec{b}_1$ and $\vec{b}_2$).
• The cross product $\vec{b}_1 \times \vec{b}_2$ gives a vector perpendicular to both directions.
• If you take any point $A$ on the first line and any point $B$ on the second line, the vector $\overrightarrow{AB}$ will have a component along this perpendicular direction.
• The length of that component is exactly the shortest distance.

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: $\vec{r} = (8 + 3\lambda)\hat{i} - (9 + 16\lambda)\hat{j} + (10 + 7\lambda)\hat{k}$

Rewrite in standard form $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$:

• $\vec{a}_1 = 8\hat{i} - 9\hat{j} + 10\hat{k}$
• $\vec{b}_1 = 3\hat{i} - 16\hat{j} + 7\hat{k}$

Second line: $\vec{r} = 15\hat{i} + 29\hat{j} + 5\hat{k} + \mu(3\hat{i} + 8\hat{j} - 5\hat{k})$

So:

• $\vec{a}_2 = 15\hat{i} + 29\hat{j} + 5\hat{k}$
• $\vec{b}_2 = 3\hat{i} + 8\hat{j} - 5\hat{k}$

2. Find the vector connecting a point on each line

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

Expand:

• $\hat{i}$ component: $(-16)(-5) - (7)(8) = 80 - 56 = 24$
• $\hat{j}$ component: $-[ (3)(-5) - (7)(3) ] = -[ -15 - 21 ] = -(-36) = 36$
• $\hat{k}$ component: $(3)(8) - (-16)(3) = 24 + 48 = 72$

So: …

Method: Shortest distance between two skew lines

Use this for the shortest distance between two lines that are neither parallel nor intersecting.

Steps

Step 1: Put both lines in point + direction form.

From each $\vec r=\vec a+\lambda\vec b$, read a point ($\vec a_1,\vec a_2$) and a direction ($\vec b_1,\vec b_2$). If a line is written as $(8+3\lambda)\hat i-\dots$, group the constant part and the $\lambda$-part component by component to recover $\vec a$ and $\vec b$.

Step 2: Build the common-perpendicular direction.

Compute $\vec b_1\times\vec b_2$; it is perpendicular to both lines. (If it is $\vec 0$ the lines are parallel and this method does not apply — use the parallel-line formula instead.)

Step 3: Project the join onto it. …

Common Mistakes

Mistake 1: Reading $\vec a$ and $\vec b$ wrongly from a line written as $(8+3\lambda)\hat i-(9+16\lambda)\hat j+\dots$.

Why it's wrong: the constant parts form $\vec a_1=(8,-9,10)$ and the $\lambda$-coefficients form $\vec b_1=(3,-16,7)$; mixing them corrupts everything downstream. Correct approach: group constant vs $\lambda$ terms component by component, minding the leading minus on $\hat j$.

Mistake 2: Dropping the modulus in the numerator.

Why it's wrong: the triple product can be negative, but a distance cannot. Correct approach: take $|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|$, giving $|1176|/84=14$. …