Three Dimensional Geometry

Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XII, Chapter 11.

Do two roads that cross on a map ever meet?

Look at a flyover on a city map. The flyover and the road beneath it cross, and the map shows an X. Yet no car on one ever meets a car on the other. Seen from above, the two lines intersect; in three dimensions they never do, and they are not parallel either. Mathematicians call such pairs skew lines, and they cannot exist in a flat plane. This is the one genuinely new idea in the chapter. Everything else is a clean way to describe a line in space (its direction, its equation, the angle between two lines) and then to measure the gap between two lines.

This chapter is about lines. The tools are the vector ideas from the previous chapter, so if the dot product or cross product feels shaky, revisit Vector Algebra first.

Direction cosines and direction ratios

A line in space is pinned down by its direction and one point on it. For the direction, imagine a directed line $L$ through the origin making angles $\alpha, \beta, \gamma$ with the positive $x$, $y$, $z$ axes. The cosines of these angles are the direction cosines (DCs):

$$l = \cos\alpha, \quad m = \cos\beta, \quad n = \cos\gamma, \qquad l^2 + m^2 + n^2 = 1$$

The identity holds because $(l, m, n)$ are the components of a unit vector along $L$, and a unit vector has length 1. Reverse the direction of the line and each angle becomes its supplement, so all three cosines change sign. A line without an arrow therefore has two sets of DCs, equal in size and opposite in sign. A line that does not pass through the origin has the same DCs as the parallel line through the origin.

DCs are awkward because they usually contain square roots. Direction ratios (DRs) $a, b, c$ are any three numbers proportional to the DCs. They are usually integers, and they convert easily:

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

For the line through $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$, the DRs are simply the differences $x_2 - x_1,\ y_2 - y_1,\ z_2 - z_1$, and the DCs are those differences divided by the length $PQ$.

Worked example: DCs of the line through two points. Find the direction cosines of the line through $P(-2, 4, -5)$ and $Q(1, 2, 3)$.

Common wrong step. Writing $3, -2, 8$ as the DCs, or dividing by $77$ instead of $\sqrt{77}$. A quick test catches both: the squares of true DCs add up to 1, and $\dfrac{9 + 4 + 64}{77} = 1$.

Pause and predict: do the direction ratios $1, 2, 3$ and $-2, -4, -6$ describe the same line?

Check your answer: They describe the same direction up to reversal, so the lines are parallel (or coincident). The second set is $-2$ times the first. Their DCs are $\pm\dfrac{1}{\sqrt{14}}(1, 2, 3)$, equal in size and opposite in sign.

Pause and predict: can a line make angles of $30°$, $45°$ and $60°$ with the three axes?

Check your answer: No. The sum $\cos^2 30° + \cos^2 45° + \cos^2 60° = \dfrac{3}{4} + \dfrac{1}{2} + \dfrac{1}{4} = \dfrac{3}{2}$, which is not 1. Three angles that appear in a CUET question must always pass this check.

Equation of a line in space

A line is fixed by (i) a point and a direction, or (ii) two points. Both reduce to the first, because the two points give a direction.

Vector form. Let the line pass through the point $A$ with position vector $\vec{a}$ and be parallel to $\vec{b}$. A point $P$ with position vector $\vec{r}$ lies on the line exactly when $\vec{AP}$ is a scalar multiple of $\vec{b}$:

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

Think of $\lambda$ as a slider: at $\lambda = 0$ you stand at $A$, positive values move you along $\vec{b}$, negative values the other way, and sweeping all real numbers traces the whole line.

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

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

Read it like an address: the numbers being subtracted in the numerators are the point, and the denominators are the direction ratios. If the denominators are DCs $l, m, n$, the same equation holds. For the line through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, use the differences as the denominators:

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

Worked example: write the equation both ways. Find the equation of the line through $(5, 2, -4)$ that is parallel to $3\hat{i} + 2\hat{j} - 8\hat{k}$.

Common wrong step. Writing $z - 4$ for the last numerator. The point has $z_1 = -4$, so $z - (-4) = z + 4$.

Worked example: read a messy equation correctly. The Cartesian equations of a line are $6x - 2 = 3y + 1 = 2z - 2$. Find its direction ratios and a point on it, and write the vector form.

Common wrong step. Reading the DRs as $6, 3, 2$ straight from the coefficients and the point as $(2, -1, 2)$. Those are the numbers in front of $x, y, z$ and the constants, not the direction ratios. Always rewrite each part as $\text{coefficient} \times (x - \text{something})$ and divide through by a common factor.

Angle between two lines

Two lines in space may not meet, so the "angle between them" means the angle between their direction vectors, and it is taken as the acute angle. If the lines are $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$, the dot product gives

$$\cos\theta = \dfrac{\lvert\vec{b_1}\cdot\vec{b_2}\rvert}{\lvert\vec{b_1}\rvert\,\lvert\vec{b_2}\rvert}$$

The modulus in the numerator is what forces $\theta$ to be acute. With direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ the same formula reads

$$\cos\theta = \dfrac{\lvert a_1 a_2 + b_1 b_2 + c_1 c_2\rvert}{\sqrt{a_1^2 + b_1^2 + c_1^2}\,\sqrt{a_2^2 + b_2^2 + c_2^2}}$$

and with DCs it collapses to $\cos\theta = \lvert l_1 l_2 + m_1 m_2 + n_1 n_2\rvert$, because both square-root factors equal 1.

Notice the parallel test: the ratios are equal to each other, not equal to 1. The vectors are proportional, not identical.

Worked example: angle between two lines. Find the angle between $\vec{r} = 3\hat{i} + 2\hat{j} - 4\hat{k} + \lambda(\hat{i} + 2\hat{j} + 2\hat{k})$ and $\vec{r} = 5\hat{i} - 2\hat{j} + \mu(3\hat{i} + 2\hat{j} + 6\hat{k})$.

Common wrong step. Using the position vectors (the constant parts) in the dot product. The point on a line has nothing to do with its direction: moving a line parallel to itself changes none of its angles.

Shortest distance between two lines

Two lines in space can be in three relations: they intersect, they are parallel, or they are skew. The shortest distance (SD) is zero for intersecting lines. For the other two cases you need a formula, and each comes from one picture.

Skew lines. Let the lines be $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$. The shortest segment between them is perpendicular to both, so it points along $\vec{b_1}\times\vec{b_2}$. The vector $\vec{a_2} - \vec{a_1}$ joins a point on one line to a point on the other, and the SD is the length of its shadow along that common perpendicular:

$$d = \dfrac{\lvert(\vec{b_1}\times\vec{b_2})\cdot(\vec{a_2} - \vec{a_1})\rvert}{\lvert\vec{b_1}\times\vec{b_2}\rvert}$$

There is a neat solid interpretation. The numerator is the volume of the parallelepiped built on $\vec{a_2} - \vec{a_1}$, $\vec{b_1}$ and $\vec{b_2}$. The denominator is the area of its base, the parallelogram on $\vec{b_1}$ and $\vec{b_2}$. Volume divided by base area is the height, which is the gap between the two parallel planes that hold the two lines. In Cartesian form, with $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ the DRs of the two lines:

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

If the numerator is zero the lines are coplanar, so (not being parallel) they intersect.

Parallel lines. For $\vec{r} = \vec{a_1} + \lambda\vec{b}$ and $\vec{r} = \vec{a_2} + \mu\vec{b}$, pick any point on one line and find its distance to the other. The distance from the point $A_2$ to the line through $A_1$ in direction $\vec{b}$ is $\lvert A_1A_2\rvert\sin\theta$, and $\lvert\vec{b}\times(\vec{a_2} - \vec{a_1})\rvert = \lvert\vec{b}\rvert\lvert A_1A_2\rvert\sin\theta$. Divide by $\lvert\vec{b}\rvert$:

$$d = \dfrac{\lvert\vec{b}\times(\vec{a_2} - \vec{a_1})\rvert}{\lvert\vec{b}\rvert}$$

Which formula do you use? One cross product decides:

Worked example: shortest distance between skew lines. Find the shortest distance between $\vec{r} = \hat{i} + \hat{j} + \lambda(2\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = 2\hat{i} + \hat{j} - \hat{k} + \mu(3\hat{i} - 5\hat{j} + 2\hat{k})$.

Common wrong step. Forgetting to divide by $\lvert\vec{b_1}\times\vec{b_2}\rvert$ and reporting $10$ as the distance. The numerator is a volume; dividing by the base area turns it into a length.

Worked example: distance between parallel lines. Find the distance between $\vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$ and $\vec{r} = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu(2\hat{i} + 3\hat{j} + 6\hat{k})$.

Common wrong step. Using the skew-line formula. Here $\vec{b_1}\times\vec{b_2} = \vec{0}$, so the denominator is zero and the formula collapses. Always test for parallel directions first.

Beyond the textbook. For the exam: skew lines are non-coplanar, neither parallel nor intersecting, and the SD formula gives the gap between the two full lines. In reality: the flyover over a road is the picture to keep in mind. The clearance under the flyover is, to a good approximation, the shortest distance between two skew lines, and a map that shows only a crossing hides it. In air traffic the same idea explains why two routes that cross on a flat radar picture can still be separated by altitude.

Beyond the textbook. For the exam: the formula always produces the distance between infinite lines. In reality: real objects are segments or rays. For two segments, the closest pair of points on the infinite lines may fall outside the segments, and then the true shortest distance is measured to an end point. Computer graphics and games write a ray as $\vec{r} = \vec{a} + \lambda\vec{b}$ with $\lambda \ge 0$ and have to respect exactly this restriction. For a CUET question, assume full lines unless the word "segment" appears.

Which tool when?

Working in vector form is usually lighter than expanding Cartesian determinants. If an equation looks messy, convert to vector form first.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Before: Vector Algebra supplies the dot product for angles and the cross product for distances.
• Class XI link: Introduction to Three Dimensional Geometry gives the axes, octants and distance formula that every line problem here assumes.
• Flat version: Straight Lines is the two dimensional cousin. The point and slope form there becomes the point and direction form here.
• Determinants: Determinants is needed for the Cartesian distance formula and the coplanarity test.