Three-Dimensional Geometry

Direction Cosines and Direction Ratios of a Line

If $\alpha$, $\beta$, and $\gamma$ are the angles which a vector $\mathbf{OP}$ makes with the positive directions of the coordinate axes $OX$, $OY$, and $OZ$ respectively, then $\cos(\alpha)$, $\cos(\beta)$, and $\cos(\gamma)$ are known as the direction cosines of $\mathbf{OP}$. These direction cosines are generally denoted by the letters $l$, $m$, and $n$, such that $l = \cos(\alpha)$, $m = \cos(\beta)$, and $n = \cos(\gamma)$. The angles $\alpha$, $\beta$, and $\gamma$ are called direction angles and satisfy the condition $0 \le \alpha, \beta, \gamma \le \pi$. From geometric considerations, the direction cosines relate to the coordinates of point $P(x, y, z)$ via:

$\cos(\alpha) = \frac{x}{|\mathbf{OP}|}$ $\cos(\beta) = \frac{y}{|\mathbf{OP}|}$ $\cos(\gamma) = \frac{z}{|\mathbf{OP}|}$

Here, $|\mathbf{OP}|$ represents the magnitude of the position vector of $P$, defined as $|\mathbf{OP}| = \sqrt{x^2 + y^2 + z^2}$.

Squaring and summing the direction cosines yields the fundamental relation:

$l^2 + m^2 + n^2 = \cos^2(\alpha) + \cos^2(\beta) + \cos^2(\gamma) = \frac{x^2 + y^2 + z^2}{|\mathbf{OP}|^2} = \frac{x^2 + y^2 + z^2}{x^2 + y^2 + z^2} = 1$

If the position vector is $\mathbf{r} = \mathbf{OP} = x\mathbf{\hat{i}} + y\mathbf{\hat{j}} + z\mathbf{\hat{k}}$, then the corresponding unit vector along $\mathbf{r}$ is given by:

$\mathbf{\hat{r}} = l\mathbf{\hat{i}} + m\mathbf{\hat{j}} + n\mathbf{\hat{k}}$

By definition, the direction cosines of the $X$-axis are $(\cos(0^\circ), \cos(90^\circ), \cos(90^\circ)) = (1, 0, 0)$. Similarly, the direction cosines of the $Y$-axis and $Z$-axis are $(0, 1, 0)$ and $(0, 0, 1)$, respectively.

Let $l$, $m$, and $n$ be the direction cosines of a vector $\mathbf{r}$. Any three numbers $a$, $b$, and $c$ that are proportional to $l$, $m$, and $n$ are called the direction ratios of the vector. Mathematically, this proportion is expressed as:

$\frac{l}{a} = \frac{m}{b} = \frac{n}{c} = k$

This implies that $(l, m, n) = (ka, kb, kc)$, establishing $(a, b, c)$ as direction ratios. For example, if $\left(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$ are direction cosines of a vector $\mathbf{r}$, then its direction ratios can be expressed as $(1, -1, 1)$, $(-1, 1, -1)$, $(2, -2, 2)$, or generally $(\lambda, -\lambda, \lambda)$. To obtain direction ratios from direction cosines, one simply multiplies the direction cosines by a common non-zero scalar. Consequently, a given vector possesses infinitely many direction ratios, whereas its direction cosines are strictly unique.

To obtain direction cosines from direction ratios $a$, $b$, and $c$, let $l = \lambda a$, $m = \lambda b$, and $n = \lambda c$ for a scalar $\lambda$. Applying the identity $l^2 + m^2 + n^2 = 1$ gives:

$a^2\lambda^2 + b^2\lambda^2 + c^2\lambda^2 = 1$ $\lambda^2(a^2 + b^2 + c^2) = 1$ $\lambda = \pm \frac{1}{\sqrt{a^2 + b^2 + c^2}}$

Substituting this back into the equations for $l$, $m$, and $n$ yields the direction cosines in terms of direction ratios:

$l = \pm \frac{a}{\sqrt{a^2 + b^2 + c^2}}$ $m = \pm \frac{b}{\sqrt{a^2 + b^2 + c^2}}$ $n = \pm \frac{c}{\sqrt{a^2 + b^2 + c^2}}$

Direction Cosines of a Line Passing Through Two Points

Since exactly one straight line passes through two distinct given points, the direction cosines of a line passing through $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ can be determined. Let $l$, $m$, and $n$ be the direction cosines of line $PQ$, making angles $\alpha$, $\beta$, and $\gamma$ with the $X$, $Y$, and $Z$ axes respectively. Dropping perpendiculars from $P$ and $Q$ to the $XY$-plane meeting at $R$ and $S$, and drawing a perpendicular from $P$ to $QS$ meeting at $N$, forms a right-angled triangle $PNQ$ where $\angle PQN = \gamma$. In right triangle $PNQ$:

$\cos(\gamma) = \frac{NQ}{PQ} = \frac{z_2 - z_1}{PQ}$

By similar geometric projections along the other coordinate axes:

$\cos(\alpha) = \frac{x_2 - x_1}{PQ}$ $\cos(\beta) = \frac{y_2 - y_1}{PQ}$

Therefore, the direction cosines of the line segment joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ are:

$\left(\frac{x_2 - x_1}{PQ}, \frac{y_2 - y_1}{PQ}, \frac{z_2 - z_1}{PQ}\right)$

where the distance $PQ$ is given by:

$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Equation of a Line in Space

A straight line in space is specified basically in two ways: passing through a given point and parallel to a given vector, or passing through two given points. Consider a line passing through a given point $A$ with position vector $\mathbf{a}$ relative to an origin $O$, and parallel to a given vector $\mathbf{b}$. Let $P$ be any point on the line with position vector $\mathbf{r}$. Since $\mathbf{AP}$ is parallel to $\mathbf{b}$, $\mathbf{AP} = \lambda \mathbf{b}$ for some scalar parameter $\lambda$. By vector addition:

$\mathbf{r} = \mathbf{OP} = \mathbf{OA} + \mathbf{AP} = \mathbf{a} + \lambda \mathbf{b}$

Thus, the vector equation of the straight line is:

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

Here, $\mathbf{r}$ is the position vector of any point $P(x, y, z)$ on the line, given by $\mathbf{r} = x\mathbf{\hat{i}} + y\mathbf{\hat{j}} + z\mathbf{\hat{k}}$. In particular, the equation of a straight line through the origin and parallel to $\mathbf{b}$ is $\mathbf{r} = \lambda \mathbf{b}$.

To derive the Cartesian form from the vector form, let the coordinates of the given point $A$ be $(x_1, y_1, z_1)$ and the direction ratios of the line be $a$, $b$, and $c$. Let the coordinates of any point $P$ be $(x, y, z)$. Writing the position vectors:

$\mathbf{r} = x\mathbf{\hat{i}} + y\mathbf{\hat{j}} + z\mathbf{\hat{k}}$ $\mathbf{a} = x_1\mathbf{\hat{i}} + y_1\mathbf{\hat{j}} + z_1\mathbf{\hat{k}}$ $\mathbf{b} = a\mathbf{\hat{i}} + b\mathbf{\hat{j}} + c\mathbf{\hat{k}}$

Substituting these into the vector equation and equating the coefficients of $\mathbf{\hat{i}}$, $\mathbf{\hat{j}}$, and $\mathbf{\hat{k}}$ gives the parametric equations of the line:

$x = x_1 + \lambda a$ $y = y_1 + \lambda b$ $z = z_1 + \lambda c$

Eliminating the parameter $\lambda$ yields the Cartesian form:

$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$

If $l$, $m$, and $n$ are the direction cosines of the line, the equation of the line is:

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

Angle Between Two Lines

Let $L_1$ and $L_2$ be two lines passing through the origin with direction ratios $a_1, b_1, c_1$ and $a_2, b_2, c_2$ respectively. Let $P$ be a point on $L_1$ and $Q$ be a point on $L_2$. The directed line segments $OP$ and $OQ$ are vectors with components $a_1, b_1, c_1$ and $a_2, b_2, c_2$. The acute angle $\theta$ between them is given by:

$\cos(\theta) = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2}\sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$

Using $\sin(\theta) = \sqrt{1 - \cos^2(\theta)}$, the angle in terms of $\sin(\theta)$ is:

$\sin(\theta) = \sqrt{1 - \frac{(a_1 a_2 + b_1 b_2 + c_1 c_2)^2}{(a_1^2 + b_1^2 + c_1^2)(a_2^2 + b_2^2 + c_2^2)}}$ $\sin(\theta) = \frac{\sqrt{(a_1^2 + b_1^2 + c_1^2)(a_2^2 + b_2^2 + c_2^2) - (a_1 a_2 + b_1 b_2 + c_1 c_2)^2}}{\sqrt{a_1^2 + b_1^2 + c_1^2}\sqrt{a_2^2 + b_2^2 + c_2^2}}$ $\sin(\theta) = \frac{\sqrt{(a_1 b_2 - a_2 b_1)^2 + (b_1 c_2 - b_2 c_1)^2 + (c_1 a_2 - c_2 a_1)^2}}{\sqrt{a_1^2 + b_1^2 + c_1^2}\sqrt{a_2^2 + b_2^2 + c_2^2}}$

If direction cosines $l_1, m_1, n_1$ for $L_1$ and $l_2, m_2, n_2$ for $L_2$ are given instead of direction ratios, using $l_1^2 + m_1^2 + n_1^2 = 1 = l_2^2 + m_2^2 + n_2^2$, the expressions take the following forms:

$\cos(\theta) = |l_1 l_2 + m_1 m_2 + n_1 n_2|$ $\sin(\theta) = \sqrt{(l_1 m_2 - l_2 m_1)^2 + (m_1 n_2 - m_2 n_1)^2 + (n_1 l_2 - n_2 l_1)^2}$

Two lines with direction ratios $a_1, b_1, c_1$ and $a_2, b_2, c_2$ satisfy key conditions for orientation. They are perpendicular ($\theta = 90^\circ$) if:

$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$

They are parallel ($\theta = 0^\circ$) if:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

If $\theta$ is the acute angle between lines given in vector form $\mathbf{r} = \mathbf{a}_1 + \lambda \mathbf{b}_1$ and $\mathbf{r} = \mathbf{a}_2 + \mu \mathbf{b}_2$, the cosine of the angle is:

$\cos(\theta) = \left| \frac{\mathbf{b}_1 \cdot \mathbf{b}_2}{|\mathbf{b}_1||\mathbf{b}_2|} \right|$

In Cartesian form, given two straight lines:

$\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}$ $\frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2}$

The direction vectors are $\mathbf{b}_1 = a_1\mathbf{\hat{i}} + b_1\mathbf{\hat{j}} + c_1\mathbf{\hat{k}}$ and $\mathbf{b}_2 = a_2\mathbf{\hat{i}} + b_2\mathbf{\hat{j}} + c_2\mathbf{\hat{k}}$. The dot product is $\mathbf{b}_1 \cdot \mathbf{b}_2 = a_1 a_2 + b_1 b_2 + c_1 c_2$, and magnitudes are $|\mathbf{b}_1| = \sqrt{a_1^2 + b_1^2 + c_1^2}, $|\mathbf{b}_2| = \sqrt{a_2^2 + b_2^2 + c_2^2}. Thus:

$\cos(\theta) = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2}\sqrt{a_2^2 + b_2^2 + c_2^2}}$

The condition for perpendicularity is $\mathbf{b}_1 \cdot \mathbf{b}_2 = 0 \implies a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$. The condition for parallelism is $\mathbf{b}_1 = \lambda \mathbf{b}_2 \implies \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

Shortest Distance Between Two Lines

If $l_1$ and $l_2$ are two skew lines, there is one and only one line perpendicular to both $l_1$ and $l_2$, known as the line of shortest distance. The distance $PQ$ along this perpendicular is the shortest distance. Let $l_1$ and $l_2$ be two lines whose vector equations are $\mathbf{r} = \mathbf{a}_1 + \lambda \mathbf{b}_1$ and $\mathbf{r} = \mathbf{a}_2 + \mu \mathbf{b}_2$. These lines pass through points $A$ and $B$ with position vectors $\mathbf{a}_1$ and $\mathbf{a}_2$ respectively, and are parallel to vectors $\mathbf{b}_1$ and $\mathbf{b}_2$. Since $PQ$ is perpendicular to both $l_1$ and $l_2$, it is parallel to $\mathbf{b}_1 \times \mathbf{b}_2$. Let $\mathbf{\hat{n}}$ be a unit vector along $PQ$:

$\mathbf{\hat{n}} = \pm \frac{\mathbf{b}_1 \times \mathbf{b}_2}{|\mathbf{b}_1 \times \mathbf{b}_2|}$

The shortest distance $PQ$ equals the magnitude of the projection of $\mathbf{AB}$ on $PQ$:

$PQ = |\mathbf{AB} \cdot \mathbf{\hat{n}}| = \left| \pm (\mathbf{a}_2 - \mathbf{a}_1) \cdot \frac{\mathbf{b}_1 \times \mathbf{b}_2}{|\mathbf{b}_1 \times \mathbf{b}_2|} \right| = \left| \frac{(\mathbf{b}_1 \times \mathbf{b}_2) \cdot (\mathbf{a}_2 - \mathbf{a}_1)}{|\mathbf{b}_1 \times \mathbf{b}_2|} \right|$

In scalar triple product form, this is expressed as:

$d = \frac{|[\mathbf{b}_1\, \mathbf{b}_2\, (\mathbf{a}_2 - \mathbf{a}_1)]|}{|\mathbf{b}_1 \times \mathbf{b}_2|}$

Two lines intersect if and only if the shortest distance between them is zero. Setting $d = 0$ gives the condition for intersection:

$(\mathbf{b}_1 \times \mathbf{b}_2) \cdot (\mathbf{a}_2 - \mathbf{a}_1) = 0$

In Cartesian form, let the two skew lines be:

$\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}$ $\frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2}$

The vector equations for these lines are $\mathbf{r} = (x_1\mathbf{\hat{i}} + y_1\mathbf{\hat{j}} + z_1\mathbf{\hat{k}}) + \lambda(a_1\mathbf{\hat{i}} + b_1\mathbf{\hat{j}} + c_1\mathbf{\hat{k}})$ and $\mathbf{r} = (x_2\mathbf{\hat{i}} + y_2\mathbf{\hat{j}} + z_2\mathbf{\hat{k}}) + \mu(a_2\mathbf{\hat{i}} + b_2\mathbf{\hat{j}} + c_2\mathbf{\hat{k}})$. Applying $\frac{(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)}{|\mathbf{b}_1 \times \mathbf{b}_2|}$ yields the shortest distance formula $d$ in Cartesian form:

$d = \left| \frac{\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}}{\sqrt{(b_1 c_2 - b_2 c_1)^2 + (c_1 a_2 - a_1 c_2)^2 + (a_1 b_2 - a_2 b_1)^2}} \right|$

Let $l_1$ and $l_2$ be two parallel lines whose equations are $\mathbf{r} = \mathbf{a}_1 + \lambda \mathbf{b}$ and $\mathbf{r} = \mathbf{a}_2 + \mu \mathbf{b}$ respectively. The lines pass through points $A$ and $B$ with position vectors $\mathbf{a}_1$ and $\mathbf{a}_2$ respectively, and both are parallel to $\mathbf{b}$. Let $BM$ be the shortest distance between $l_1$ and $l_2$, and let $\theta$ be the angle between $\mathbf{AB}$ and $l_1$. In right triangle $AMB$:

$\sin(\theta) = \frac{BM}{AB} \implies BM = AB \sin(\theta) = |\mathbf{AB}| \sin(\theta)$

Using the definition of cross product:

$|\mathbf{AB} \times \mathbf{b}| = |\mathbf{AB}||\mathbf{b}| \sin(\pi - \theta) = |\mathbf{AB}||\mathbf{b}| \sin(\theta) = (|\mathbf{AB}| \sin(\theta))|\mathbf{b}| = BM |\mathbf{b}|$

Solving for $BM$ yields:

$BM = \frac{|\mathbf{AB} \times \mathbf{b}|}{|\mathbf{b}|} = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \times \mathbf{b}|}{|\mathbf{b}|}$

Therefore, the shortest distance $d$ between two parallel lines $\mathbf{r} = \mathbf{a}_1 + \lambda \mathbf{b}$ and $\mathbf{r} = \mathbf{a}_2 + \mu \mathbf{b}$ is:

$d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \times \mathbf{b}|}{|\mathbf{b}|}$