In three-dimensional space, a directed line's orientation is uniquely determined by the angles it makes with the positive coordinate axes. These angles are fundamental to vector and Cartesian representations of spatial geometry.
Historical Chronicle / Epigraph: Leonhard Euler (1707-1783)
If a directed line $L$ passing through the origin makes angles $\alpha$, $\beta$, and $\gamma$ with the positive $x$, $y$, and $z$-axes respectively, then:
Any three numbers $a$, $b$, and $c$ that are proportional to the direction cosines $l$, $m$, and $n$ of a line are called the direction ratios (d.r.'s) or direction numbers of the line.
For a unique line passing through two given points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$:
A line in three-dimensional space can be uniquely determined under two primary geometric conditions: passing through a point parallel to a given vector, or passing through two distinct points.
Let the line pass through a point $A$ with position vector $\vec{a}$ and be parallel to a non-zero vector $\vec{b}$.
Let the line pass through two points $A$ and $B$ with position vectors $\vec{a}$ and $\vec{b}$ respectively.
The angle between two lines in space is defined as the angle between their parallel vectors passing through the origin.
In three-dimensional space, lines can be intersecting, parallel, or skew. Skew lines are lines that are neither parallel nor intersecting; they lie in different planes.
The shortest distance between two skew lines is the length of the common perpendicular segment joining them.
If two lines are parallel, they are coplanar and have the same direction vector $\vec{b}$.
| Geometric Concept | Vector Representation | Cartesian Representation | Key Conditions / Properties |
|---|---|---|---|
| Direction Cosines (d.c.'s) | $\hat{u} = l\hat{i} + m\hat{j} + n\hat{k}$ | $l = \cos\alpha, m = \cos\beta, n = \cos\gamma$ | $l^2 + m^2 + n^2 = 1$ |
| Line through $\vec{a}$ parallel to $\vec{b}$ | $\vec{r} = \vec{a} + \lambda \vec{b}$ | $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ | $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$ |
| Line through two points | $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$ | $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$ | Direction ratios: $x_2-x_1, y_2-y_1, z_2-z_1$ |
| Angle between two lines | $\cos\theta = \left|\frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1||\vec{b}_2|}\right|$ | $\cos\theta = \left|\frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{\sum a_1^2}\sqrt{\sum a_2^2}}\right|$ | Perpendicular if $\vec{b}_1 \cdot \vec{b}_2 = 0$ |
| Shortest Distance (Skew Lines) | $d = \left|\frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|}\right|$ | Determinant ratio form | $d = 0$ if lines intersect |
| Shortest Distance (Parallel Lines) | $d = \left|\frac{\vec{b} \times (\vec{a}_2 - \vec{a}_1)}{|\vec{b}|}\right|$ | N/A | Lines share direction vector $\vec{b}$ |
Mathematical Problem Statement: Find the direction cosines of the line passing through the two points $P(-2, 4, -5)$ and $Q(1, 2, 3)$.
Step-by-Step Algorithmic Solution Trace:
- Identify the coordinates of the given points:$$P(x_1, y_1, z_1) = (-2, 4, -5) \quad \text{and} \quad Q(x_2, y_2, z_2) = (1, 2, 3)$$
- Compute the direction ratios of the line segment $PQ$:$$x_2 - x_1 = 1 - (-2) = 3$$$$y_2 - y_1 = 2 - 4 = -2$$$$z_2 - z_1 = 3 - (-5) = 8$$
- Calculate the distance $PQ$ using the distance formula:$$PQ = \sqrt{(3)^2 + (-2)^2 + (8)^2} = \sqrt{9 + 4 + 64} = \sqrt{77}$$
- Determine the direction cosines ($l, m, n$) using the ratio of direction ratios to the distance $PQ$:$$l = \frac{3}{\sqrt{77}}, \quad m = \frac{-2}{\sqrt{77}}, \quad n = \frac{8}{\sqrt{77}}$$
Final Evaluation Output: The direction cosines of the line joining the two points are $\left(\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}}\right)$.
Mathematical Problem Statement: Find the shortest distance between the skew lines $l_1$ and $l_2$ whose vector equations are:$$\vec{r} = (\hat{i} + \hat{j}) + \lambda(2\hat{i} - \hat{j} + \hat{k})$$$$\vec{r} = (2\hat{i} + \hat{j} - \hat{k}) + \mu(3\hat{i} - 5\hat{j} + 2\hat{k})$$
Step-by-Step Algorithmic Solution Trace:
- Identify the position vectors $\vec{a}_1, \vec{a}_2$ and direction vectors $\vec{b}_1, \vec{b}_2$:$$\vec{a}_1 = \hat{i} + \hat{j}, \quad \vec{b}_1 = 2\hat{i} - \hat{j} + \hat{k}$$$$\vec{a}_2 = 2\hat{i} + \hat{j} - \hat{k}, \quad \vec{b}_2 = 3\hat{i} - 5\hat{j} + 2\hat{k}$$
- Compute the difference vector $\vec{a}_2 - \vec{a}_1$:$$\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (1-1)\hat{j} + (-1-0)\hat{k} = \hat{i} - \hat{k}$$
- Calculate the cross product $\vec{b}_1 \times \vec{b}_2$:$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix}$$$$\vec{b}_1 \times \vec{b}_2 = \hat{i}(-2 - (-5)) - \hat{j}(4 - 3) + \hat{k}(-10 - (-3)) = 3\hat{i} - \hat{j} - 7\hat{k}$$
- Calculate the magnitude of the cross product $|\vec{b}_1 \times \vec{b}_2|$:$$|\vec{b}_1 \times \vec{b}_2| = \sqrt{3^2 + (-1)^2 + (-7)^2} = \sqrt{9 + 1 + 49} = \sqrt{59}$$
- Compute the dot product $(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)$:$$(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = (3\hat{i} - \hat{j} - 7\hat{k}) \cdot (\hat{i} - \hat{k}) = 3(1) + (-1)(0) + (-7)(-1) = 3 + 7 = 10$$
- Apply the shortest distance formula:$$d = \left| \frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|} \right| = \frac{10}{\sqrt{59}}$$
Final Evaluation Output: The shortest distance between the two lines is $\frac{10}{\sqrt{59}}$ units.