Lines in Space

Q: What is the difference between direction cosines and direction ratios?

Direction cosines l, m, n are the cosines of the angles a line makes with the axes, so they satisfy l² + m² + n² = 1 and are unique apart from sign. Direction ratios a, b, c are any three numbers proportional to them, so a line has infinitely many sets of direction ratios.

Q: Why do direction cosines satisfy l² + m² + n² = 1?

For a point P(x, y, z) at distance r from the origin on the line, right triangles with the axes give x = lr, y = mr and z = nr. Substituting in x² + y² + z² = r² gives r²(l² + m² + n²) = r², so the sum of squares is 1.

Q: How do you find the angle between two lines in 3D?

Use cos θ = l1l2 + m1m2 + n1n2 with direction cosines, or divide a1a2 + b1b2 + c1c2 by the product of the magnitudes when you have direction ratios. Take the modulus for the acute angle. The same formula works for skew lines because only the directions matter.

Q: How do you find the shortest distance between two skew lines?

Write both lines in vector form and use d = |(a2 − a1) · (b1 × b2)| / |b1 × b2|. If you also need the line of shortest distance, take general points on both lines, make their join perpendicular to both directions and solve for the two parameters.

MathsThree Dimensional GeometryFor JEE aspirants

A straight line in space is fixed by a point on it and a direction. That direction is measured by direction cosines $l, m, n$ or direction ratios $a, b, c$ . This page on lines in space covers how to find them, the angle between two lines, projection of a segment, the symmetric and vector equations of a line, the foot and image of a point, the line of intersection of two planes, and the shortest distance between skew lines. Lines in space are tested every year in JEE Main and JEE Advanced.

Key Formulas: Quick Reference

Direction cosines: $l = cos α$ , $m = cos β$ , $n = cos γ$ , with $l^{2} + m^{2} + n^{2} = 1$
From direction ratios: $l = \frac{a}{a ^{2} + b ^{2} + c ^{2}}$ (same pattern for $m$ and $n$ )
Direction ratios of $P Q$ : $x_{2} - x_{1}$ , $y_{2} - y_{1}$ , $z_{2} - z_{1}$
Angle between lines: $cos θ = l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}$
Perpendicular: $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$ ; parallel: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$
Projection of $P Q$ on a line: $∣ l (x_{2} - x_{1}) + m (y_{2} - y_{1}) + n (z_{2} - z_{1}) ∣$
Symmetric form: $\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c}$ ; vector form: $r = a + λ b$
Line of intersection of two planes is parallel to $n_{1} \times n_{2}$
Skew lines: $d = \frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣}$
Parallel lines: $d = \frac{∣ ( a _{2} - a _{1} ) \times b ∣}{∣ b ∣}$

1. Direction Cosines of a Line

If $α$ , $β$ , $γ$ are the angles which a directed line makes with the positive directions of the x, y and z axes, then $cos α$ , $cos β$ , $cos γ$ are called the direction cosines of the line. They are denoted by $l$ , $m$ , $n$ :

l = cos α, m = cos β, n = cos γ

Figure 1: Direction angles

α

β

γ

of a directed line. The direction cosines are

l = cos α

m = cos β

n = cos γ

, and a parallel line

A B

shares them.

If the direction of the line is reversed, the angles become $π - α$ , $π - β$ , $π - γ$ and the direction cosines become $- l$ , $- m$ , $- n$ . So an undirected line has two sets of direction cosines that differ only in sign.

Axis	Angles with x, y, z axes	Direction cosines
x-axis	$0^{\circ}, 9 0^{\circ}, 9 0^{\circ}$	$(1, 0, 0)$
y-axis	$9 0^{\circ}, 0^{\circ}, 9 0^{\circ}$	$(0, 1, 0)$
z-axis	$9 0^{\circ}, 9 0^{\circ}, 0^{\circ}$	$(0, 0, 1)$

Relation between the direction cosines

Let $O P$ be any line through the origin $O$ with direction cosines $l$ , $m$ , $n$ . Let $P = (x, y, z)$ and $O P = r$ . Then

O P^{2} = x^{2} + y^{2} + z^{2} = r^{2} ...(1)

From $P$ draw $P A$ , $P B$ , $P C$ perpendicular to the coordinate axes, so that $O A = x$ , $O B = y$ , $O C = z$ and $∠ P O A = α$ , $∠ P O B = β$ , $∠ P O C = γ$ .

From the right triangle $O C P$ (Figure 2), $n = cos γ = \frac{z}{r}$ , so $z = n r$ . In the same way, triangles $O A P$ and $O B P$ give $x = l r$ and $y = m r$ . Putting these in (1):

r^{2} (l^{2} + m^{2} + n^{2}) = x^{2} + y^{2} + z^{2} = r^{2} \Rightarrow l^{2} + m^{2} + n^{2} = 1

Figure 2: In right triangle

O C P

cos γ = \frac{z}{r}

, so

z = n r

; likewise

x = l r

y = m r

, which gives

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

If $P$ lies on a line through $O$ with direction cosines $l, m, n$ and $O P = r$ , then $P = (l r, m r, n r)$ .
The point at distance $r$ from $A (x_{1}, y_{1}, z_{1})$ along a line with direction cosines $l, m, n$ is $(x_{1} + l r, y_{1} + m r, z_{1} + n r)$ .
$l \hat{i} + m \hat{j} + n \hat{k}$ is a unit vector parallel to the line.

Solved Example 1

If a line makes angles

α

β

γ

with the coordinate axes, prove that

sin^{2} α + sin^{2} β + sin^{2} γ = 2

Solution:

Since the line makes angles $α$ , $β$ , $γ$ with the axes, $cos α$ , $cos β$ , $cos γ$ are its direction cosines, so $cos^{2} α + cos^{2} β + cos^{2} γ = 1$ .

$(1 - sin^{2} α) + (1 - sin^{2} β) + (1 - sin^{2} γ) = 1 \Rightarrow sin^{2} α + sin^{2} β + sin^{2} γ = 2$

In the same way, using $cos 2 θ = 2 cos^{2} θ - 1$ , we get $cos 2 α + cos 2 β + cos 2 γ = 2 - 3 = - 1$ .

2. Direction Ratios of a Line

If $a$ , $b$ , $c$ are three numbers proportional to the direction cosines $l$ , $m$ , $n$ of a straight line, then $a$ , $b$ , $c$ are called the direction ratios of the line. They are also called direction numbers or direction components.

By definition, $\frac{l}{a} = \frac{m}{b} = \frac{n}{c} = k$ (say), so $l = ak$ , $m = bk$ , $n = c k$ . Then

k^{2} (a^{2} + b^{2} + c^{2}) = l^{2} + m^{2} + n^{2} = 1 \Rightarrow k = \pm \frac{1}{a ^{2} + b ^{2} + c ^{2}}

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

The same sign, either positive or negative, is taken throughout.

Direction cosines of a line are unique (apart from reversing the sign), but direction ratios are not: $k a$ , $k b$ , $k c$ is also a set of direction ratios for any $k \neq = 0$ .
If $a, b, c$ are direction ratios of a line $L$ , then $a \hat{i} + b \hat{j} + c \hat{k}$ is a vector parallel to $L$ .
Parallel lines have the same direction cosines. For a line that does not pass through the origin, draw a parallel line through the origin to find its direction cosines.

Solved Example 2

2, - 3, 6

are the direction ratios of a line, find its direction cosines.

Solution:

$a^{2} + b^{2} + c^{2} = 4 + 9 + 36 = 7$ .

So the direction cosines are $\frac{2}{7}, - \frac{3}{7}, \frac{6}{7}$ , or, for the opposite direction, $- \frac{2}{7}, \frac{3}{7}, - \frac{6}{7}$ .

3. Direction Cosines of the Line Joining Two Points

The direction ratios of the line $P Q$ joining $P (x_{1}, y_{1}, z_{1})$ and $Q (x_{2}, y_{2}, z_{2})$ are $a = x_{2} - x_{1}$ , $b = y_{2} - y_{1}$ and $c = z_{2} - z_{1}$ . Dividing by $P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$ gives the direction cosines:

l = \frac{x _{2} - x _{1}}{P Q}, m = \frac{y _{2} - y _{1}}{P Q}, n = \frac{z _{2} - z _{1}}{P Q}

Figure 3: Direction ratios of

P Q

are the axis steps

x_{2} - x_{1}

y_{2} - y_{1}

z_{2} - z_{1}

; dividing by

P Q

gives the direction cosines.

Solved Example 3

Find the direction ratios and direction cosines of the line joining the points

A (6, - 7, - 1)

and

B (2, - 3, 1)

Solution:

Direction ratios of $A B$ are $(6 - 2, - 7 + 3, - 1 - 1) = (4, - 4, - 2)$ , or simply $(2, - 2, - 1)$ .

$a^{2} + b^{2} + c^{2} = 4 + 4 + 1 = 9$ , so $a^{2} + b^{2} + c^{2} = 3$ .

Direction cosines are $(\pm \frac{2}{3}, \mp \frac{2}{3}, \mp \frac{1}{3})$ .

Solved Example 4

Find the direction cosines of the two lines which are connected by the relations

l - 5 m + 3 n = 0

and

7 l^{2} + 5 m^{2} - 3 n^{2} = 0

Solution:

From the first relation, $l = 5 m - 3 n$ ...(1). Putting this in $7 l^{2} + 5 m^{2} - 3 n^{2} = 0$ ...(2):

$7 (5 m - 3 n)^{2} + 5 m^{2} - 3 n^{2} = 0 \Rightarrow 180 m^{2} - 210 mn + 60 n^{2} = 0 \Rightarrow (2 m - n) (3 m - 2 n) = 0$

So $\frac{m}{n} = \frac{1}{2}$ or $\frac{m}{n} = \frac{2}{3}$ .

Case 1: $n = 2 m$ . Then $l = 5 m - 6 m = - m$ , so $\frac{l}{- 1} = \frac{m}{1} = \frac{n}{2} = \frac{l ^{2} + m ^{2} + n ^{2}}{1 + 1 + 4} = \frac{1}{6}$ .

Direction cosines of one line are $- \frac{1}{6}, \frac{1}{6}, \frac{2}{6}$ .

Case 2: $n = \frac{3 m}{2}$ . Then $l = 5 m - \frac{9 m}{2} = \frac{m}{2}$ , so $\frac{l}{1} = \frac{m}{2} = \frac{n}{3} = \frac{1}{1 + 4 + 9} = \frac{1}{14}$ .

Direction cosines of the other line are $\frac{1}{14}, \frac{2}{14}, \frac{3}{14}$ .

4. Angle Between Two Lines

Let $θ$ be the angle between two lines whose direction cosines are $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ . Then

cos θ = l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}

If the direction ratios are $a_{1}, b_{1}, c_{1}$ and $a_{2}, b_{2}, c_{2}$ , the angle between the lines is given by

cos θ = \frac{a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2}}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}

For lines that do not meet, the angle is the angle between their directions (Figure 4). For the acute angle, take the modulus of the right-hand side.

Figure 4: The angle between two lines, even non-intersecting ones, is the angle

θ

between their directions. Draw a parallel to

L_{1}

through a point of

L_{2}

and measure

θ

there.

Sine of the angle

sin^{2} θ = (l_{1}^{2} + m_{1}^{2} + n_{1}^{2}) (l_{2}^{2} + m_{2}^{2} + n_{2}^{2}) - (l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2})^{2}

= (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}

\Rightarrow sin θ = \pm \sum (l_{1} m_{2} - l_{2} m_{1})^{2}

Conditions for perpendicular and parallel lines

Lines are	Using direction cosines	Using direction ratios
Perpendicular ( $θ = 9 0^{\circ}$ )	$l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0$	$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$
Parallel ( $θ = 0^{\circ}$ )	$\frac{l _{1}}{l _{2}} = \frac{m _{1}}{m _{2}} = \frac{n _{1}}{n _{2}}$	$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$

The parallel condition comes from $sin θ = 0$ : the sum $(l_{1} m_{2} - l_{2} m_{1})^{2} + (m_{1} n_{2} - m_{2} n_{1})^{2} + (n_{1} l_{2} - n_{2} l_{1})^{2} = 0$ holds only when $l_{1} m_{2} - l_{2} m_{1} = 0$ , $m_{1} n_{2} - m_{2} n_{1} = 0$ and $n_{1} l_{2} - n_{2} l_{1} = 0$ . Two parallel lines therefore have the same direction cosines.

A line perpendicular to both lines has direction ratios proportional to

m_{1} n_{2} - m_{2} n_{1}

n_{1} l_{2} - n_{2} l_{1}

l_{1} m_{2} - l_{2} m_{1}

. These are the components of the cross product of the two direction vectors, and they are used again for the line of intersection of two planes and for shortest distance.

Solved Example 5

Show that the two lines having direction ratios

- 1, 3, 2

and

2, 2, - 2

are perpendicular.

Solution:

$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = (- 1) (2) + (3) (2) + (2) (- 2) = - 2 + 6 - 4 = 0$

Therefore the lines are perpendicular.

Solved Example 6

Find the angle between the lines whose direction cosines are

\frac{3}{4}, \frac{1}{4}, \frac{3}{2}

and

\frac{3}{4}, \frac{1}{4}, - \frac{3}{2}

Solution:

$cos θ = l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = \frac{3}{16} + \frac{1}{16} - \frac{3}{4} = \frac{4 - 12}{16} = - \frac{1}{2}$

So $θ = 12 0^{\circ}$ , and the acute angle between the lines is $6 0^{\circ}$ .

Solved Example 7

Find the angle between any two diagonals of a cube.

Solution:

Take a cube of side $a$ with one vertex at $O$ and the edges through $O$ along the axes: $A (a, 0, 0)$ , $B (0, a, 0)$ , $C (0, 0, a)$ , $D (0, a, a)$ , $E (a, a, a)$ , $F (a, a, 0)$ , $G (a, 0, a)$ . The four diagonals are $O E$ , $A D$ , $C F$ and $GB$ .

Figure 11: The four diagonals of a cube have direction ratios

(\pm 1, 1, \pm 1)

. Any two of them meet at an angle

θ = cos^{- 1} (\frac{1}{3})

Direction ratios of $O E$ are $a, a, a$ , or $1, 1, 1$ , so its direction cosines are $\frac{1}{3}, \frac{1}{3}, \frac{1}{3}$ .

Direction ratios of $A D$ are $- a, a, a$ , or $- 1, 1, 1$ , so its direction cosines are $- \frac{1}{3}, \frac{1}{3}, \frac{1}{3}$ .

Similarly, $C F$ and $GB$ have direction ratios $1, 1, - 1$ and $- 1, 1, - 1$ .

For $O E$ and $A D$ : $cos θ = - \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{1}{3}$ , so $θ = cos^{- 1} (\frac{1}{3})$ . Any other pair gives $∣ cos θ ∣ = \frac{1}{3}$ as well.

Solved Example 8

If two pairs of opposite edges of a tetrahedron are mutually perpendicular, show that the third pair is also mutually perpendicular.

Solution:

Let $O A B C$ be the tetrahedron with $O$ at the origin and $A (x_{1}, y_{1}, z_{1})$ , $B (x_{2}, y_{2}, z_{2})$ , $C (x_{3}, y_{3}, z_{3})$ . Let $O A ⊥ B C$ and $O B ⊥ C A$ . We must prove $O C ⊥ B A$ .

Direction ratios of $O A$ are $x_{1}, y_{1}, z_{1}$ and of $B C$ are $x_{3} - x_{2}, y_{3} - y_{2}, z_{3} - z_{2}$ . Since $O A ⊥ B C$ :

$x_{1} (x_{3} - x_{2}) + y_{1} (y_{3} - y_{2}) + z_{1} (z_{3} - z_{2}) = 0$ ...(1)

Since $O B ⊥ C A$ : $x_{2} (x_{1} - x_{3}) + y_{2} (y_{1} - y_{3}) + z_{2} (z_{1} - z_{3}) = 0$ ...(2)

Adding (1) and (2), the terms $x_{1} x_{2}$ , $y_{1} y_{2}$ , $z_{1} z_{2}$ cancel and we get

$x_{3} (x_{1} - x_{2}) + y_{3} (y_{1} - y_{2}) + z_{3} (z_{1} - z_{2}) = 0$

Direction ratios of $O C$ are $x_{3}, y_{3}, z_{3}$ and of $B A$ are $x_{1} - x_{2}, y_{1} - y_{2}, z_{1} - z_{2}$ . Hence $O C ⊥ B A$ .

5. Projection of a Line Segment on a Line

The projection of the segment joining $P (x_{1}, y_{1}, z_{1})$ and $Q (x_{2}, y_{2}, z_{2})$ on a line whose direction cosines are $l, m, n$ is

P^{'} Q^{'} = ∣ l (x_{2} - x_{1}) + m (y_{2} - y_{1}) + n (z_{2} - z_{1}) ∣

This is $P Q cos θ$ , where $θ$ is the angle between $P Q$ and the line, because the direction cosines of $P Q$ are $\frac{x _{2} - x _{1}}{P Q}$ , $\frac{y _{2} - y _{1}}{P Q}$ , $\frac{z _{2} - z _{1}}{P Q}$ .

Figure 5: Projection of

P Q

on a line with direction cosines

l, m, n

P^{'} Q^{'} = P Q cos θ = ∣ l (x_{2} - x_{1}) + m (y_{2} - y_{1}) + n (z_{2} - z_{1}) ∣

Vector form

The projection of $a$ on $b$ is $a \cdot \hat{b} = \frac{a \cdot b}{∣ b ∣}$ . Here take $a = P Q = (x_{2} - x_{1}) \hat{i} + (y_{2} - y_{1}) \hat{j} + (z_{2} - z_{1}) \hat{k}$ and $b = l \hat{i} + m \hat{j} + n \hat{k}$ .

$l ∣ r ∣$ , $m ∣ r ∣$ and $n ∣ r ∣$ are the projections of $r$ on $O X$ , $O Y$ and $O Z$ .
$r = ∣ r ∣ (l \hat{i} + m \hat{j} + n \hat{k})$ .

Solved Example 9

Find the projection of the line segment joining

(1, 2, 3)

and

(- 1, 4, 2)

on the line having direction ratios

2, 3, - 6

Solution:

Let $A (1, 2, 3)$ and $B (- 1, 4, 2)$ . Since $2^{2} + 3^{2} + (- 6)^{2} = 7$ , the direction cosines of the given line are $\frac{2}{7}, \frac{3}{7}, - \frac{6}{7}$ .

Projection of $A B$ $= \frac{2}{7} (- 1 - 1) + \frac{3}{7} (4 - 2) - \frac{6}{7} (2 - 3) = \frac{- 4 + 6 + 6}{7} = \frac{8}{7}$

6. Equation of a Straight Line in Space

Symmetric (Cartesian) form

The equation of the straight line passing through $P (x_{1}, y_{1}, z_{1})$ with direction ratios $a, b, c$ is

\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c} = r

The general point on this line is $(x_{1} + a r, y_{1} + b r, z_{1} + cr)$ . When the denominators are the direction cosines $l, m, n$ , the point is $(x_{1} + l r, y_{1} + m r, z_{1} + n r)$ and $r$ is its actual distance from $P$ .

Two-point form

The line passing through $P (x_{1}, y_{1}, z_{1})$ and $Q (x_{2}, y_{2}, z_{2})$ is

\frac{x - x _{1}}{x _{2} - x _{1}} = \frac{y - y _{1}}{y _{2} - y _{1}} = \frac{z - z _{1}}{z _{2} - z _{1}}

Vector form

The line through the point with position vector $a$ and parallel to $b$ is $r = a + λ b$ , where $λ$ is a scalar. The line through two points with position vectors $a$ and $b$ is $r = a + λ (b - a)$ .

Figure 7: Vector equation of a line,

r = a + λ b

. Each real value of

λ

gives one point of the line;

λ = 0

gives

A

Converting between the two forms:

\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c} ⟺ r = (x_{1} \hat{i} + y_{1} \hat{j} + z_{1} \hat{k}) + λ (a \hat{i} + b \hat{j} + c \hat{k})

Special lines

Line	Equation
x-axis	$y = 0, z = 0$
y-axis	$x = 0, z = 0$
z-axis	$x = 0, y = 0$
Parallel to the x-axis	$y = p, z = q$
Parallel to the y-axis	$x = h, z = q$
Parallel to the z-axis	$x = h, y = p$
Through the origin	$\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$

A zero denominator is allowed.

\frac{x - 2}{0} = \frac{y + 1}{3} = \frac{z}{4}

means

x = 2

together with

\frac{y + 1}{3} = \frac{z}{4}

: the line lies in the plane

x = 2

Solved Example 10

Find the equations of the straight lines through the point

(a, b, c)

which are (a) parallel to the z-axis, (b) perpendicular to the z-axis.

Solution:

(a) A line parallel to the z-axis has $α = 9 0^{\circ}$ , $β = 9 0^{\circ}$ , $γ = 0^{\circ}$ , so $l = 0$ , $m = 0$ , $n = 1$ . The line through $(a, b, c)$ is

\frac{x - a}{0} = \frac{y - b}{0} = \frac{z - c}{1}, that is x = a, y = b

(b) A line perpendicular to the z-axis has $γ = 9 0^{\circ}$ , so $n = 0$ and $l^{2} + m^{2} = 1$ . If it makes angle $α$ with the x-axis, then $l = cos α$ and $m = sin α$ (the sign is absorbed by the choice of $α$ ). The line is

\frac{x - a}{cos α} = \frac{y - b}{sin α} = \frac{z - c}{0}, that is \frac{x - a}{l} = \frac{y - b}{m} = \frac{z - c}{0}

Solved Example 11

Find the equation of the line through the points

(3, 4, - 7)

and

(1, - 1, 6)

in vector form as well as in Cartesian form.

Solution:

Let $A (3, 4, - 7)$ and $B (1, - 1, 6)$ . Then $a = 3 \hat{i} + 4 \hat{j} - 7 \hat{k}$ and $b = \hat{i} - \hat{j} + 6 \hat{k}$ , so $b - a = - 2 \hat{i} - 5 \hat{j} + 13 \hat{k}$ .

Vector form: $r = 3 \hat{i} + 4 \hat{j} - 7 \hat{k} + t (- 2 \hat{i} - 5 \hat{j} + 13 \hat{k})$ .

Cartesian form: $\frac{x - 3}{1 - 3} = \frac{y - 4}{- 1 - 4} = \frac{z + 7}{6 + 7}$ , that is $\frac{x - 3}{2} = \frac{y - 4}{5} = \frac{z + 7}{- 13}$ .

Solved Example 12

Find the coordinates of the points on the line

\frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{6}

which are at a distance of

3

units from the point

(1, - 2, 3)

Solution:

The point $P (1, - 2, 3)$ lies on the line. Direction ratios are $2, 3, 6$ and $4 + 9 + 36 = 7$ , so the direction cosines are $\frac{2}{7}, \frac{3}{7}, \frac{6}{7}$ . Writing the line with direction cosines:

\frac{x - 1}{2/7} = \frac{y + 2}{3/7} = \frac{z - 3}{6/7} = r

Any point is $Q (1 + \frac{2 r}{7}, - 2 + \frac{3 r}{7}, 3 + \frac{6 r}{7})$ and $P Q = ∣ r ∣$ . Given $∣ r ∣ = 3$ , $r = \pm 3$ .

For $r = 3$ : $Q = (\frac{13}{7}, - \frac{5}{7}, \frac{39}{7})$ . For $r = - 3$ : $Q = (\frac{1}{7}, - \frac{23}{7}, \frac{3}{7})$ .

Solved Example 13

Find the coordinates of the point where the line joining the points

(2, - 3, 1)

and

(3, - 4, - 5)

cuts the plane

2 x + y + z = 7

Solution:

Direction ratios of the line are $3 - 2$ , $- 4 + 3$ , $- 5 - 1$ , that is $1, - 1, - 6$ . Its equation is

\frac{x - 2}{1} = \frac{y + 3}{- 1} = \frac{z - 1}{- 6} = r

Any point on the line is $(r + 2, - r - 3, - 6 r + 1)$ . If it lies on the plane,

$2 (r + 2) + (- r - 3) + (- 6 r + 1) = 7 \Rightarrow - 5 r + 2 = 7 \Rightarrow r = - 1$

The point is $(- 1 + 2, 1 - 3, 6 + 1) = (1, - 2, 7)$ .

Solved Example 14

Show that the lines

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}

and

\frac{x - 4}{5} = \frac{y - 1}{2} = z

intersect. Also find their point of intersection.

Solution:

Any point on the first line is $P (2 r + 1, 3 r + 2, 4 r + 3)$ and any point on the second line is $Q (5 λ + 4, 2 λ + 1, λ)$ . The lines intersect if $P$ and $Q$ coincide for some $r$ and $λ$ :

$2 r + 1 = 5 λ + 4 \Rightarrow 2 r - 5 λ = 3$ ...(1)

$3 r + 2 = 2 λ + 1 \Rightarrow 3 r - 2 λ = - 1$ ...(2)

$4 r + 3 = λ \Rightarrow 4 r - λ = - 3$ ...(3)

Solving (1) and (2) gives $r = - 1$ , $λ = - 1$ . These values also satisfy (3): $- 4 + 1 = - 3$ .

So the lines intersect, at $P (- 1, - 1, - 1)$ .

7. Perpendicular from a Point to a Line: Foot, Length and Image

Using projection

Let $A B$ be a straight line passing through $A (a, b, c)$ with direction cosines $l, m, n$ , and let $P (x, y, z)$ be a point. If $N$ is the foot of the perpendicular from $P$ , then

A N = projection of A P on A B = l (x - a) + m (y - b) + n (z - c)

A P = (x - a)^{2} + (y - b)^{2} + (z - c)^{2}, P N = A P^{2} - A N^{2}

Figure 6: Foot

N

, perpendicular distance

P N = A P^{2} - A N^{2}

and image

P^{'}

of a point in a line.

N

is the midpoint of

P P^{'}

Using the general point of the line

Write any point of the line $\frac{x - a}{l} = \frac{y - b}{m} = \frac{z - c}{n} = r$ as $N (a + l r, b + m r, c + n r)$ .
Write the direction ratios of $P N$ and use $P N ⊥$ line: $l (a + l r - x) + m (b + m r - y) + n (c + n r - z) = 0$ . Solve for $r$ .
Put $r$ back to get the foot $N$ . Then $P N$ is the length of the perpendicular and the line $P N$ is the perpendicular itself.
For the image $P^{'}$ , use the fact that $N$ is the midpoint of $P P^{'}$ , so $P^{'} = (2 x_{N} - x, 2 y_{N} - y, 2 z_{N} - z)$ .

Vector form

The foot of the perpendicular from the point with position vector $p$ to the line $r = a + λ b$ , and the image $q$ of that point, are

f = a + \frac{( p - a ) \cdot b}{∣ b ∣ ^{2}} b, q = 2 f - p

Solved Example 15

Find the perpendicular distance of the point

P (0, - 1, 3)

from the straight line passing through

A (1, - 3, 2)

and having direction ratios

1, 2, 2

Solution:

Direction cosines of the line are $\frac{1}{1 + 4 + 4}, \frac{2}{3}, \frac{2}{3}$ , that is $\frac{1}{3}, \frac{2}{3}, \frac{2}{3}$ .

$A N = l (x - a) + m (y - b) + n (z - c) = \frac{1}{3} (0 - 1) + \frac{2}{3} (- 1 + 3) + \frac{2}{3} (3 - 2) = \frac{5}{3}$

$A P = (0 - 1)^{2} + (- 1 + 3)^{2} + (3 - 2)^{2} = 6$

Perpendicular distance $P N = A P^{2} - A N^{2} = 6 - \frac{25}{9} = \frac{29}{3}$ .

Solved Example 16

Find the length of the perpendicular from

P (2, - 3, 1)

to the line

\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{- 1}

Solution:

Method 1 (general point). Any point on the line is $Q (2 r - 1, 3 r + 3, - r - 2)$ . Direction ratios of $P Q$ are $2 r - 3, 3 r + 6, - r - 3$ . Since $P Q$ is perpendicular to the line with direction ratios $2, 3, - 1$ :

$2 (2 r - 3) + 3 (3 r + 6) - (- r - 3) = 0 \Rightarrow 14 r + 15 = 0 \Rightarrow r = - \frac{15}{14}$

So $Q = (- \frac{22}{7}, - \frac{3}{14}, - \frac{13}{14})$ and $P Q^{2} = (\frac{72}{14})^{2} + (\frac{39}{14})^{2} + (\frac{27}{14})^{2} = \frac{7434}{196} = \frac{531}{14}$ .

Method 2 (projection). $R (- 1, 3, - 2)$ lies on the line and the direction cosines are $\frac{2}{14}, \frac{3}{14}, - \frac{1}{14}$ .

$R Q =$ projection of $R P$ $= \frac{2 ( 3 ) + 3 ( - 6 ) - 1 ( 3 )}{14} = - \frac{15}{14}$ , and $P R^{2} = 3^{2} + 6^{2} + 3^{2} = 54$ .

$P Q = P R^{2} - R Q^{2} = 54 - \frac{225}{14} = \frac{531}{14}$ units, the same as Method 1.

Solved Example 17

Find the equation of the line drawn through the point

(1, 0, 2)

to meet at right angles the line

\frac{x + 1}{3} = \frac{y - 2}{- 2} = \frac{z + 1}{- 1}

Solution:

Let $P (1, 0, 2)$ . Any point on the given line is $Q (3 r - 1, - 2 r + 2, - r - 1)$ , so the direction ratios of $P Q$ are $3 r - 2, - 2 r + 2, - r - 3$ .

Since $P Q$ is perpendicular to the line with direction ratios $3, - 2, - 1$ :

$3 (3 r - 2) - 2 (- 2 r + 2) - (- r - 3) = 0 \Rightarrow 14 r - 7 = 0 \Rightarrow r = \frac{1}{2}$

Direction ratios of $P Q$ are $- \frac{1}{2}, 1, - \frac{7}{2}$ , or $1, - 2, 7$ . The required line is $\frac{x - 1}{1} = \frac{y}{- 2} = \frac{z - 2}{7}$ .

Solved Example 18

Find the image of the point

(1, 6, 3)

in the line

\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}

Solution:

Let $P (1, 6, 3)$ . Any point on the line is $N (λ, 2 λ + 1, 3 λ + 2)$ , so the direction ratios of $P N$ are $λ - 1, 2 λ - 5, 3 λ - 1$ .

For the foot, $P N ⊥$ line: $(λ - 1) + 2 (2 λ - 5) + 3 (3 λ - 1) = 0 \Rightarrow 14 λ - 14 = 0 \Rightarrow λ = 1$

So the foot is $N (1, 3, 5)$ . Since $N$ is the midpoint of $P$ and its image $P^{'}$ :

$P^{'} = (2 \cdot 1 - 1, 2 \cdot 3 - 6, 2 \cdot 5 - 3) = (1, 0, 7)$

8. Line of Intersection of Two Planes

A straight line in space can be written as the intersection of two non-parallel planes

a_{1} x + b_{1} y + c_{1} z + d_{1} = 0, a_{2} x + b_{2} y + c_{2} z + d_{2} = 0

This is called the non-symmetric (general) form of the line. To write it in symmetric form we need its direction ratios and one point on it.

Figure 8: A line in non-symmetric form is the intersection of two planes. It is perpendicular to both normals, so its direction is

n_{1} \times n_{2}

Step 1: Direction ratios

The line lies in both planes, so it is perpendicular to both normals: $a_{1} l + b_{1} m + c_{1} n = 0$ and $a_{2} l + b_{2} m + c_{2} n = 0$ . By cross-multiplication,

\frac{l}{b _{1} c _{2} - b _{2} c _{1}} = \frac{m}{c _{1} a _{2} - c _{2} a _{1}} = \frac{n}{a _{1} b _{2} - a _{2} b _{1}}

These are the components of

n_{1} \times n_{2} = \hat{i} \hat{j} \hat{k} a_{1} b_{1} c_{1} a_{2} b_{2} c_{2}

Step 2: A point on the line

If $a_{1} b_{2} - a_{2} b_{1} \neq = 0$ , the line is not parallel to the XY-plane, so it meets it. Put $z = 0$ and solve $a_{1} x + b_{1} y + d_{1} = 0$ and $a_{2} x + b_{2} y + d_{2} = 0$ for $x$ and $y$ . If this fails, put $y = 0$ or $x = 0$ instead.

Solved Example 19

Find in symmetric form the equations of the line

3 x + 2 y - z - 4 = 0 = 4 x + y - 2 z + 3

Solution:

The line is $3 x + 2 y - z - 4 = 0$ , $4 x + y - 2 z + 3 = 0$ ...(1). Let $l, m, n$ be its direction cosines. The line is perpendicular to the normals of both planes:

$3 l + 2 m - n = 0$ and $4 l + m - 2 n = 0$

By cross-multiplication, $\frac{l}{- 4 + 1} = \frac{m}{- 4 + 6} = \frac{n}{3 - 8}$ , that is $\frac{l}{- 3} = \frac{m}{2} = \frac{n}{- 5} = \frac{1}{9 + 4 + 25} = \frac{1}{38}$ .

So the direction cosines are $- \frac{3}{38}, \frac{2}{38}, - \frac{5}{38}$ .

To find a point, put $z = 0$ in (1): $3 x + 2 y - 4 = 0$ and $4 x + y + 3 = 0$ . Solving, $x = - 2$ , $y = 5$ . So $(- 2, 5, 0)$ lies on the line.

The symmetric form is $\frac{x + 2}{- 3} = \frac{y - 5}{2} = \frac{z}{- 5}$ .

Solved Example 20

Find the angle between the lines

x - 3 y - 4 = 0, 4 y - z + 5 = 0

and

x + 3 y - 11 = 0, 2 y - z + 6 = 0

Solution:

Line 1 is perpendicular to the normals $(1, - 3, 0)$ and $(0, 4, - 1)$ : $l_{1} - 3 m_{1} = 0$ and $4 m_{1} - n_{1} = 0$ . So $\frac{l _{1}}{3} = \frac{m _{1}}{1} = \frac{n _{1}}{4}$ .

Line 2 is perpendicular to the normals $(1, 3, 0)$ and $(0, 2, - 1)$ : $l_{2} + 3 m_{2} = 0$ and $2 m_{2} - n_{2} = 0$ . So $\frac{l _{2}}{- 3} = \frac{m _{2}}{1} = \frac{n _{2}}{2}$ .

$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = (3) (- 3) + (1) (1) + (4) (2) = - 9 + 1 + 8 = 0$

Hence $θ = 9 0^{\circ}$ .

9. Skew Lines and Shortest Distance

Relative position of two lines

Two lines in space are parallel (coplanar and never meeting), intersecting (coplanar and meeting at one point) or skew. Skew lines are lines that are neither parallel nor intersecting, so no plane contains both of them.

Figure 9: Relative positions of two lines in space. Parallel and intersecting lines are coplanar; skew lines are neither parallel nor intersecting.

Line of shortest distance

For two skew lines, the straight line that is perpendicular to each of them is called the line of shortest distance, and the length it intercepts between the two lines is the shortest distance.

Write the lines as $\frac{x - x _{1}}{l _{1}} = \frac{y - y _{1}}{m _{1}} = \frac{z - z _{1}}{n _{1}} = r_{1}$ ...(1) and $\frac{x - x _{2}}{l _{2}} = \frac{y - y _{2}}{m _{2}} = \frac{z - z _{2}}{n _{2}} = r_{2}$ ...(2).
Take $P (x_{1} + l_{1} r_{1}, y_{1} + m_{1} r_{1}, z_{1} + n_{1} r_{1})$ on (1) and $Q (x_{2} + l_{2} r_{2}, y_{2} + m_{2} r_{2}, z_{2} + n_{2} r_{2})$ on (2).
Direction ratios of $P Q$ are $(l_{1} r_{1} + x_{1} - x_{2} - l_{2} r_{2}, m_{1} r_{1} + y_{1} - y_{2} - m_{2} r_{2}, n_{1} r_{1} + z_{1} - z_{2} - n_{2} r_{2})$ .
Use $P Q ⊥$ (1) and $P Q ⊥$ (2) to get two equations in $r_{1}$ and $r_{2}$ . Solve them to get $P$ and $Q$ ; then $P Q$ is the shortest distance.

Figure 10: The shortest distance between skew lines

L_{1}

and

L_{2}

is the length

d

of their common perpendicular

P Q

Shortest distance formula (Cartesian)

D = x_{2} - x_{1} y_{2} - y_{1} z_{2} - z_{1} l_{1} m_{1} n_{1} l_{2} m_{2} n_{2}

d = \frac{∣ D ∣}{( m _{1} n _{2} - m _{2} n _{1} ) ^{2} + ( n _{1} l _{2} - n _{2} l _{1} ) ^{2} + ( l _{1} m _{2} - l _{2} m _{1} ) ^{2}}

The same formula works with direction ratios in place of direction cosines.

Shortest distance formula (vector)

For $r = a_{1} + λ b_{1}$ and $r = a_{2} + μ b_{2}$ :

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

The lines are skew if $(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) \neq = 0$ , that is $[b_{1} b_{2} (a_{2} - a_{1})] \neq = 0$ .
If this scalar triple product is $0$ (equivalently $D = 0$ ) and the lines are not parallel, the lines intersect.
If a line is given in general form, first convert it to symmetric form and then apply the formula.

Distance between parallel lines

For the parallel lines $r = a_{1} + λ b$ and $r = a_{2} + μ b$ , $b_{1} \times b_{2} = 0$ , so the skew-lines formula cannot be used. Instead,

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

Solved Example 21

Find the shortest distance between the lines

\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}

and

\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}

. Also find the equation of the line of shortest distance.

Solution:

Any point on the first line is $P (3 r_{1} + 3, 8 - r_{1}, r_{1} + 3)$ and on the second line is $Q (- 3 - 3 r_{2}, 2 r_{2} - 7, 4 r_{2} + 6)$ .

Direction ratios of $P Q$ are $(3 r_{1} + 3 r_{2} + 6, - r_{1} - 2 r_{2} + 15, r_{1} - 4 r_{2} - 3)$ .

$P Q ⊥$ first line: $3 (3 r_{1} + 3 r_{2} + 6) - (- r_{1} - 2 r_{2} + 15) + (r_{1} - 4 r_{2} - 3) = 0 \Rightarrow 11 r_{1} + 7 r_{2} = 0$ ...(1)

$P Q ⊥$ second line: $- 3 (3 r_{1} + 3 r_{2} + 6) + 2 (- r_{1} - 2 r_{2} + 15) + 4 (r_{1} - 4 r_{2} - 3) = 0 \Rightarrow 7 r_{1} + 29 r_{2} = 0$ ...(2)

Solving (1) and (2), $r_{1} = 0$ and $r_{2} = 0$ . So $P (3, 8, 3)$ and $Q (- 3, - 7, 6)$ .

Shortest distance $P Q = (3 + 3)^{2} + (8 + 7)^{2} + (3 - 6)^{2} = 270 = 330$ units.

Direction ratios of $P Q$ are $6, 15, - 3$ , or $2, 5, - 1$ . The line of shortest distance is $\frac{x - 3}{2} = \frac{y - 8}{5} = \frac{z - 3}{- 1}$ , or in vector form $r = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} + t (2 \hat{i} + 5 \hat{j} - \hat{k})$ .

Solved Example 22

Using the formula, find the shortest distance between the lines

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}

and

\frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}

Solution:

Here $a_{1} = (1, 2, 3)$ , $b_{1} = (2, 3, 4)$ , $a_{2} = (2, 4, 5)$ , $b_{2} = (3, 4, 5)$ , so $a_{2} - a_{1} = (1, 2, 2)$ .

$b_{1} \times b_{2} = (3 \cdot 5 - 4 \cdot 4, 4 \cdot 3 - 2 \cdot 5, 2 \cdot 4 - 3 \cdot 3) = (- 1, 2, - 1)$ , and $∣ b_{1} \times b_{2} ∣ = 6$ .

$(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = - 1 + 4 - 2 = 1$

Shortest distance $d = \frac{∣1∣}{6} = \frac{1}{6}$ units.

Solved Example 23

Find the distance between the parallel lines

r = \hat{i} + 2 \hat{j} - 4 \hat{k} + λ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})

and

r = 3 \hat{i} + 3 \hat{j} - 5 \hat{k} + μ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})

Solution:

Here $b = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ , $∣ b ∣ = 7$ and $a_{2} - a_{1} = 2 \hat{i} + \hat{j} - \hat{k}$ .

$(a_{2} - a_{1}) \times b = (1 \cdot 6 - (- 1) \cdot 3) \hat{i} - (2 \cdot 6 - (- 1) \cdot 2) \hat{j} + (2 \cdot 3 - 1 \cdot 2) \hat{k} = 9 \hat{i} - 14 \hat{j} + 4 \hat{k}$

$d = \frac{81 + 196 + 16}{7} = \frac{293}{7}$ units.

Solved Example 24

A tetrahedron is formed by the planes

y + z = 0

x + z = 0

x + y = 0

and

x + y + z = 3 a

. Find the shortest distance between a pair of its opposite edges.

Solution:

The first three planes meet at $O (0, 0, 0)$ . Take the pair of opposite edges $O A$ (given by $y + z = 0$ , $x + z = 0$ ) and $B C$ (given by $x + y = 0$ , $x + y + z = 3 a$ ).

In symmetric form, $O A$ is $\frac{x}{1} = \frac{y}{1} = \frac{z}{- 1}$ and $B C$ is $\frac{x}{1} = \frac{y}{- 1} = \frac{z - 3 a}{0}$ . So the direction ratios are $1, 1, - 1$ and $1, - 1, 0$ , and $C^{'} (0, 0, 3 a)$ is a point on $B C$ .

Let the line of shortest distance have direction cosines $l, m, n$ . It is perpendicular to both edges: $l + m - n = 0$ and $l - m = 0$ . So $\frac{l}{1} = \frac{m}{1} = \frac{n}{2}$ , giving $l = \frac{1}{6}$ , $m = \frac{1}{6}$ , $n = \frac{2}{6}$ .

Shortest distance = projection of $O C^{'}$ on this line $= 0 \cdot \frac{1}{6} + 0 \cdot \frac{1}{6} + 3 a \cdot \frac{2}{6} = \frac{6 a}{6} = 6 a$ .

By symmetry, every pair of opposite edges of this tetrahedron is $6 a$ apart.

Common Mistakes to Avoid

Watch out

Using direction ratios as direction cosines. Numbers like $1, - 1, - 1$ are direction ratios; direction cosines must satisfy $l^{2} + m^{2} + n^{2} = 1$ .
Writing $cos θ = a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}$ without dividing by the two magnitudes.
Treating $r$ in $\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c} = r$ as a distance when $a, b, c$ are only direction ratios.
Reading the point from the equation with the wrong sign: $\frac{x + 3}{2}$ belongs to a line through $x = - 3$ .
Calling $\frac{z - c}{0}$ undefined. A zero denominator simply means $z = c$ all along the line.
Using the projection formula with direction ratios instead of direction cosines.
Applying the skew-lines formula to parallel lines, where $b_{1} \times b_{2} = 0$ . Use the parallel-lines formula instead.
Forgetting to check the third equation when testing whether two lines intersect.

Frequently Asked Questions

What is the difference between direction cosines and direction ratios?

Direction cosines $l, m, n$ are the cosines of the angles a line makes with the axes, so they satisfy $l^{2} + m^{2} + n^{2} = 1$ and are unique apart from sign. Direction ratios $a, b, c$ are any three numbers proportional to them, so a line has infinitely many sets of direction ratios.

Why do direction cosines satisfy $l^{2} + m^{2} + n^{2} = 1$ ?

For a point $P (x, y, z)$ at distance $r$ from the origin on the line, right triangles with the axes give $x = l r$ , $y = m r$ and $z = n r$ . Substituting in $x^{2} + y^{2} + z^{2} = r^{2}$ gives $r^{2} (l^{2} + m^{2} + n^{2}) = r^{2}$ , so the sum of squares is $1$ .

How do you find the angle between two lines in 3D?

Use $cos θ = l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}$ with direction cosines, or divide $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}$ by the product of the magnitudes when you have direction ratios. Take the modulus for the acute angle. The same formula works for skew lines because only the directions matter.

What are skew lines?

Skew lines are two lines in space that are neither parallel nor intersecting. Because they never meet and do not point the same way, no single plane can contain both. They exist only in three dimensions; in a plane, any two non-parallel lines must meet.

How do you find the shortest distance between two skew lines?

Write both lines in vector form and use $d = \frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣}$ . If you also need the line of shortest distance, take general points on both lines, make their join perpendicular to both directions and solve for the two parameters.

How do you convert a line from non-symmetric form to symmetric form?

A line given by two plane equations has direction ratios equal to the cross product of the two normals, found by cross-multiplication. For a point, put $z = 0$ (or $y = 0$ or $x = 0$ ) and solve the two plane equations. Then write $\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c}$ .

How important are lines in space for JEE Main?

Direction ratios, angle between lines, equation of a line, skew lines and shortest distance are listed explicitly in the JEE Main syllabus, and questions on the foot or image of a point and on intersecting lines are frequent. Most can be solved in under two minutes once the general-point method is automatic.

What should JEE Advanced aspirants practise in this concept?

JEE Advanced combines ideas: lines given as intersections of planes, shortest distance between edges of a solid, conditions for lines to intersect with unknown parameters, and images of a point in a line. Practise converting between Cartesian and vector forms quickly, since multi-correct options often hide the same line in different forms.

Previous year questions on Lines in Space

60 questions from past papers, each with a step-by-step solution.

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1. Direction Cosines of a Line

Relation between the direction cosines

2. Direction Ratios of a Line

3. Direction Cosines of the Line Joining Two Points

4. Angle Between Two Lines

Sine of the angle

Conditions for perpendicular and parallel lines

5. Projection of a Line Segment on a Line

Vector form

6. Equation of a Straight Line in Space

Symmetric (Cartesian) form

Two-point form

Vector form

Special lines

7. Perpendicular from a Point to a Line: Foot, Length and Image

Using projection

Using the general point of the line

Vector form

8. Line of Intersection of Two Planes

Step 1: Direction ratios

Step 2: A point on the line

9. Skew Lines and Shortest Distance

Relative position of two lines

Line of shortest distance

Shortest distance formula (Cartesian)

Shortest distance formula (vector)

Distance between parallel lines

Common Mistakes to Avoid

Frequently Asked Questions

What is the difference between direction cosines and direction ratios?

Why do direction cosines satisfy l2+m2+n2=1?

How do you find the angle between two lines in 3D?

What are skew lines?

How do you find the shortest distance between two skew lines?

How do you convert a line from non-symmetric form to symmetric form?

How important are lines in space for JEE Main?

What should JEE Advanced aspirants practise in this concept?

Previous year questions on Lines in Space

Ready to master Three Dimensional Geometry?

Why do direction cosines satisfy $l^{2} + m^{2} + n^{2} = 1$ ?