Equation and Shortest Distance of Skew Lines

Skew lines are lines in 3D space that do not intersect and are not parallel. They lie in different planes. The vector equation of a line is given by r = a + λb, where a is a position vector and b is the direction vector. To find the shortest distance between two skew lines, we use the formula: D = |(a₂ - a₁) · (b₁ × b₂)| / |b₁ × b₂|, where a₁, a₂ are position vectors, and b₁, b₂ are direction vectors. This distance is along the line perpendicular to both.

1.0What are Skew Lines?

Definition: Skew lines are two lines that do not intersect and are not parallel. They exist in different planes in 3D space.

2.0Characteristics of Skew Lines

• They never meet.
• They are not coplanar.
• Their direction ratios (or direction vectors) are not proportional.
• They do not have a common point of intersection.

Example:

Line 1 passes through point A(1, 2, 3) and has direction vector d1 = $\hat{i}+\hat{j}+\hat{k}$

Line 2 passes through point B(4, 0, -1) and has direction vector d2 = $\hat{i}-2\hat{j}+2\hat{k}$

These lines are not parallel (their direction vectors are not proportional) and do not intersect, hence they are skew.

3.0Equation of a Line in 3D

A line in 3D space can be written in vector form or Cartesian form.

Vector Form

$\vec{r}=\vec{a}+\lambda \vec{b}$

Where:

• $\vec{a}$ is the position vector of a point on the line.
• $\vec{b}$ is the direction vector.
• $\lambda$ is a scalar parameter.

Cartesian Form

$\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$

Where:

• $(x_1,y_1,z_1)$ is a point on the line.
• (l, m, n) are the direction ratios of the line.

4.0Shortest Distance of Skew Lines

Let line 1 be ${\vec{r}}_1={\vec{a}}_1+\lambda {\vec{b}}_1$, and line 2 be ${\vec{r}}_2={\vec{a}}_2+\mu {\vec{b}}_2$

Formula: The shortest distance (SD) between two skew lines is: $SD=\frac{|({\vec{a}}_2-{\vec{a}}_1)\cdot ({\vec{b}}_1\times {\vec{b}}_2)|}{|{\vec{b}}_1\times {\vec{b}}_2|}$

Derivation (Brief Overview)

• ${\vec{a}}_2-{\vec{a}}_1$ gives a vector connecting any two points on the lines.
• ${\vec{b}}_1\times {\vec{b}}_2$ is a vector perpendicular to both lines.
• The scalar projection of the connecting vector onto the perpendicular gives the shortest distance.

5.0Solved Example on Shortest Distance of Skew Lines

Example 1: Find the shortest distance between the lines:

Line 1: ${\vec{r}}_1=2\hat{i}+3\hat{j}-\hat{k}+\lambda (\hat{i}+2\hat{j}+\hat{k})$

Line 2: ${\vec{r}}_2=-\hat{i}+4\hat{j}+2\hat{k}+\mu (2\hat{i}-\hat{j}+\hat{k})$

Solution:

• ${\vec{a}}_1=2\hat{i}+3\hat{j}-\hat{k}$
• ${\vec{a}}_2=-\hat{i}+4\hat{j}+2\hat{k}$
• ${\vec{b}}_1=\hat{i}+2\hat{j}+\hat{k}$
• ${\vec{b}}_2=2\hat{i}-\hat{j}+\hat{k}$

Step 1: Compute ${\vec{a}}_2-{\vec{a}}_1:$ ${\vec{a}}_2-{\vec{a}}_1=-3\hat{i}+\hat{j}+3\hat{k}$

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

${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 1\\ 2& -1& 1\end{array}\right|=3\hat{i}+\hat{j}-5\hat{k}$

Step 3: Compute dot product:

$({\vec{a}}_2-{\vec{a}}_1)\cdot ({\vec{b}}_1\times {\vec{b}}_2)=(-3)(3)+(1)(1)+(3)(-5)=-9+1-15=-23$

Step 4: Magnitude of the cross product:

${\vec{b}}_1\times {\vec{b}}_2|=\sqrt{3^2+1^2+(-5{)}^2}=\sqrt{35}$

Step 5: Final Answer:

$SD=\frac{|-23|}{\sqrt{35}}=\frac{23}{\sqrt{35}}$

Example 2: Find the shortest distance between:

Line 1: ${\vec{r}}_1=\hat{i}+2\hat{j}+\lambda (2\hat{i}-\hat{j}+3\hat{k})$

Line 2: ${\vec{r}}_2=3\hat{i}-\hat{j}+4\hat{k}+\mu (\hat{i}+\hat{j}+2\hat{k})$

Solution:

• ${\vec{a}}_1=\hat{i}+2\hat{j}$
• ${\vec{b}}_1=2\hat{i}-\hat{j}+3\hat{k}$
• ${\vec{a}}_2=3\hat{i}-\hat{j}+4\hat{k}$
• ${\vec{b}}_2=\hat{i}+\hat{j}+2\hat{k}$

${\vec{a}}_2-{\vec{a}}_1=2\hat{i}-3\hat{j}+4\hat{k}$

${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -1& 3\\ 1& 1& 2\end{array}\right|=(-5)\hat{i}-1\hat{j}+3\hat{k}$

Dot product: $({\vec{a}}_2-{\vec{a}}_1)\cdot ({\vec{b}}_1\times {\vec{b}}_2)=2(-5)+(-3)(-1)+4(3)=-10+3+12=5$

Magnitude: $\sqrt{(-5{)}^2+(-1{)}^2+3^2}=\sqrt{35}$

SD = $\frac{|5|}{\sqrt{35}}=\frac{5}{\sqrt{35}}$

Example 3: Given:

Line 1: Passes through A(1, 0, -1) and parallel to ${\vec{d}}_1=2\hat{i}+3\hat{j}+\hat{k}$

Line 2: Passes through B(2, -1, 1) and parallel to ${\vec{d}}_2=-\hat{i}+4\hat{j}+2\hat{k}$

Find the shortest distance.

Solution:

• ${\vec{a}}_1=\hat{i}+0\hat{j}-\hat{k}$
• ${\vec{a}}_2=2\hat{i}-\hat{j}+\hat{k}$
• ${\vec{b}}_1=2\hat{i}+3\hat{j}+\hat{k}$
• ${\vec{b}}_2=-\hat{i}+4\hat{j}+2\hat{k}$

${\vec{a}}_2-{\vec{a}}_1=\hat{i}-\hat{j}+2\hat{k}$

6.0Practice Questions on Shortest Distance of Skew Lines

1. Find the shortest distance between the lines:

• $\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda (\hat{i}+\hat{j}+\hat{k})$
• $\vec{r}=4\hat{i}-\hat{j}+\mu (2\hat{i}-3\hat{j}+\hat{k})$

2. Determine whether the following lines are skew:

• $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{1}$
• $\frac{x}{1}=\frac{y-2}{-3}=\frac{z+4}{-1}$

3. Find the angle between the lines:

• ${\vec{r}}_1=2\hat{i}+\hat{j}-3\hat{k}+\lambda (\hat{i}-2\hat{j}+\hat{k})$
• ${\vec{r}}_2=-\hat{i}+4\hat{j}+2\hat{k}+\mu (3\hat{i}+\hat{j}-2\hat{k})$

4. Write the vector equation of a line passing through point A(1, 2, 3) and parallel to vector $\vec{v}=2\hat{i}-\hat{j}+3\hat{k}$