Unit 3: The Straight Line in Space & Skew Lines

A straight line in $\mathbb{R}^3$ is uniquely determined either by:

1. A point through which it passes and its direction in space (specified by a direction vector or direction cosines).
2. The intersection of two non-parallel planes.

1. Vector and Parametric Equations

Let a line $\mathcal{L}$ pass through a fixed point $A$ with position vector $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ and be parallel to a direction vector $\vec{d} = l\hat{i} + m\hat{j} + n\hat{k}$.

If $P$ with position vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ is any arbitrary point on the line, the vector $\vec{AP} = \vec{r} - \vec{a}$ is collinear with $\vec{d}$. Hence, there exists a scalar parameter $t \in \mathbb{R}$ such that:

Equating components along the standard basis vectors:

These are the parametric equations of the straight line.

2. Symmetrical (Standard) Cartesian Form

Eliminating the scalar parameter $t$ from the parametric equations (assuming $l, m, n \neq 0$):

This is the canonical symmetrical form of a line passing through $(x_1, y_1, z_1)$ with direction ratios $(l, m, n)$.

Convention when a direction ratio vanishes: If one direction ratio is zero, say $n = 0$, the line lies in a plane parallel to the $xy$-plane ($z = z_1$). We write:

If $l, m, n$ are normalized to actual direction cosines $(\cos\alpha, \cos\beta, \cos\gamma)$, then the parameter $r = t$ represents the actual directed algebraic distance along the line from $(x_1, y_1, z_1)$ to $(x, y, z)$.

3. Two-Point Form of a Line

If the line passes through two distinct points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$, its direction vector is $\vec{d} = \vec{AB} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$. The symmetric equations become:

4. Non-Symmetric Form (General Equation as Two Planes)

A straight line in $\mathbb{R}^3$ can also be represented as the simultaneous intersection of two non-parallel planes:

where the normal vectors $\vec{n}_1 = (a_1, b_1, c_1)$ and $\vec{n}_2 = (a_2, b_2, c_2)$ are not proportional ($\vec{n}_1 \times \vec{n}_2 \neq \vec{0}$).

Algorithm: Reduction from General Form to Symmetrical Form

To convert the two-plane system into symmetrical form $\frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n}$:

1. Find the Direction Ratios $(l, m, n)$:

Since the line lies entirely in both $\Pi_1$ and $\Pi_2$, its direction vector $\vec{d}$ must be perpendicular to both normals $\vec{n}_1$ and $\vec{n}_2$. Thus:

Hence, $(l, m, n) = (b_1 c_2 - b_2 c_1, \; c_1 a_2 - c_2 a_1, \; a_1 b_2 - a_2 b_1)$.

2. Find a Specific Point $(x_0, y_0, z_0)$ on the Line:

Set one coordinate to a convenient constant (frequently $z = 0$, provided $a_1 b_2 - a_2 b_1 \neq 0$) and solve the resulting system of two linear equations in two variables:

Using Cramer's Rule:

The symmetrical equation is then immediately established.

1. Angle Between Two Straight Lines

Let two lines $\mathcal{L}_1$ and $\mathcal{L}_2$ have direction ratios $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ respectively. The angle $\theta$ between them is the angle between their direction vectors:

• Perpendicularity: $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.
• Parallelism: $\frac{l_1}{l_2} = \frac{m_1}{m_2} = \frac{n_1}{n_2}$.

2. Coplanarity Criterion for Two Lines

Consider two lines in space given in symmetric form:

Let $A(x_1, y_1, z_1)$ lie on $\mathcal{L}_1$ and $B(x_2, y_2, z_2)$ lie on $\mathcal{L}_2$. Their direction vectors are $\vec{d}_1 = (l_1, m_1, n_1)$ and $\vec{d}_2 = (l_2, m_2, n_2)$.

Two straight lines are coplanar (lie in a single shared plane) if and only if they either intersect at a unique point or are strictly parallel. In both cases, the displacement vector connecting points on the two lines, $\vec{AB} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$, must lie in the plane spanned by $\vec{d}_1$ and $\vec{d}_2$.

Consequently, the scalar triple product of $\vec{AB}$, $\vec{d}_1$, and $\vec{d}_2$ must vanish:

Expanding this into determinant form yields the fundamental condition of coplanarity:

3. Equation of the Plane Containing Coplanar Lines

If the coplanarity determinant vanishes and the lines are not parallel ($\vec{d}_1 \times \vec{d}_2 \neq \vec{0}$), they span a unique plane $\Pi$. Since the plane contains point $(x_1, y_1, z_1)$ and both direction vectors $\vec{d}_1$ and $\vec{d}_2$, the equation of the plane is:

Alternatively, using the reference point $(x_2, y_2, z_2)$ yields an identical plane.

In two dimensions, any two non-parallel lines must intersect. In three dimensions, this is no longer true.

1. Definition of Skew Lines

Two straight lines in $\mathbb{R}^3$ are defined as skew lines if they are neither parallel nor intersecting. Skew lines do not lie in any common plane; they exist in distinct, non-parallel planes.

2. Derivation of the Shortest Distance (S.D.) Formula

Let $\mathcal{L}_1$ pass through $A(\vec{a}_1)$ with direction $\vec{d}_1$, and $\mathcal{L}_2$ pass through $B(\vec{a}_2)$ with direction $\vec{d}_2$.

The shortest distance between $\mathcal{L}_1$ and $\mathcal{L}_2$ is measured along their common perpendicular — the unique line that intersects both $\mathcal{L}_1$ and $\mathcal{L}_2$ at right angles.

Let $\vec{n}$ be a vector perpendicular to both lines. By definition:

The unit vector in this common normal direction is:

Consider the connecting vector $\vec{AB} = \vec{a}_2 - \vec{a}_1$. The shortest distance $d$ is precisely the absolute length of the orthogonal projection of $\vec{AB}$ onto the common normal $\hat{n}$:

3. Cartesian Determinant Form of Shortest Distance

Substituting $\vec{a}_1 = (x_1, y_1, z_1)$, $\vec{a}_2 = (x_2, y_2, z_2)$, $\vec{d}_1 = (l_1, m_1, n_1)$, and $\vec{d}_2 = (l_2, m_2, n_2)$:

The numerator is the determinant:

The cross product in the denominator is:

Its magnitude is:

Thus, the exact Cartesian formula for the shortest distance is:

4. Equations of the Line of Shortest Distance

The line of shortest distance (the common perpendicular $\mathcal{L}_{SD}$) can be represented as the intersection of two planes:

1. The plane containing $\mathcal{L}_1$ and parallel to the common normal $\vec{n} = \vec{d}_1 \times \vec{d}_2$.
2. The plane containing $\mathcal{L}_2$ and parallel to the common normal $\vec{n} = \vec{d}_1 \times \vec{d}_2$.

Let $(l, m, n)$ be the direction ratios of the common perpendicular $\vec{n} = \vec{d}_1 \times \vec{d}_2$. Then the two defining planes are:

The simultaneous solution of these two plane equations defines the exact straight line of shortest distance in $\mathbb{R}^3$.