Three Dimensional Geometry: Class 12 Maths · Chapter 11

Lines in space: which way they point, how to write them, and how far apart they pass

Mathematics Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

75 min
29 practice questions
18 MCQs

What you will learn in Three Dimensional Geometry

Find the direction cosines of a line from its direction angles, from any set of direction ratios, or from two points on it
Use proportional direction ratios to decide whether three points are collinear or two lines are parallel
Write the line through a known point with a known direction as r = a + λb, and move between its vector, parametric and Cartesian forms
Find the acute angle between two lines from their direction ratios or vector equations, and test whether they are perpendicular or parallel
Tell intersecting, parallel and skew lines apart, and explain why the shortest gap between skew lines meets both at right angles
Compute the shortest distance between two skew lines, and between two parallel lines, with dot and cross products

Original companion text aligned to NCERT Mathematics Class 12, Chapter 11, pp. 377–393. Not affiliated with or endorsed by NCERT.

11.1 Introduction

Stand under a metro viaduct and look up. The track runs one way above you and the road runs another way under your feet. They never meet, yet they are not parallel either. How far apart are they where they come closest? Answering that needs a precise language for lines in space.

We first describe which way a line points, by its direction cosines and direction ratios. Then come the equation of a line in vector and Cartesian form, the angle between two lines, and the shortest distance between two lines that do not meet.

You already own the tools: coordinates of points in space from Class XI, and vectors with their dot and cross products from Chapter 10. Vectors keep the algebra short; coordinates show what each formula means.

Quick check

1 of 3

Warm-up from Class XI: how far is the point

(3, 1, - 3)

from the point

(1, - 2, 3)

units

Show answer

Answer 7 units

The differences are 2, 3 and

- 6

, so the distance is

\sqrt{4 + 9 + 36} = \sqrt{49} = 7

11.2 Direction Cosines and Direction Ratios of a Line

Take a line $L$ through the origin $O$ and put an arrow on it, choosing one of its two directions; it is now a directed line. Its direction is pinned down by three angles: $α$ with the positive $x$ -axis, $β$ with the positive $y$ -axis and $γ$ with the positive $z$ -axis, each between $0^{\circ}$ and $180^{\circ}$ . These are its direction angles.

Definition

Direction cosines

For a directed line with direction angles $α$ , $β$ , $γ$ , the numbers $l = \cos α$ , $m = \cos β$ and $n = \cos γ$ are its direction cosines.

ExampleThe positive $z$ -axis has direction angles $90^{\circ}, 90^{\circ}, 0^{\circ}$ and direction cosines $0, 0, 1$ .

Why the cosines rather than the angles? Mark the point $P$ on $L$ one unit from $O$ , on the side the arrow points to. The $x$ -coordinate of $P$ is the projection of $O P$ on the $x$ -axis, which is $1 \times \cos α$ . The same holds for $y$ and $z$ (the figure shows it for $y$ ), so $P$ is exactly the point $(l, m, n)$ and the unit vector along $L$ is $l \hat{i} + m \hat{j} + n \hat{k}$ (Fig. 11.1).

3/4The same works for every axis: $P = (l, m, n)$ is the far corner of a box, and $O P = 1$ gives $l^{2} + m^{2} + n^{2} = 1$ .

Fig. 11.1Direction cosines are the coordinates of the unit point on a directed line.NCERT Fig. 11.1 · p. 378

Three axes meet at O, with x coming towards the viewer, y to the right and z up. A directed line L passes through O, and P is the point one unit along it. In the second step the right triangle formed by O, P and the foot of the perpendicular from P to the y-axis is shaded, the angle beta is marked at O, and the edge along the y-axis is labelled m, equal to cos beta. In the third step a dashed box with edges l, m and n along the three axes is drawn, with OP as its diagonal, and the labels read P = (l, m, n) and l squared plus m squared plus n squared equals 1. In the last step the arrow is reversed: the point P prime at (−l, −m, −n) appears on the other side of O, each angle such as alpha becomes 180 degrees minus alpha, and each cosine changes sign.

$O P$ is the diagonal of a box whose edges are $l$ , $m$ and $n$ , and $O P = 1$ . The distance formula therefore gives the one rule every set of direction cosines obeys.

l^{2} + m^{2} + n^{2} = 1

(11.1)

where

$l, m, n$: the direction cosines of any directed line

A line without an arrow can be given either of two arrows, and the two choices give different direction cosines. Turning the arrow round changes every direction angle $θ$ into $180^{\circ} - θ$ , and $\cos (180^{\circ} - θ) = - \cos θ$ , so the second set is simply $(- l, - m, - n)$ . That is why we fix an arrow before talking about the direction cosines of a line.

Nothing here needs the line to pass through $O$ . Direction angles depend only on which way a line points, so a line anywhere in space has the same direction cosines as its parallel copy through the origin, drawn with the same arrow.

Direction ratios

Direction cosines are tied together by (11.1), which makes them awkward to read straight off a problem. Usually a direction is known only up to scale. "Walk 2 east, 1 north and 2 up" points the same way as "walk 4 east, 2 north and 4 up". Numbers like these, proportional to $l, m, n$ , are the direction ratios of the line; some books call them direction numbers.

Definition

Direction ratios

Three numbers $a, b, c$ , not all zero, are direction ratios of a line with direction cosines $l, m, n$ when $a = k l$ , $b = k m$ and $c = k n$ for some real $k \neq 0$ .

Example $2, - 1, 2$ and $- 4, 2, - 4$ are direction ratios of the same line.

DerivationFrom direction ratios to direction cosines

Suppose we know three direction ratios $a, b, c$ for some line. Build the vector $\vec{v} = a \hat{i} + b \hat{j} + c \hat{k}$ from them; it points along the line, either with the arrow or against it.

Dividing a vector by its length gives the unit vector in its direction, and the components of the unit vector along the line are its direction cosines. So the direction cosines are the components of $\pm \vec{v} / ∣ \vec{v} ∣$ .

\begin{aligned} (l, m, n) \\ = \pm \frac{(a, b, c)}{r} \end{aligned}

(11.2)

where

$r$: the length $\sqrt{a^{2} + b^{2} + c^{2}}$ of the vector with components $a, b, c$
$\pm$: one sign for all three: plus if the line is directed along $\vec{v}$ , minus if against it

So a line has just two sets of direction cosines but infinitely many sets of direction ratios: scaling a set by any non-zero number gives another, and any two sets are scalings of each other.

InteractiveDirection lab

A three-dimensional picture of the axes with an arrow from the origin to the point (a, b, c), the dashed box with that point as its far corner, and arcs for the angles the arrow makes with the three axes. Three sliders set the direction ratios a, b and c from minus 6 to 6, and a Reverse button turns the arrow round. A panel shows the length of the vector, the direction cosines as fractions and decimals, the three direction angles in degrees, and the sum of the squares of the cosines. Two more sliders, or dragging on the picture, turn and tilt the view. Nothing moves until the learner acts.

Fig. 11.2Set three direction ratios and read off the direction cosines and angles. Drag the picture (or use the sliders) to turn it.

In the direction lab, set $a, b, c$ to $2, - 1, 2$ and then to $4, - 2, 4$ : the arrow grows but the cosines stay the same. Press Reverse and watch every cosine change sign while each angle becomes its supplement.

Quick check

1 of 3

Could a line make an angle of

45^{\circ}

with each of the three axes?

Yes; any three angles from

0^{\circ}

180^{\circ}

are possibleYes, but only if the line passes through the originNo; the squares of the cosines would add to

\frac{3}{2}

, not 1No; a line can make at most two equal angles with the axes

Show answer

Answer (C) No; the squares of the cosines would add to $\frac{3}{2}$ , not 1

Each cosine would be

\frac{1}{\sqrt{2}}

, and three halves add to

\frac{3}{2} \neq 1

. Position does not matter (option B), and a line can make three equal angles: about

{54.7}^{\circ}

each, as you will see in Practice 11.1.

11.2.1 Direction cosines of a line passing through two points

Two points fix a line and, once we decide to travel from the first to the second, a direction. Let the points $P$ and $Q$ have coordinates $(x_{1}, y_{1}, z_{1})$ and $(x_{2}, y_{2}, z_{2})$ .

The vector $\vec{P Q}$ has components $x_{2} - x_{1}$ , $y_{2} - y_{1}$ and $z_{2} - z_{1}$ . Dividing it by its length $P Q$ gives the unit vector along the line, and the components of a unit vector are exactly the direction cosines.

3/3In the right triangle $P M Q$ , $\cos γ = \frac{z_{2} - z_{1}}{P Q}$ .

Fig. 11.3The direction of

P Q

is read off the box that has

P

and

Q

at opposite corners.NCERT Fig. 11.2 · p. 379

Small axes at the origin O sit at the lower left. Points P and Q are joined by a segment. In the second step a dashed box with edges parallel to the axes is drawn so that P and Q are opposite corners; three edges forming a path from P to Q are highlighted and labelled x2 minus x1, y2 minus y1 and z2 minus z1. In the third step the corner M directly above P is marked, the right triangle P, M, Q is shaded with its right angle at M, the angle gamma is marked at P between PM and PQ, and the label reads cos gamma equals z2 minus z1 over PQ.

Fig. 11.3 shows the same thing as a picture. The three components are the edges of the box that has $P$ and $Q$ at opposite corners, and $P Q$ is its long diagonal. In the shaded right triangle the vertical edge $z_{2} - z_{1}$ lies next to the angle $γ$ and $P Q$ is the hypotenuse, so $\cos γ = \frac{z_{2} - z_{1}}{P Q}$ . Signs look after themselves: if $Q$ is lower than $P$ , then $n < 0$ and $γ$ is obtuse.

\begin{aligned} l & = \frac{x_{2} - x_{1}}{P Q} \\ m & = \frac{y_{2} - y_{1}}{P Q} \\ n & = \frac{z_{2} - z_{1}}{P Q} \end{aligned}

(11.3)

where

$P Q$: the distance between the points: the square root of the sum of the squares of the three differences

In particular $x_{2} - x_{1}$ , $y_{2} - y_{1}$ , $z_{2} - z_{1}$ are direction ratios of the line $P Q$ . So are $x_{1} - x_{2}$ , $y_{1} - y_{2}$ , $z_{1} - z_{2}$ , which describe the same line directed from $Q$ to $P$ .

Example 1NCERT Example 1 · p. 380

From angles to cosines

The direction angles of a line are $120^{\circ}$ , $45^{\circ}$ and $60^{\circ}$ . What are its direction cosines? Check them against (11.1).

Try it yourself first. Then reveal the solution one step at a time.

Example 2NCERT Example 2 · p. 380

From direction ratios to cosines

Turn the direction ratios $4, - 4, 7$ into direction cosines, and pick the set for which the line makes an acute angle with the positive $z$ -axis.

Try it yourself first. Then reveal the solution one step at a time.

Example 3NCERT Example 3 · p. 380

A zip line

An adventure camp measures positions in metres. Its straight zip line runs from the top of a tower at $A (10, 20, 90)$ to a landing platform at $B (70, 50, 70)$ further down the hill. Find the direction cosines of the line directed from $A$ towards $B$ and the length of the cable.

Try it yourself first. Then reveal the solution one step at a time.

Example 4NCERT Example 4 · p. 381

The axes themselves

Find the direction cosines of the three coordinate axes (directed the positive way), and of the line through $O$ in the $x y$ -plane that bisects the angle between the positive $x$ - and $y$ -axes.

Try it yourself first. Then reveal the solution one step at a time.

Example 5NCERT Example 5 · p. 381

Three points on one line

Do the points $A (- 1, 4, 2)$ , $B (1, 1, 3)$ and $C (5, - 5, 5)$ lie on one straight line, that is, are they collinear? Decide with direction ratios.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

The line from

(2, - 1, 4)

(5, 3, 4)

makes an angle

β

with the

y

-axis. Find

\cos β

Show answer

Answer 0.8

The differences are

3, 4, 0

and the distance is 5, so

\cos β = \frac{4}{5} = 0.8

. The line stays at height 4, so its angle with the

z

-axis is

90^{\circ}

Practice · 6 questions

Practice 11.1: questions with answers

Short answerEasyNCERT Exercise 11.1, Q1
A directed line makes angles of $135^{\circ}$ , $120^{\circ}$ and $60^{\circ}$ with the positive $x$ -, $y$ - and $z$ -axes. Write down its direction cosines and verify (11.1).
Show solution
Answer $- \frac{1}{\sqrt{2}}$ , $- \frac{1}{2}$ , $\frac{1}{2}$
$\cos 135^{\circ} = - \cos 45^{\circ} = - \frac{1}{\sqrt{2}}$ , $\cos 120^{\circ} = - \frac{1}{2}$ and $\cos 60^{\circ} = \frac{1}{2}$ .
Check: $\frac{1}{2} + \frac{1}{4} + \frac{1}{4} = 1$ .
Short answerMediumNCERT Exercise 11.1, Q2
A cube has one corner at the origin and three edges along the positive axes. Find the direction cosines of the cube's diagonal from $O$ to the opposite corner, and the angle it makes with each edge at $O$ .
Show solution
Answer $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ ; about ${54.7}^{\circ}$ with each edge
The diagonal goes from $(0, 0, 0)$ to $(s, s, s)$ , so $s, s, s$ (or simply $1, 1, 1$ ) are direction ratios.
Their length is $\sqrt{3}$ , so the direction cosines are $\frac{1}{\sqrt{3}}$ each, directed away from $O$ .
Equal cosines mean equal angles, so the diagonal makes the same angle with every edge: $\cos^{- 1} \frac{1}{\sqrt{3}} \approx {54.7}^{\circ}$ .
Directed the other way, the cosines are all $- \frac{1}{\sqrt{3}}$ .
Short answerEasyNCERT Exercise 11.1, Q3
Find the direction cosines of a line whose direction ratios are $12, - 15, - 16$ .
Show solution
Answer $\pm (\frac{12}{25}, - \frac{3}{5}, - \frac{16}{25})$
$144 + 225 + 256 = 625$ , whose square root is 25.
By (11.2) the direction cosines are $\pm (\frac{12}{25}, - \frac{15}{25}, - \frac{16}{25})$ , that is, $\pm (\frac{12}{25}, - \frac{3}{5}, - \frac{16}{25})$ .
ProofMediumNCERT Exercise 11.1, Q4
Prove that the points $(3, - 1, 0)$ , $(5, 2, - 4)$ and $(- 1, - 7, 8)$ lie on one line.
Show solution
Call the points $A$ , $B$ , $C$ . Direction ratios of $A B$ are $2, 3, - 4$ .
Direction ratios of $A C$ are $- 4, - 6, 8$ , which is $- 2$ times the first set.
So $A B ∥ A C$ , and the two segments share $A$ . Hence $A$ , $B$ , $C$ are collinear, with $A$ between $B$ and $C$ .
Short answerMediumNCERT Exercise 11.1, Q5
A triangle has vertices $A (4, 2, 3)$ , $B (- 2, - 1, 1)$ and $C (0, 0, - 1)$ . Find the direction cosines of its sides, directed $A \to B$ , $B \to C$ and $C \to A$ .
Show solution
Answer $A B$ : $- \frac{6}{7}, - \frac{3}{7}, - \frac{2}{7}$ ; $B C$ : $\frac{2}{3}, \frac{1}{3}, - \frac{2}{3}$ ; $C A$ : $\frac{2}{3}, \frac{1}{3}, \frac{2}{3}$
$\vec{A B}$ has components $- 6, - 3, - 2$ and length $\sqrt{36 + 9 + 4} = 7$ .
$\vec{B C}$ has components $2, 1, - 2$ and length 3.
$\vec{C A}$ has components $4, 2, 4$ and length 6.
Dividing each by its length gives the answers. Check: the three vectors add to the zero vector, as they must for a closed triangle.
Quick answerEasy
Which of these can be the direction cosines of a line?
$\frac{1}{2}, \frac{1}{2}, \frac{1}{2}$ $\frac{1}{3}, \frac{2}{3}, \frac{2}{3}$ $1, 1, 0$ $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{3}$
Show answer
Answer (B) $\frac{1}{3}, \frac{2}{3}, \frac{2}{3}$
Only $\frac{1}{9} + \frac{4}{9} + \frac{4}{9}$ equals 1. The others give $\frac{3}{4}$ , 2 and $\frac{7}{9}$ ; they can be direction ratios, but not direction cosines.

11.3 Equation of a Line in Space

In a plane, a point and a slope fix a line. In space one slope is not enough, but two other pieces of data are: a point on the line together with its direction, or two points on it. Both lead to the same kind of equation.

11.3.1 Equation of a line through a given point and parallel to a given vector

Fix the point $A$ , with position vector $\vec{a}$ , and a direction $\vec{b} \neq \vec{0}$ . When does a point $P$ , with position vector $\vec{r}$ , belong to the line through $A$ along $\vec{b}$ ? Exactly when the arrow from $A$ to $P$ is a multiple of $\vec{b}$ . That arrow is $\vec{r} - \vec{a}$ , so the test is $\vec{r} - \vec{a} = λ \vec{b}$ for some real number $λ$ .

Vector equation of a line

\vec{r} = \vec{a} + λ \vec{b}

(11.4)

where

$\vec{a}$: position vector of a known point $A$ on the line
$\vec{b}$: a vector parallel to the line
$λ$: a real number, the parameter

Read it as a recipe: start at $A$ , then take $λ$ steps of $\vec{b}$ . Each real $λ$ gives one point of the line, and each point of the line comes from exactly one $λ$ , which is why $λ$ is called a parameter. $λ = 0$ gives $A$ itself, $λ = 1$ one full step, and a negative $λ$ walks backwards (Fig. 11.4).

3/4With $λ = 2.5$ , $\vec{A P} = 2.5 \vec{b}$ .

Fig. 11.4Every point of the line is

\vec{a} + λ \vec{b}

for one value of

λ

.NCERT Fig. 11.3 · p. 382

Axes meet at the origin O. A position-vector arrow labelled a runs from O to the point A, and a short thick arrow labelled b starts at A along the line. A purple point P moves along the line through A. In the first step lambda is 0 and P sits on A. In the second step lambda is 1, P is at the tip of b and an arrow labelled r runs from O to P. In the third step lambda is 2.5 and P is further along, with the stretch from A to P highlighted. In the last step lambda is minus 1.2 and P lies behind A, on the other side from b. The label r equals a plus lambda b stays on the picture throughout.

The components of $\vec{b}$ are direction ratios of the line, and any set of direction ratios can be used as the components of $\vec{b}$ . One warning about letters: the number $b$ among $a, b, c$ , the vector $\vec{b}$ and its length $∣ \vec{b} ∣$ are three different things.

DerivationFrom the vector form to the Cartesian form

Read (11.4) one coordinate at a time. If $A = (x_{1}, y_{1}, z_{1})$ and $\vec{b}$ has components $a, b, c$ , then the $x$ -coordinate of $\vec{a} + λ \vec{b}$ is $x_{1} + λ a$ , and the same pattern holds for $y$ and $z$ .

So the point $(x, y, z)$ lies on the line exactly when, for one and the same $λ$ , the three parametric equations below hold.

\begin{aligned} x & = x_{1} + λ a \\ y & = y_{1} + λ b \\ z & = z_{1} + λ c \end{aligned}

(11.5)

If none of $a, b, c$ is zero, each equation can be solved for $λ$ . One $λ$ serves all three, so the three expressions must be equal, which removes the parameter altogether. We call the result the Cartesian form of the line's equation.

\begin{aligned} \frac{x - x_{1}}{a} & = \frac{y - y_{1}}{b} \\ = \frac{z - z_{1}}{c} \end{aligned}

(11.6)

where

$(x_{1}, y_{1}, z_{1})$: a point on the line
$a, b, c$: direction ratios of the line (direction cosines $l, m, n$ work too)

Two remarks make (11.6) safe to use. First, the numerators must be "variable minus constant" with coefficient 1 on the variable; rewrite a form such as $\frac{3 - x}{2}$ as $\frac{x - 3}{- 2}$ before reading off anything. Second, if a direction ratio is 0, say $c = 0$ , the line stays at the height $z = z_{1}$ . We still write $\frac{z - z_{1}}{0}$ as shorthand, meaning only that $z - z_{1} = 0$ .

The second way to fix a line, by two points, reduces to the first. If $A$ and $B$ have position vectors $\vec{a}$ and $\vec{b}$ , the line $A B$ passes through $A$ and is parallel to $\vec{b} - \vec{a}$ .

Line through two points

\vec{r} = \vec{a} + λ (\vec{b} - \vec{a})

(11.7)

Example 6NCERT Example 6 · p. 382

The path of a drone

A drone leaves the point $(- 1, 3, 2)$ and flies in a straight line in the direction of $4 \hat{i} - \hat{j} + 5 \hat{k}$ . Write the vector and Cartesian equations of its path, and find the point with $λ = 2$ .

Try it yourself first. Then reveal the solution one step at a time.

Example

A line written in disguise

Find a point on the line $\frac{3 - x}{2} = \frac{2 y + 1}{4} = \frac{z}{- 3}$ , its direction ratios, and its vector equation.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Which point lies on

\vec{r} = (\hat{i} - 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + \hat{j} - \hat{k})

(2, 1, - 1)

(3, 1, 2)

(1, - 2, 4)

(5, 0, 1)

Show answer

Answer (D) $(5, 0, 1)$

λ = 2

gives

(1 + 4, - 2 + 2, 3 - 2) = (5, 0, 1)

(2, 1, - 1)

is the direction vector, not a point of the line;

(3, 1, 2)

needs

λ = 1

for

x

but then

y

would be

- 1

11.4 Angle between Two Lines

Two lines in space need not meet, but they still have an angle between them: the angle between their directions. Slide a copy of each line, without turning it, until both copies pass through one point $C$ , and measure the angle there (Fig. 11.5). The copies form two angles, $θ$ and $180^{\circ} - θ$ ; by agreement we report the one that is not obtuse.

3/3The copies make angles $θ$ and $180^{\circ} - θ$ ; we take the acute one.

Fig. 11.5The angle between two lines is the angle between their directions, taken acute.NCERT Fig. 11.4 · p. 383

Axes meet at the origin. Line L1, solid blue, and line L2, dashed orange, pass each other without meeting; arrows labelled b1 and b2 show their directions. In the second step dotted copies of both lines are drawn through a single point C, keeping their directions. In the third step the original lines fade, the angle theta between the copies is marked with a solid arc, and the other angle, 180 degrees minus theta, is marked with a dashed arc on the opposite side.

Give the first line direction ratios $a_{1}, b_{1}, c_{1}$ and the second $a_{2}, b_{2}, c_{2}$ , so that ${\vec{b}}_{1} = a_{1} \hat{i} + b_{1} \hat{j} + c_{1} \hat{k}$ and ${\vec{b}}_{2} = a_{2} \hat{i} + b_{2} \hat{j} + c_{2} \hat{k}$ run along them. The dot product gives the angle between these two vectors. Taking the absolute value turns an obtuse answer into its acute supplement, which is the angle between the lines.

Angle between two lines

\cos θ = \frac{∣ {\vec{b}}_{1} \cdot {\vec{b}}_{2} ∣}{∣ {\vec{b}}_{1} ∣ ∣ {\vec{b}}_{2} ∣}

(11.8)

where

${\vec{b}}_{1} \cdot {\vec{b}}_{2}$: $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}$
$∣ {\vec{b}}_{1} ∣$: $\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}}$ , and similarly for $∣ {\vec{b}}_{2} ∣$
$θ$: the acute (or right) angle between the lines

The formula needs only the directions. For lines $\vec{r} = {\vec{a}}_{1} + λ {\vec{b}}_{1}$ and $\vec{r} = {\vec{a}}_{2} + μ {\vec{b}}_{2}$ use ${\vec{b}}_{1}$ and ${\vec{b}}_{2}$ as they stand; for lines in Cartesian form, read the direction ratios from the denominators (after rewriting into the standard form if needed). When the direction cosines are known, the lengths in (11.8) are 1.

\cos θ = ∣ s ∣

(11.9)

where

$s$: $l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}$ , the dot product of the two unit vectors

The sine comes from the cross product in the same way: $\sin θ = \frac{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}{∣ {\vec{b}}_{1} ∣ ∣ {\vec{b}}_{2} ∣}$ . The components of ${\vec{b}}_{1} \times {\vec{b}}_{2}$ are $b_{1} c_{2} - b_{2} c_{1}$ , $c_{1} a_{2} - c_{2} a_{1}$ and $a_{1} b_{2} - a_{2} b_{1}$ . With direction cosines the lengths are again 1.

\sin^{2} θ = u^{2} + v^{2} + w^{2}

(11.10)

where

$u$: $l_{1} m_{2} - l_{2} m_{1}$
$v$: $m_{1} n_{2} - m_{2} n_{1}$
$w$: $n_{1} l_{2} - n_{2} l_{1}$ , the components of the cross product of the two unit vectors

Example 7NCERT Example 7 · p. 384

Angle from vector equations

At what angle are the lines $\vec{r} = (\hat{i} + \hat{j}) + λ (2 \hat{i} - 2 \hat{j} + \hat{k})$ and $\vec{r} = (3 \hat{i} - \hat{k}) + μ (4 \hat{i} + 4 \hat{j} - 7 \hat{k})$ inclined to each other?

Try it yourself first. Then reveal the solution one step at a time.

Example 8NCERT Example 8 · p. 385

Angle from Cartesian equations

Two lines are given in Cartesian form, $\frac{x - 1}{2} = \frac{y + 3}{1} = \frac{z - 4}{- 1}$ and $\frac{x}{1} = \frac{y - 2}{- 2} = \frac{z + 5}{- 3}$ . How large is the angle between them?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

For which value of

k

are lines with direction ratios

k, 2, - 1

and

3, k, 4

perpendicular?

Show answer

Answer 0.8

3 k + 2 k - 4 = 0

, so

k = \frac{4}{5} = 0.8

11.5 Shortest Distance between Two Lines

Two lines in a plane either meet or are parallel. In space there is a third possibility, the one under the metro viaduct: the road below and the track above are not parallel, yet they never meet (Fig. 11.6). Such lines are skew lines. No single plane contains both of them, so they are not coplanar.

3/3A flyover over a road: the lines neither meet nor run parallel. The pillar is perpendicular to both.

Fig. 11.6Two lines in space meet, run parallel, or are skew.NCERT Fig. 11.5 · p. 385

A flat green ground sheet seen at a slant carries a straight road drawn as a solid blue line. In the first step a dashed orange road crosses it at a marked point, labelled they meet, distance 0. In the second step the orange road runs alongside the blue one without crossing, labelled parallel road, same gap everywhere. In the third step the orange line is the edge of a flyover deck raised above the ground and running across the road; a purple vertical pillar joins the deck to the road with right-angle marks at both ends, and the note reads never meet, not parallel.

Definition

Skew lines

A pair of lines in space with no common point and no common direction. No single plane can hold them both.

ExampleA straight flyover deck and the road it crosses.

The shortest distance between two lines is the length of the shortest segment that has one end on each line. For lines that meet it is 0. For parallel lines it is the perpendicular gap between them, the same wherever you measure it. For skew lines there is exactly one shortest segment, and it meets both lines at right angles. It is called their common perpendicular; in the flyover picture it is the pillar.

Try this

Two pencils and a ruler

You’ll need

two long pencils
a lump of clay or an eraser
a ruler

Predict first

Where along the pencils do you expect the smallest gap to be?

At the ends of the pencilsWhere a ruler between them is square to both pencilsAnywhere: the gap is the same all alongWhere the pencils look as if they cross

Lay one pencil flat on the table.
Fix the second pencil higher up with the clay so that it crosses over the first without touching it and without being parallel to it.
Hold the ruler between the pencils at several places and note the gap each time.
Find the place where the gap is smallest, and check with a set-square whether the ruler there is square to both pencils.

11.5.1 Distance between two skew lines

DerivationShortest distance between skew lines

Let the lines be $l_{1} : \vec{r} = {\vec{a}}_{1} + λ {\vec{b}}_{1}$ and $l_{2} : \vec{r} = {\vec{a}}_{2} + μ {\vec{b}}_{2}$ , passing through $A_{1}$ and $A_{2}$ . Let $M N$ be their common perpendicular, with $M$ on $l_{1}$ and $N$ on $l_{2}$ , and let $d = M N$ (Fig. 11.7).

$M N$ is perpendicular to both ${\vec{b}}_{1}$ and ${\vec{b}}_{2}$ , so it lies along their cross product. With the unit vector $\hat{n} = \frac{{\vec{b}}_{1} \times {\vec{b}}_{2}}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}$ we get $\vec{M N} = \pm d \hat{n}$ .

Now go from $A_{1}$ to $A_{2}$ the long way round: along $l_{1}$ to $M$ , across to $N$ , then along $l_{2}$ to $A_{2}$ . In vectors, ${\vec{a}}_{2} - {\vec{a}}_{1} = s {\vec{b}}_{1} \pm d \hat{n} + t {\vec{b}}_{2}$ for some numbers $s$ and $t$ .

Now take the dot product of both sides with $\hat{n}$ and remember that $\hat{n}$ is perpendicular to ${\vec{b}}_{1}$ and to ${\vec{b}}_{2}$ as well, so the two along-the-line pieces drop out and only $\pm d$ is left. So $d$ is the length of the projection of $\vec{A_{1} A_{2}}$ on $\hat{n}$ . Writing $\hat{n}$ out in full gives the formula below.

Skew lines

d = \frac{∣ \vec{N} \cdot ({\vec{a}}_{2} - {\vec{a}}_{1}) ∣}{∣ \vec{N} ∣}

(11.11)

where

$\vec{N}$: the cross product ${\vec{b}}_{1} \times {\vec{b}}_{2}$ of the two directions
${\vec{a}}_{1}, {\vec{a}}_{2}$: position vectors of points $A_{1}$ on $l_{1}$ and $A_{2}$ on $l_{2}$

In full, $d$ is the absolute value of $({\vec{b}}_{1} \times {\vec{b}}_{2}) \cdot ({\vec{a}}_{2} - {\vec{a}}_{1})$ divided by the length $∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣$ of the cross product.

4/4Along $l_{1}$ , across $M N$ , along $l_{2}$ : only the middle stretch survives a dot product with $\hat{n}$ .

Fig. 11.7The common perpendicular

M N

is the projection of

\vec{A_{1} A_{2}}

{\vec{b}}_{1} \times {\vec{b}}_{2}

.NCERT Fig. 11.6 · p. 386

Small axes sit at the lower left. A solid blue line l1 runs low across the picture through the point A1, with an arrow b1 along it, and a dashed orange line l2 runs higher up in a different direction through A2, with an arrow b2. In the second step a purple arrow joins A1 to A2 and is labelled a2 minus a1. In the third step the common perpendicular MN is drawn between the lines with right-angle marks at M on l1 and N on l2, labelled d, and the note says n hat is parallel to b1 cross b2. In the last step a broad green path runs from A1 along l1 to M, up to N and along l2 to A2, and the label reads d equals the absolute value of (a2 minus a1) dot n hat.

Formula (11.11) also classifies the pair. If ${\vec{b}}_{1} \times {\vec{b}}_{2} = \vec{0}$ the lines are parallel and the formula does not apply (see Section 11.5.2). Otherwise a zero numerator means $d = 0$ : the lines meet. A non-zero numerator means they are skew.

Cartesian form. For $l_{1}$ through $(x_{1}, y_{1}, z_{1})$ with direction ratios $a_{1}, b_{1}, c_{1}$ , and $l_{2}$ through $(x_{2}, y_{2}, z_{2})$ with $a_{2}, b_{2}, c_{2}$ , the numerator of (11.11) is a scalar triple product, which is a determinant. Writing $Δ x = x_{2} - x_{1}$ and so on:

D = ∣ \begin{array}{ccc} Δ x & Δ y & Δ z \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} ∣

(11.12)

d = \frac{∣ D ∣}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}

(11.13)

where

$∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣$: the square root of the sum of the squares of $b_{1} c_{2} - b_{2} c_{1}$ , $c_{1} a_{2} - c_{2} a_{1}$ and $a_{1} b_{2} - a_{2} b_{1}$

Example 9NCERT Example 9 · p. 388

Shortest distance between skew lines

How close do the lines $\vec{r} = (2 \hat{i} - \hat{j}) + λ (\hat{i} + 2 \hat{j} + 2 \hat{k})$ and $\vec{r} = (5 \hat{i} - 2 \hat{j} + 4 \hat{k}) + μ (2 \hat{i} + \hat{j} - 2 \hat{k})$ come to each other?

Try it yourself first. Then reveal the solution one step at a time.

Example

Zero distance: the lines meet

Show that the lines $\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - \hat{j} + 2 \hat{k})$ and $\vec{r} = (- \hat{j} + 7 \hat{k}) + μ (\hat{i} + \hat{j} - \hat{k})$ intersect, and find the common point.

Try it yourself first. Then reveal the solution one step at a time.

11.5.2 Distance between parallel lines

When two lines are parallel we may use one direction vector $\vec{b}$ for both, rescaling one equation if necessary: $l_{1} : \vec{r} = {\vec{a}}_{1} + λ \vec{b}$ and $l_{2} : \vec{r} = {\vec{a}}_{2} + μ \vec{b}$ . They lie in a common plane, and the gap between them does not change along their length.

DerivationDistance between parallel lines

Draw the parallelogram with one side $\vec{b}$ laid along $l_{1}$ from $A_{1}$ and the other side $\vec{A_{1} A_{2}} = {\vec{a}}_{2} - {\vec{a}}_{1}$ (Fig. 11.8). From Chapter 10, its area is $∣ \vec{b} \times ({\vec{a}}_{2} - {\vec{a}}_{1}) ∣$ .

Its base is $∣ \vec{b} ∣$ , and its height is the perpendicular gap $d$ between the two lines, because the opposite side lies along $l_{2}$ . So its area is also $∣ \vec{b} ∣ d$ . Equating the two expressions gives the distance.

Parallel lines

d = \frac{∣ \vec{b} \times ({\vec{a}}_{2} - {\vec{a}}_{1}) ∣}{∣ \vec{b} ∣}

(11.14)

3/3Its height is the gap $d$ , so the area is also $∣ \vec{b} ∣ d$ .

Fig. 11.8One parallelogram, two ways to find its area.NCERT Fig. 11.7 · p. 387

Two parallel slanting lines, l1 solid blue below and l2 dashed orange above. Point A1 on l1 carries an arrow b along l1, and a purple arrow runs from A1 to the point A2 on l2. In the second step the parallelogram with sides b and A1A2 is shaded and labelled area equals the absolute value of b cross (a2 minus a1). In the third step a dashed perpendicular drops from A2 to its foot F on l1, with a right-angle mark, labelled d, and the label changes to area equals the absolute value of b times d.

Example 10NCERT Example 10 · p. 388

Distance between parallel lines

Show that the lines $\vec{r} = (\hat{i} + \hat{j}) + λ (\hat{i} - 2 \hat{j} + 2 \hat{k})$ and $\vec{r} = (4 \hat{i} + 2 \hat{j} - 2 \hat{k}) + μ (2 \hat{i} - 4 \hat{j} + 4 \hat{k})$ are parallel, and find the gap between them.

Try it yourself first. Then reveal the solution one step at a time.

InteractiveShortest distance explorer

A three-dimensional view of two lines, one solid blue and one dashed orange, with four ready-made pairs: a road and a flyover, the skew lines of Example 9, the meeting lines of the second example, and the parallel lines of Example 10. A point P on the first line and a point Q on the second are set by two sliders, lambda and mu, and a purple segment joins them. A panel shows the coordinates of P and Q, the length PQ, and the dot products of PQ with the two direction vectors, alongside the distance given by the chapter formula. A Closest points button moves P and Q to the ends of the common perpendicular, where both dot products are zero. Dragging the picture or using two view sliders turns and tilts it. Nothing moves until the learner acts.

Fig. 11.9Pick a pair of lines, then move a point on each line and try to make the gap as small as you can. Drag the picture to turn it.

In the explorer, choose Example 9 and slide $λ$ and $μ$ until $P Q$ is as short as you can make it. The smallest value you can reach is 4, and there both dot products $\vec{P Q} \cdot {\vec{b}}_{1}$ and $\vec{P Q} \cdot {\vec{b}}_{2}$ are 0: the shortest segment is perpendicular to both lines. Try the same with the parallel pair, where every $λ$ has its own closest $μ$ .

Quick check

1 of 4

The lines

\vec{r} = λ (\hat{i} + \hat{j})

and

\vec{r} = 3 \hat{k} + μ (\hat{i} - \hat{j})

are

skew, 3 units apartintersectingparallelskew,

\sqrt{2}

units apart

Show answer

Answer (A) skew, 3 units apart

The directions are not proportional, and the lines lie in the parallel planes

z = 0

and

z = 3

, so they never meet: they are skew. By (11.11),

{\vec{b}}_{1} \times {\vec{b}}_{2} = - 2 \hat{k}

and its dot product with

3 \hat{k}

- 6

, so

d = \frac{6}{2} = 3

Practice · 17 questions

Practice 11.2: questions with answers

ProofEasyNCERT Exercise 11.2, Q1
Lines $L_{1}$ , $L_{2}$ , $L_{3}$ have direction cosines $\frac{2}{3}, \frac{2}{3}, \frac{1}{3}$ ; $- \frac{2}{3}, \frac{1}{3}, \frac{2}{3}$ and $\frac{1}{3}, - \frac{2}{3}, \frac{2}{3}$ . Prove that each is perpendicular to the other two.
Show solution
First and second: $\frac{- 4 + 2 + 2}{9} = 0$ .
First and third: $\frac{2 - 4 + 2}{9} = 0$ .
Second and third: $\frac{- 2 - 2 + 4}{9} = 0$ .
Every pair is perpendicular. (Each set also has squares adding to 1, so these really are direction cosines.)
ProofEasyNCERT Exercise 11.2, Q2
Line $A B$ joins $A (2, 0, - 1)$ to $B (4, 3, 1)$ , and line $C D$ joins $C (1, 1, 5)$ to $D (5, - 1, 4)$ . Prove that $A B ⊥ C D$ .
Show solution
Direction ratios are $2, 3, 2$ for the first line and $4, - 2, - 1$ for the second.
Their sum of products is $8 - 6 - 2 = 0$ , so the lines are perpendicular.
ProofEasyNCERT Exercise 11.2, Q3
Is $E F$ parallel to $G H$ for the points $E (1, 5, - 2)$ , $F (4, 3, 2)$ , $G (0, - 1, 6)$ and $H (- 6, 3, - 2)$ given here? Prove your answer.
Show solution
Direction ratios are $3, - 2, 4$ for the first line and $- 6, 4, - 8$ for the second.
The second set is $- 2$ times the first, so the lines are parallel.
Short answerEasyNCERT Exercise 11.2, Q4
A laser beam starts at $(- 2, 0, 5)$ and travels in the direction $\hat{i} - 4 \hat{j} + 3 \hat{k}$ . Write the vector equation of the line it travels along.
Show solution
Answer $\vec{r} = (- 2 \hat{i} + 5 \hat{k}) + λ (\hat{i} - 4 \hat{j} + 3 \hat{k})$
Here the point has position vector $\vec{a} = - 2 \hat{i} + 5 \hat{k}$ and the direction $\vec{b} = \hat{i} - 4 \hat{j} + 3 \hat{k}$ is given.
So the line is $\vec{r} = (- 2 \hat{i} + 5 \hat{k}) + λ (\hat{i} - 4 \hat{j} + 3 \hat{k})$ .
Short answerEasyNCERT Exercise 11.2, Q5
Give both forms, vector and Cartesian, for the line that starts from the point $3 \hat{i} + \hat{j} - 2 \hat{k}$ (a position vector) and runs along $2 \hat{i} - \hat{j} + 3 \hat{k}$ .
Show solution
Answer $\vec{r} = (3 \hat{i} + \hat{j} - 2 \hat{k}) + λ (2 \hat{i} - \hat{j} + 3 \hat{k})$ ; $\frac{x - 3}{2} = \frac{y - 1}{- 1} = \frac{z + 2}{3}$
Vector form: $\vec{r} = (3 \hat{i} + \hat{j} - 2 \hat{k}) + λ (2 \hat{i} - \hat{j} + 3 \hat{k})$ .
Cartesian form, from the point $(3, 1, - 2)$ and ratios $2, - 1, 3$ : $\frac{x - 3}{2} = \frac{y - 1}{- 1} = \frac{z + 2}{3}$
Short answerMediumNCERT Exercise 11.2, Q6
The line $\frac{2 - x}{4} = \frac{y + 1}{3} = \frac{z - 7}{2}$ is slid, without turning, until it passes through $(1, - 3, 4)$ . Write the Cartesian equation of the new line.
Show solution
Answer $\frac{x - 1}{- 4} = \frac{y + 3}{3} = \frac{z - 4}{2}$
$\frac{2 - x}{4} = \frac{x - 2}{- 4}$ , so the given line has direction ratios $- 4, 3, 2$ .
A parallel line has the same ratios, so through $(1, - 3, 4)$ it is $\frac{x - 1}{- 4} = \frac{y + 3}{3} = \frac{z - 4}{2}$
Short answerEasyNCERT Exercise 11.2, Q7
Write the line $\frac{x + 2}{5} = \frac{y - 7}{- 2} = \frac{z + 1}{4}$ in vector form.
Show solution
Answer $\vec{r} = (- 2 \hat{i} + 7 \hat{j} - \hat{k}) + λ (5 \hat{i} - 2 \hat{j} + 4 \hat{k})$
The line passes through $(- 2, 7, - 1)$ and has direction ratios $5, - 2, 4$ .
So $\vec{r} = (- 2 \hat{i} + 7 \hat{j} - \hat{k}) + λ (5 \hat{i} - 2 \hat{j} + 4 \hat{k})$ .
8
Short answerMediumNCERT Exercise 11.2, Q8
What angle does each pair of lines make?
1. $\vec{r} = (\hat{i} + 2 \hat{j}) + λ (\hat{i} + 4 \hat{j} + 8 \hat{k})$ and $\vec{r} = (3 \hat{j} - \hat{k}) + μ (2 \hat{i} - \hat{j} + 2 \hat{k})$
  Show solution
  Answer $\cos^{- 1} \frac{14}{27} \approx {58.8}^{\circ}$
  ${\vec{b}}_{1} \cdot {\vec{b}}_{2} = 2 - 4 + 16 = 14$ , $∣ {\vec{b}}_{1} ∣ = 9$ and $∣ {\vec{b}}_{2} ∣ = 3$ .
  So $\cos θ = \frac{14}{27}$ and $θ \approx {58.8}^{\circ}$ .
2. $\vec{r} = (\hat{i} - \hat{j} + \hat{k}) + λ (2 \hat{i} + \hat{j} - 2 \hat{k})$ and $\vec{r} = (2 \hat{i} + 3 \hat{k}) + μ (6 \hat{i} - 2 \hat{j} + 3 \hat{k})$
  Show solution
  Answer $\cos^{- 1} \frac{4}{21} \approx {79.0}^{\circ}$
  ${\vec{b}}_{1} \cdot {\vec{b}}_{2} = 12 - 2 - 6 = 4$ , $∣ {\vec{b}}_{1} ∣ = 3$ and $∣ {\vec{b}}_{2} ∣ = 7$ .
  So $\cos θ = \frac{4}{21}$ and $θ \approx {79.0}^{\circ}$ .
9
Short answerMediumNCERT Exercise 11.2, Q9
These pairs are in Cartesian form. Find the angle in each case.
1. $\frac{x - 1}{3} = \frac{y + 2}{- 1} = \frac{z}{2}$ and $\frac{x + 3}{1} = \frac{y - 1}{2} = \frac{z - 4}{- 4}$
  Show solution
  Answer $\cos^{- 1} \frac{1}{\sqrt{6}} \approx {65.9}^{\circ}$
  Direction ratios $3, - 1, 2$ and $1, 2, - 4$ . The sum of products is $3 - 2 - 8 = - 7$ .
  The lengths are $\sqrt{14}$ and $\sqrt{21}$ , whose product is $7 \sqrt{6}$ .
  So $\cos θ = \frac{7}{7 \sqrt{6}} = \frac{1}{\sqrt{6}}$ and $θ \approx {65.9}^{\circ}$ .
2. $\frac{x}{1} = \frac{y - 3}{2} = \frac{z + 1}{2}$ and $\frac{x - 2}{4} = \frac{y + 5}{4} = \frac{z - 1}{7}$
  Show solution
  Answer $\cos^{- 1} \frac{26}{27} \approx {15.6}^{\circ}$
  Direction ratios $1, 2, 2$ (length 3) and $4, 4, 7$ (length 9).
  The sum of products is $4 + 8 + 14 = 26$ , so $\cos θ = \frac{26}{27}$ and $θ \approx {15.6}^{\circ}$ .
Short answerHardNCERT Exercise 11.2, Q10
Which value of $p$ makes the lines $\frac{2 - x}{2} = \frac{3 y - 6}{p} = \frac{z + 1}{4}$ and $\frac{1 - x}{p} = \frac{y + 2}{6} = \frac{3 - z}{2}$ perpendicular?
Show solution
Answer $p = 2$
First line: $\frac{x - 2}{- 2} = \frac{y - 2}{p / 3} = \frac{z + 1}{4}$ , with direction ratios $- 2, \frac{p}{3}, 4$ .
Second line: $\frac{x - 1}{- p} = \frac{y + 2}{6} = \frac{z - 3}{- 2}$ , with direction ratios $- p, 6, - 2$ .
Perpendicular: $2 p + 2 p - 8 = 0$ , so $p = 2$ .
ProofEasyNCERT Exercise 11.2, Q11
Are the lines $\frac{x - 3}{4} = \frac{y + 1}{- 1} = \frac{z - 2}{- 2}$ and $\frac{x}{2} = \frac{y - 5}{6} = \frac{z}{1}$ perpendicular? Justify.
Show solution
The direction ratios are $4, - 1, - 2$ and $2, 6, 1$ .
Their sum of products is $8 - 6 - 2 = 0$ , so the lines are perpendicular.
NumericalMediumNCERT Exercise 11.2, Q12
How close do the lines $\vec{r} = (\hat{i} + 2 \hat{k}) + λ (\hat{i} + 2 \hat{j} - \hat{k})$ and $\vec{r} = (3 \hat{j} - \hat{k}) + μ (2 \hat{i} + \hat{j} + \hat{k})$ come to each other?
Show solution
Answer $\frac{1}{\sqrt{3}} \approx 0.577$ units
${\vec{a}}_{2} - {\vec{a}}_{1} = - \hat{i} + 3 \hat{j} - 3 \hat{k}$ .
${\vec{b}}_{1} \times {\vec{b}}_{2} = 3 \hat{i} - 3 \hat{j} - 3 \hat{k}$ , of length $\sqrt{27} = 3 \sqrt{3}$ .
The dot product is $- 3 - 9 + 9 = - 3$ .
So $d = \frac{3}{3 \sqrt{3}} = \frac{1}{\sqrt{3}} \approx 0.577$ .
NumericalMediumNCERT Exercise 11.2, Q13
Compute the gap, at their closest, between the lines $\frac{x - 2}{3} = \frac{y + 1}{- 1} = \frac{z - 4}{2}$ and $\frac{x + 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{- 1}$ .
Show solution
Answer $\frac{25}{\sqrt{83}} \approx 2.74$ units
The points are $(2, - 1, 4)$ and $(- 1, 2, - 3)$ , so $Δ x, Δ y, Δ z = - 3, 3, - 7$ . The direction ratios are $3, - 1, 2$ and $1, 2, - 1$ .
Expanding the determinant along its first row gives $D = - 3 (1 - 4) - 3 (- 3 - 2) - 7 (6 + 1) = 9 + 15 - 49 = - 25$ .
The cross product of the directions has components $- 3, 5, 7$ , so its length is $\sqrt{83}$ .
Hence $d = \frac{25}{\sqrt{83}} \approx 2.74$ .
NumericalMediumNCERT Exercise 11.2, Q14
Find the length of the common perpendicular of $\vec{r} = (\hat{i} + \hat{j} + \hat{k}) + λ (2 \hat{i} + \hat{j} - 2 \hat{k})$ and $\vec{r} = (4 \hat{i} - \hat{j} + 2 \hat{k}) + μ (2 \hat{i} - 2 \hat{j} + \hat{k})$ .
Show solution
Answer $\frac{1}{3}$ unit
${\vec{a}}_{2} - {\vec{a}}_{1} = 3 \hat{i} - 2 \hat{j} + \hat{k}$ .
${\vec{b}}_{1} \times {\vec{b}}_{2} = - 3 \hat{i} - 6 \hat{j} - 6 \hat{k}$ , of length 9.
The dot product is $- 9 + 12 - 6 = - 3$ , so $d = \frac{3}{9} = \frac{1}{3}$ .
NumericalHardNCERT Exercise 11.2, Q15
Two lines are given in parametric form: $\vec{r} = (2 + t) \hat{i} + (1 - 2 t) \hat{j} + 3 t \hat{k}$ and $\vec{r} = (2 s - 1) \hat{i} + (s + 4) \hat{j} + (1 - s) \hat{k}$ . How far apart are they at their closest?
Show solution
Answer $\frac{29}{5 \sqrt{3}} = \frac{29 \sqrt{3}}{15} \approx 3.35$ units
The first line is $(2 \hat{i} + \hat{j}) + t (\hat{i} - 2 \hat{j} + 3 \hat{k})$ and the second is $(- \hat{i} + 4 \hat{j} + \hat{k}) + s (2 \hat{i} + \hat{j} - \hat{k})$ .
So ${\vec{a}}_{2} - {\vec{a}}_{1} = - 3 \hat{i} + 3 \hat{j} + \hat{k}$ and ${\vec{b}}_{1} \times {\vec{b}}_{2} = - \hat{i} + 7 \hat{j} + 5 \hat{k}$ , of length $\sqrt{75} = 5 \sqrt{3}$ .
The dot product is $3 + 21 + 5 = 29$ , so $d = \frac{29}{5 \sqrt{3}} = \frac{29 \sqrt{3}}{15} \approx 3.35$ .
Quick answerMedium
The lines $\vec{r} = (\hat{i} + \hat{j}) + λ (2 \hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = (3 \hat{i} + 2 \hat{k}) + μ (- 4 \hat{i} + 2 \hat{j} - 2 \hat{k})$ are
the same lineintersecting linesskew linesdistinct parallel lines
Show answer
Answer (D) distinct parallel lines
The second direction is $- 2$ times the first, so the lines are parallel. The join $2 \hat{i} - \hat{j} + 2 \hat{k}$ is not along $2 \hat{i} - \hat{j} + \hat{k}$ , so they are not the same line; by (11.14) they are $\sqrt{\frac{5}{6}} \approx 0.91$ apart.
Quick answerMedium
Find the distance between the parallel lines $\vec{r} = λ (2 \hat{i} - 2 \hat{j} + \hat{k})$ and $\vec{r} = (3 \hat{i} + 3 \hat{j}) + μ (2 \hat{i} - 2 \hat{j} + \hat{k})$ , to two decimal places.
Show answer
Answer 4.24
$\vec{b} \times (3 \hat{i} + 3 \hat{j}) = - 3 \hat{i} + 3 \hat{j} + 12 \hat{k}$ , of length $\sqrt{162} = 9 \sqrt{2}$ , so $d = 3 \sqrt{2} \approx 4.24$ .

Recap

Three Dimensional Geometry: key points and summary

A directed line makes angles $α, β, γ$ with the positive axes; their cosines are its direction cosines $l, m, n$ and reversing the arrow flips all three signs — Section 11.2
Every set of direction cosines satisfies $l^{2} + m^{2} + n^{2} = 1$ — (11.1)
Scaling the direction cosines by any non-zero number gives direction ratios; to go back, divide $a, b, c$ by $\sqrt{a^{2} + b^{2} + c^{2}}$ — (11.2)
For the line through two points the differences of coordinates are direction ratios, and dividing by the distance gives the cosines — Section 11.2.1
The line through the point $\vec{a}$ parallel to $\vec{b}$ is $\vec{r} = \vec{a} + λ \vec{b}$ ; through the points $\vec{a}$ and $\vec{b}$ it is $\vec{r} = \vec{a} + λ (\vec{b} - \vec{a})$ — (11.4)
In coordinates the same line is $\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c}$ — (11.6)
The angle between two lines is the acute angle between their directions, $\cos θ = \frac{∣ {\vec{b}}_{1} \cdot {\vec{b}}_{2} ∣}{∣ {\vec{b}}_{1} ∣ ∣ {\vec{b}}_{2} ∣}$ — Section 11.4
Lines are perpendicular when $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$ and parallel when their direction ratios are proportional — the test
Two lines in space meet, run parallel, or are skew (neither); for skew lines the shortest gap lies along the common perpendicular — Section 11.5
Skew lines: $d = \frac{∣ ({\vec{b}}_{1} \times {\vec{b}}_{2}) \cdot ({\vec{a}}_{2} - {\vec{a}}_{1}) ∣}{∣ {\vec{b}}_{1} \times {\vec{b}}_{2} ∣}$ — (11.11)
Parallel lines: $d = \frac{∣ \vec{b} \times ({\vec{a}}_{2} - {\vec{a}}_{1}) ∣}{∣ \vec{b} ∣}$ — (11.14)

Flash cards · 6

Formulas to remember

1 / 6

Miscellaneous Exercise on Chapter 11

Practice · 6 questions

Practice questions with answers (NCERT “Miscellaneous exercise”)

Short answerMediumNCERT Miscellaneous Exercise, Q1
For real numbers $p, q, r$ , two lines have direction ratios $p + q, q + r, r + p$ and $p - q, q - r, r - p$ (neither set all zero). Find the angle between them.
Show solution
Answer $90^{\circ}$
Multiply matching ratios: each product is a difference of two squares, $(p + q) (p - q) = p^{2} - q^{2}$ , and likewise for the other two.
The sum is $(p^{2} - q^{2}) + (q^{2} - r^{2}) + (r^{2} - p^{2}) = 0$ , so $\cos θ = 0$ and the lines are perpendicular, whatever $p$ , $q$ , $r$ are.
Short answerEasyNCERT Miscellaneous Exercise, Q2
Write the vector and Cartesian equations of the line through $(0, 2, - 3)$ parallel to the $z$ -axis.
Show solution
Answer $\vec{r} = (2 \hat{j} - 3 \hat{k}) + λ \hat{k}$ ; $\frac{x}{0} = \frac{y - 2}{0} = \frac{z + 3}{1}$ , that is, $x = 0$ and $y = 2$
The direction is $\hat{k}$ , so $\vec{r} = (2 \hat{j} - 3 \hat{k}) + λ \hat{k}$ .
In Cartesian form $\frac{x}{0} = \frac{y - 2}{0} = \frac{z + 3}{1}$ , where the zero denominators mean $x = 0$ and $y = 2$ for every point, while $z$ is free.
Short answerMediumNCERT Miscellaneous Exercise, Q3
The lines $\frac{x + 2}{k} = \frac{y - 1}{3} = \frac{z + 4}{- 2}$ and $\frac{x - 5}{2} = \frac{y + 3}{k} = \frac{z - 1}{4}$ are known to be perpendicular. What is $k$ ?
Show solution
Answer $k = \frac{8}{5}$
Perpendicular: $2 k + 3 k - 8 = 0$ , so $5 k = 8$ and $k = \frac{8}{5}$ .
NumericalMediumNCERT Miscellaneous Exercise, Q4
Find how far apart these lines are at their closest: $\vec{r} = (3 \hat{i} - \hat{k}) + λ (2 \hat{i} + 2 \hat{j} + \hat{k})$ and $\vec{r} = (4 \hat{i} - 2 \hat{j} + \hat{k}) + μ (- 2 \hat{i} + \hat{j} + 2 \hat{k})$ .
Show solution
Answer3 units
${\vec{a}}_{2} - {\vec{a}}_{1} = \hat{i} - 2 \hat{j} + 2 \hat{k}$ .
${\vec{b}}_{1} \times {\vec{b}}_{2} = 3 \hat{i} - 6 \hat{j} + 6 \hat{k}$ , of length 9.
The dot product is $3 + 12 + 12 = 27$ , so $d = \frac{27}{9} = 3$ .
Short answerHardNCERT Miscellaneous Exercise, Q5
A line through $(2, - 1, 3)$ must be at right angles to two given lines, $\frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 4}{- 2}$ and $\vec{r} = (\hat{i} + 4 \hat{j}) + μ (3 \hat{i} + \hat{k})$ . Write its vector equation.
Show solution
Answer $\vec{r} = (2 \hat{i} - \hat{j} + 3 \hat{k}) + λ (2 \hat{i} - 7 \hat{j} - 6 \hat{k})$
The directions are $\hat{i} + 2 \hat{j} - 2 \hat{k}$ and $3 \hat{i} + \hat{k}$ .
Their cross product is the determinant with rows $\hat{i}, \hat{j}, \hat{k}$ ; $1, 2, - 2$ ; $3, 0, 1$ .
This cross product is $2 \hat{i} - 7 \hat{j} - 6 \hat{k}$ . Check: its dot products with the two directions are $2 - 14 + 12 = 0$ and $6 - 6 = 0$ .
So the line is $\vec{r} = (2 \hat{i} - \hat{j} + 3 \hat{k}) + λ (2 \hat{i} - 7 \hat{j} - 6 \hat{k})$ .
Quick answerMedium
If ${\vec{b}}_{1} \times {\vec{b}}_{2} \neq \vec{0}$ and $({\vec{b}}_{1} \times {\vec{b}}_{2}) \cdot ({\vec{a}}_{2} - {\vec{a}}_{1}) = 0$ , the lines $\vec{r} = {\vec{a}}_{1} + λ {\vec{b}}_{1}$ and $\vec{r} = {\vec{a}}_{2} + μ {\vec{b}}_{2}$ intersect.
TrueFalse
Show answer
Answer True
The directions are not parallel, and (11.11) gives $d = 0$ , so the lines share a point.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 11.1, Q1	Practice 11.1 · Q1
NCERT Exercise 11.1, Q2	Practice 11.1 · Q2
NCERT Exercise 11.1, Q3	Practice 11.1 · Q3
NCERT Exercise 11.1, Q4	Practice 11.1 · Q4
NCERT Exercise 11.1, Q5	Practice 11.1 · Q5
NCERT Exercise 11.2, Q1	Practice 11.2 · Q1
NCERT Exercise 11.2, Q2	Practice 11.2 · Q2
NCERT Exercise 11.2, Q3	Practice 11.2 · Q3
NCERT Exercise 11.2, Q4	Practice 11.2 · Q4
NCERT Exercise 11.2, Q5	Practice 11.2 · Q5
NCERT Exercise 11.2, Q6	Practice 11.2 · Q6
NCERT Exercise 11.2, Q7	Practice 11.2 · Q7
NCERT Exercise 11.2, Q8	Practice 11.2 · Q8
NCERT Exercise 11.2, Q9	Practice 11.2 · Q9
NCERT Exercise 11.2, Q10	Practice 11.2 · Q10
NCERT Exercise 11.2, Q11	Practice 11.2 · Q11
NCERT Exercise 11.2, Q12	Practice 11.2 · Q12
NCERT Exercise 11.2, Q13	Practice 11.2 · Q13
NCERT Exercise 11.2, Q14	Practice 11.2 · Q14
NCERT Exercise 11.2, Q15	Practice 11.2 · Q15
NCERT Miscellaneous Exercise, Q1	Miscellaneous exercise · Q1
NCERT Miscellaneous Exercise, Q2	Miscellaneous exercise · Q2
NCERT Miscellaneous Exercise, Q3	Miscellaneous exercise · Q3
NCERT Miscellaneous Exercise, Q4	Miscellaneous exercise · Q4
NCERT Miscellaneous Exercise, Q5	Miscellaneous exercise · Q5

Quick check: MCQs on Three Dimensional Geometry

11.1 Introduction3 questions
Direction ratios3 questions
11.2.1 Direction cosines of a line passing through two points2 questions
11.3.1 Equation of a line through a given point and parallel to a given vector3 questions
11.4 Angle between Two Lines3 questions
11.5.2 Distance between parallel lines4 questions

Glossary · 11

Key terms (English–Hindi)

Cartesian equation of a lineरेखा का कार्तीय समीकरणrekha ka kaarteey sameekaran: The line written in coordinates, $\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c}$ , through $(x_{1}, y_{1}, z_{1})$ with direction ratios $a, b, c$ .First seen in the chapter →
collinearसंरेखsanrekh: Lying on one straight line.First seen in the chapter →
common perpendicularउभयनिष्ठ लंबubhayanishth lamb: The one segment that meets two skew lines and is perpendicular to both; its length is their shortest distance.First seen in the chapter →
coplanarसमतलीयsamtaleey: Lying in one plane. Two lines are coplanar exactly when they meet or are parallel.First seen in the chapter →

directed lineदिष्ट रेखाdisht rekha: A line together with a chosen one of its two directions, shown by an arrow.First seen in the chapter →
direction anglesदिक्-कोणdik-kon: The angles $α, β, γ$ (each from $0^{\circ}$ to $180^{\circ}$ ) that a directed line makes with the positive $x$ -, $y$ - and $z$ -axes.First seen in the chapter →
direction cosinesदिक्-कोसाइनdik-cosine: The cosines $l, m, n$ of the direction angles; they satisfy $l^{2} + m^{2} + n^{2} = 1$ .First seen in the chapter →
direction ratiosदिक्-अनुपातdik-anupaat: Any three numbers, not all zero, in the same ratio as the direction cosines of a line.First seen in the chapter →

parameterप्राचलpraachal: The real number $λ$ in $\vec{r} = \vec{a} + λ \vec{b}$ ; each value picks out one point of the line.First seen in the chapter →

shortest distanceन्यूनतम दूरीnyoontam doori: The length of the shortest segment joining a point of one line to a point of the other.First seen in the chapter →
skew linesविषमतलीय रेखाएँvishamtaleey rekhaayen: A pair of lines in space that never meet and are not parallel; no plane holds both.First seen in the chapter →

11.1 Introduction

11.2 Direction Cosines and Direction Ratios of a Line

Direction ratios

11.2.1 Direction cosines of a line passing through two points

11.3 Equation of a Line in Space

11.3.1 Equation of a line through a given point and parallel to a given vector

11.4 Angle between Two Lines

11.5 Shortest Distance between Two Lines

11.5.1 Distance between two skew lines

11.5.2 Distance between parallel lines

Miscellaneous Exercise on Chapter 11

Find your NCERT exercise question

Quick check: MCQs on Three Dimensional Geometry

You reached the end