Three Dimensional Geometry

1. Check this before you revise anything

Planes have been removed from this chapter entirely — but the introduction still promises them. The chapter's opening paragraph says it will "discuss about the equations of lines and planes in space under different conditions, angle between two lines, two planes, a line and a plane, shortest distance between two skew lines and distance of a point from a plane."

Not one of those plane topics exists. The actual section list is:

The word "plane" appears exactly six times in the whole 17-page chapter: three times inside that one introductory sentence, once as "XY-plane" in a diagram label, once as "the plane of the lines" inside a formula derivation, and once as "they lie in different planes" describing skew lines. There is no plane equation, no normal form, no angle-between-planes formula, no point-to-plane distance formula, and no exercise question involving a plane anywhere.

This matches the syllabus line exactly: "Direction cosines and direction ratios of a line joining two points. Cartesian equation and vector equation of a line, skew lines, shortest distance between two lines. Angle between two lines." Nothing about planes.

The old stub taught planes as a major section — the equation of a plane in normal form, angle between two planes, distance of a point from a plane. All removed in this rebuild.

This is also one of the smallest chapters in the book: Exercise 11.1 (5 questions), Exercise 11.2 (15), and a Miscellaneous Exercise (5) — 25 questions in total, against 261 in Chapter 7. Only Chapter 8 (9 questions) and Chapter 12 (10) are smaller.

2. Direction Cosines and Direction Ratios (Textbook 11.2)

A line in space has no single "slope" the way a line in the plane does. Instead its direction is recorded by the angles it makes with the three coordinate axes.

Direction cosines. If a directed line through the origin makes angles $\alpha ,\beta ,\gamma$ with the positive $x$, $y$ and $z$ axes, then $l=\cos \alpha$, $m=\cos \beta$, $n=\cos \gamma$ are its direction cosines, written $(l,m,n)$.

A subtlety the textbook flags: a line has two directions, so reversing it replaces $(l,m,n)$ by $(-l,-m,-n)$. To make the triple unique you must fix a direction along the line first.

The fundamental relation. Take a point $P(x,y,z)$ on the line at distance $r$ from the origin. Dropping perpendiculars to the axes gives $\cos \alpha =x/r$, $\cos \beta =y/r$, $\cos \gamma =z/r$. Therefore:

$l^2+m^2+n^2=\frac{x^2+y^2+z^2}{r^2}=\frac{r^2}{r^2}=1$

This is the test that separates direction cosines from direction ratios. If the squares of your three numbers do not add to $1$, they are not direction cosines.

Direction ratios. Any triple $(a,b,c)$ proportional to $(l,m,n)$ is a set of direction ratios. So $a=\lambda l$, $b=\lambda m$, $c=\lambda n$ for some non-zero $\lambda$.

Because $\lambda$ is arbitrary, direction ratios are not unique — a line has infinitely many sets. That is precisely what makes them convenient: you can clear fractions and cancel common factors freely. To recover the direction cosines, normalise:

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

Direction cosines of a line through two points (11.2.1). For $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$, the direction ratios of $PQ$ are simply the coordinate differences:

$(x_2-x_1,y_2-y_1,z_2-z_1)$

and dividing each by $|PQ|=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$ gives the direction cosines.

Collinearity. This gives the standard method for the last question of Exercise 11.1. Three points $A,B,C$ are collinear when the direction ratios of $AB$ and $BC$ are proportional — the shared point $B$ then forces all three onto one line.

3. Equation of a Line in Space (Textbook 11.3)

A line is pinned down by one point on it and one direction along it. That single sentence generates both forms below.

Vector form (11.3.1). Let $\vec{a}$ be the position vector of a known point $A$ on the line, and $\vec{b}$ a vector parallel to the line. For any point $P$ on the line with position vector $\vec{r}$, the segment $\overrightarrow{AP}$ is parallel to $\vec{b}$, so $\overrightarrow{AP}=\lambda \vec{b}$ for some scalar $\lambda$. Since $\vec{r}=\vec{a}+\overrightarrow{AP}$:

$\vec{r}=\vec{a}+\lambda \vec{b},\lambda \in \mathbb{R}$

As $\lambda$ runs over all reals, $P$ sweeps out the entire line. Each $\lambda$ names exactly one point.

Cartesian form. Write $\vec{a}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$ and $\vec{b}=a\hat{i}+b\hat{j}+c\hat{k}$, and $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$. Comparing components in $\vec{r}=\vec{a}+\lambda \vec{b}$ gives $x=x_1+\lambda a$, $y=y_1+\lambda b$, $z=z_1+\lambda c$. Solving each for $\lambda$ and equating:

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

Here $(a,b,c)$ are direction ratios of the line and $(x_1,y_1,z_1)$ is a point on it. If the direction cosines $(l,m,n)$ are used instead of direction ratios, the common value of the three fractions is the actual distance from $(x_1,y_1,z_1)$.

Line through two points. Given $A$ and $B$, take $\vec{b}={\vec{b}}_{\text{pos}}-{\vec{a}}_{\text{pos}}$, giving $\vec{r}=\vec{a}+\lambda ({\vec{b}}_{\text{pos}}-{\vec{a}}_{\text{pos}})$, or in Cartesian form:

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

The rewriting trap. Exam questions rarely present the Cartesian form cleanly. You will meet $\frac{1-x}{3}$ or $\frac{4-y}{5}$ or $\frac{7y-14}{2p}$, and you must convert each to the standard shape before reading anything off:

$\frac{1-x}{3}=\frac{x-1}{-3},\frac{4-y}{5}=\frac{y-4}{-5},\frac{7y-14}{2p}=\frac{y-2}{2p/7}$

The sign of the direction ratio flips whenever the variable is subtracted from a constant, and a coefficient on the variable must be divided out. Skipping this rewrite is the single largest source of lost marks in this chapter.

4. Angle between Two Lines (Textbook 11.4)

The angle between two lines is defined as the angle between their direction vectors. Since the dot product gives ${\vec{b}}_1\cdot {\vec{b}}_2=|{\vec{b}}_1||{\vec{b}}_2|\cos \theta$:

$\cos \theta =\frac{|{\vec{b}}_1\cdot {\vec{b}}_2|}{|{\vec{b}}_1||{\vec{b}}_2|}$

Why the absolute value. Each line has two possible direction vectors, $\vec{b}$ and $-\vec{b}$, which would give supplementary angles $\theta$ and $\pi -\theta$. The convention is to report the acute angle, and taking the modulus in the numerator enforces that automatically. Omitting it and reporting an obtuse angle is a standard error.

In terms of direction ratios $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$:

$\cos \theta =\frac{|a_1{a}_2+b_1{b}_2+c_1{c}_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$

and if direction cosines are used, the denominator is $1$ and the formula collapses to $\cos \theta =|l_1{l}_2+m_1{m}_2+n_1{n}_2|$.

Two special cases account for most of Exercise 11.2:

The perpendicularity test is what "find $p$ such that the lines are at right angles" questions reduce to: rewrite both equations into standard form, extract direction ratios, set the dot product to zero, solve for the parameter.

5. Shortest Distance between Two Lines (Textbook 11.5)

In a plane, two lines either meet or are parallel. In space there is a third possibility, and it is the one this section exists for.

Skew lines are lines that are neither parallel nor intersecting — they lie in different planes. Two edges of a room meeting at neither a corner nor running alongside each other are skew. For such lines the natural question is not "where do they meet" but "how close do they get".

The shortest distance between two lines is measured along the segment perpendicular to both.

Skew lines (11.5.1). For $\vec{r}={\vec{a}}_1+\lambda {\vec{b}}_1$ and $\vec{r}={\vec{a}}_2+\mu {\vec{b}}_2$, the vector ${\vec{b}}_1\times {\vec{b}}_2$ is perpendicular to both directions, so it points along the common perpendicular. The shortest distance is the projection of the joining vector ${\vec{a}}_2-{\vec{a}}_1$ onto that direction:

$d=\left|\frac{({\vec{b}}_1\times {\vec{b}}_2)\cdot ({\vec{a}}_2-{\vec{a}}_1)}{|{\vec{b}}_1\times {\vec{b}}_2|}\right|$

Read that as: numerator measures how far apart the lines are along the common perpendicular, denominator normalises it to a true length.

A useful corollary. The two lines intersect exactly when $d=0$, that is when $({\vec{b}}_1\times {\vec{b}}_2)\cdot ({\vec{a}}_2-{\vec{a}}_1)=0$. This is the coplanarity condition, and it is how "show these lines intersect" questions are answered.

Parallel lines (11.5.2). If the lines are parallel then ${\vec{b}}_1$ and ${\vec{b}}_2$ point the same way, so ${\vec{b}}_1\times {\vec{b}}_2=\vec{0}$ and the formula above divides by zero. Writing both lines with the same direction $\vec{b}$, the separate formula is:

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

Always test before you divide. Compute ${\vec{b}}_1\times {\vec{b}}_2$ first. If it is the zero vector the lines are parallel and need the second formula; otherwise they are skew and the first applies. Exercise 11.2 deliberately includes one of each, so applying the skew formula blindly will fail on at least one question.

Worked, mirroring the textbook's own technique. Find $p$ so that $\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}$ and $\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}$ are perpendicular.

Rewrite each into standard form first. The first line becomes $\frac{x-1}{-3}=\frac{y-2}{2p/7}=\frac{z-3}{2}$, with direction ratios $\left(-3,\frac{2p}{7},2\right)$. The second becomes $\frac{x-1}{-3p/7}=\frac{y-5}{1}=\frac{z-6}{-5}$, with direction ratios $\left(-\frac{3p}{7},1,-5\right)$.

Setting the dot product to zero: $\frac{9p}{7}+\frac{2p}{7}-10=0$, so $\frac{11p}{7}=10$, giving $p=\frac{70}{11}$.

Summary

• Direction cosines $(l,m,n)$ are the cosines of the angles with the axes and satisfy $l^2+m^2+n^2=1$; reversing the line's direction negates all three.
• Direction ratios are any triple proportional to $(l,m,n)$, so they are not unique; normalise by $\sqrt{a^2+b^2+c^2}$ to recover direction cosines.
• For a line through two points the direction ratios are the coordinate differences; three points are collinear when two such triples are proportional and share a point.
• Line: $\vec{r}=\vec{a}+\lambda \vec{b}$, or $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$ — one point plus one direction determines it.
• Always rewrite terms like $\frac{1-x}{3}$ into $\frac{x-1}{-3}$ before reading direction ratios; the sign flips, and coefficients on the variable must be divided out.
• Angle between lines: $\cos \theta =\frac{|{\vec{b}}_1\cdot {\vec{b}}_2|}{|{\vec{b}}_1||{\vec{b}}_2|}$ — the modulus forces the acute angle.
• Perpendicular when $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$; parallel when the direction ratios are proportional.
• Skew lines: $d=\left|\frac{({\vec{b}}_1\times {\vec{b}}_2)\cdot ({\vec{a}}_2-{\vec{a}}_1)}{|{\vec{b}}_1\times {\vec{b}}_2|}\right|$, and $d=0$ is exactly the condition for the lines to intersect.
• Parallel lines need the separate formula $\frac{|\vec{b}\times ({\vec{a}}_2-{\vec{a}}_1)|}{|\vec{b}|}$; test ${\vec{b}}_1\times {\vec{b}}_2$ before choosing.
• Planes are not part of this chapter at all in the current edition, despite the introduction promising plane equations, angles between planes, and point-to-plane distance.