JEE Mains Maths · Three Dimensional Geometry

Direction Cosines and Equations of Lines

Describing a direction in space by its direction cosines or direction ratios, the angle between two lines, and writing a line through a point, through two points, or perpendicular to two given lines.

All Three Dimensional Geometry notes JEE Mains Maths strategy

Why this matters

Twenty-seven PYQs. One step carries most of them: the cross product of two directions gives a third direction perpendicular to both. Five ideas cover the page.

Concept 1 of 5: Direction cosines and direction ratios

A line making angles

\alpha

\beta

\gamma

with the three axes has direction cosines

l=\cos\alpha

m=\cos\beta

n=\cos\gamma

, and these always satisfy

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

. Any triple in the same proportion, such as

(2,3,6)

, is a set of direction ratios. To turn ratios into cosines, divide by their length.

Definition

Direction cosines: $(\cos\alpha,\cos\beta,\cos\gamma)$ , with $l^2+m^2+n^2=1$ .
Direction ratios: any $(a,b,c)$ proportional to them; $(l,m,n)=\pm\frac{(a,b,c)}{\sqrt{a^2+b^2+c^2}}$ .
Equivalently $\sin^2\alpha+\sin^2\beta+\sin^2\gamma=2$ .
The line joining $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ has direction ratios $(x_2-x_1,\,y_2-y_1,\,z_2-z_1)$ .

Direction cosines from direction ratios

(l,m,n)=\frac{(a,b,c)}{\sqrt{a^2+b^2+c^2}},\qquad l^2+m^2+n^2=1

Worked example

A line makes

60^\circ

with the

x

-axis and

45^\circ

with the

y

-axis. Find its angle with the

z

-axis.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q127Moderate

Example 1 · Three Dimensional Geometry · Direction Cosines and Equations of Lines

Each of the angles

\beta

and

\gamma

that a given line makes with the positive

y-

and

z

-axes, respectively, is half of the angle that this line makes with the positive

x

-axes. Then the sum of all possible values of the angle

\beta

is.

Ratios are not cosines until you divide

(2,3,6)

gives

4+9+36=49

, not

1

. Divide by

7

first; only then do the squares add to

1

Drill 1 more on direction cosines and direction ratios

Concept 2 of 5: Two lines from a pair of equations in l, m, n

When the direction cosines of two lines satisfy one linear and one quadratic equation, use the linear one to remove a variable. The quadratic then becomes homogeneous in the other two, so it factorises into two ratios. Each ratio is one line's direction, and the angle between them follows from the dot product.

Definition

From the linear equation, write one of $l,m,n$ in terms of the others.
Substitute into the quadratic; divide by the square of one variable to get a quadratic in the ratio.
Its two roots give the two directions.
Equal roots mean the lines are parallel; roots with product $-1$ (in the right ratio) often signal perpendicular lines, but always check with the dot product.

Angle between the two directions

\cos\theta=\frac{|a_1a_2+b_1b_2+c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}}

Worked example

The direction cosines of two lines satisfy

l+m+n=0

and

2lm+2ln-mn=0

. Find the angle between the lines.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 1 · Q70Moderate

Example 2 · Three Dimensional Geometry · Direction Cosines and Equations of Lines

Let the direction cosines of two lines satisfy the equations:

4\mathcal{l +}m - n = 0

and

2mn + 10n\mathcal{l +}3\mathcal{l}m = 0

. Then the cosine of the acute angle between these lines is :

Remove a variable with the LINEAR equation first

Only after substituting from the linear equation is the quadratic homogeneous in two variables. Dividing the original quadratic by

l^2

straight away leaves three unknowns.

Drill 3 more on two lines from a pair of equations in l, m, n

Concept 3 of 5: The angle between two lines, perpendicular and parallel lines

The angle between two lines is the angle between their direction vectors, so it comes from the dot product. Perpendicular lines have zero dot product; parallel lines have proportional direction ratios. The one care point is reading the direction ratios: the symmetric form needs

x

y

z

with coefficient

+1

in each numerator.

Definition

$\cos\theta=\frac{|\vec d_1\cdot\vec d_2|}{|\vec d_1||\vec d_2|}$ .
Perpendicular: $a_1a_2+b_1b_2+c_1c_2=0$ . Parallel: $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ .
Rewrite first: $\frac{2-x}{3}$ is $\frac{x-2}{-3}$ ; $\frac{3y-2}{k}$ is $\frac{y-\frac23}{k/3}$ .
The angle between two sides of a triangle at $P$ uses $\vec{PQ}$ and $\vec{PR}$ .

Angle between two lines

\cos\theta=\frac{|\vec d_1\cdot\vec d_2|}{|\vec d_1|\,|\vec d_2|}

Worked example

Find the angle between

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

and

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 5 Apr 2024 · Q80Moderate

Example 3 · Three Dimensional Geometry · Direction Cosines and Equations of Lines

If the line

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

makes a right angle with the line

\frac{x + 3}{3\mu}=\frac{1 - 2y}{6}=\frac{5 - z}{7}

, then

4\lambda + 9\mu

is equal to :

Coefficient of $x$ , $y$ , $z$ must be $+1$

\frac{2-x}{3}

the direction ratio for

x

-3

, and in

\frac{3y-2}{k}

it is

\frac{k}{3}

for

y

. Reading the denominators as they stand gives a wrong angle.

Drill 7 more on the angle between two lines, perpendicular and parallel lines

Concept 4 of 5: Writing a line: through a point, two points, or perpendicular to two lines

A line needs a point and a direction. In vector form it is

\vec r=\vec a+\lambda\vec d

; in symmetric form

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

. When a line must be perpendicular to two given lines, its direction is the cross product of their directions. To step a known distance along a line, use the unit direction.

Definition

Point and direction: $\vec r=\vec a+\lambda\vec d$ .
Two points $A$ , $B$ : direction $\vec{AB}$ .
Perpendicular to $\vec d_1$ and $\vec d_2$ : direction $\vec d_1\times\vec d_2$ .
Points at distance $k$ from $A$ on the line: $A\pm k\,\hat d$ .
Where it meets a coordinate plane: set that coordinate to $0$ in the parametric point.

Parametric point and a common perpendicular direction

(x_1+a\lambda,\ y_1+b\lambda,\ z_1+c\lambda),\qquad \vec d=\vec d_1\times\vec d_2

Worked example

A line through

(1,2,3)

is perpendicular to the directions

(1,1,0)

and

(0,1,1)

. Where does it meet the

xy

-plane?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q131Moderate

Example 4 · Three Dimensional Geometry · Direction Cosines and Equations of Lines

Let a straight line

L

pass through the point

P(2, - 1,3)

and be perpendicular to the lines

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

and

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

. If the line L intersects the yz -plane at the point Q , then the distance between the points P and Q is :

Step with the UNIT vector

A+k\vec d

moves

k|\vec d|

, not

k

. For a stated distance, divide

\vec d

by its length first.

Drill 9 more on writing a line: through a point, two points, or perpendicular to two lines

Concept 5 of 5: Triangles and tetrahedra with coordinates

Two edges from one vertex span a triangle: half the length of their cross product is its area. Three edges from one vertex span a tetrahedron: a sixth of their scalar triple product is its volume. Right angles show up as zero dot products.

Definition

Area: $\frac12|\vec{AB}\times\vec{AC}|$ .
Volume of a tetrahedron: $\frac16\left|[\vec{AB}\ \vec{AC}\ \vec{AD}]\right|$ .
Right angle at $A$ : $\vec{AB}\cdot\vec{AC}=0$ .
Three mutually perpendicular edges at $A$ : the opposite face's area squared is the sum of the other three faces' areas squared.

Area and volume

[ABC]=\tfrac12|\vec{AB}\times\vec{AC}|,\qquad V=\tfrac16\left|[\vec{AB}\ \vec{AC}\ \vec{AD}]\right|

Worked example

Find the area of the triangle with vertices

(1,0,0)

(0,2,0)

(0,0,3)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 28 Jan 2025 · Q56Moderate

Example 5 · Three Dimensional Geometry · Direction Cosines and Equations of Lines

Let

A(x,y,z)

be a point in xy-plane, which is equidistant from three points

(0,3,2),(2,0,3)

and

(0,0,1)

. Let

B = (1,4, - 1)

and

C = (2,0, - 2)

. Then among the statements (S1):

\bigtriangleup ABC

is an isosceles right angled triangle and (S2) : the area of

\bigtriangleup ABC

\frac{9\sqrt{2}}{2}

Half for a triangle, a sixth for a tetrahedron

The cross product's length is a parallelogram's area and the triple product is a parallelepiped's volume. Forgetting the

\frac12

or the

\frac16

doubles or sextuples the answer.

Drill 2 more on triangles and tetrahedra with coordinates

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Direction cosines and direction ratios
Direction cosines from direction ratios
$(l,m,n)=\frac{(a,b,c)}{\sqrt{a^2+b^2+c^2}},\qquad l^2+m^2+n^2=1$
Two lines from a pair of equations in l, m, n
Angle between the two directions
$\cos\theta=\frac{|a_1a_2+b_1b_2+c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}}$
The angle between two lines, perpendicular and parallel lines
Angle between two lines
$\cos\theta=\frac{|\vec d_1\cdot\vec d_2|}{|\vec d_1|\,|\vec d_2|}$
Writing a line: through a point, two points, or perpendicular to two lines
Parametric point and a common perpendicular direction
$(x_1+a\lambda,\ y_1+b\lambda,\ z_1+c\lambda),\qquad \vec d=\vec d_1\times\vec d_2$
Triangles and tetrahedra with coordinates
Area and volume
$[ABC]=\tfrac12|\vec{AB}\times\vec{AC}|,\qquad V=\tfrac16\left|[\vec{AB}\ \vec{AC}\ \vec{AD}]\right|$

Watch out for (5)

Ratios are not cosines until you divide
→ Direction cosines and direction ratios
Remove a variable with the LINEAR equation first
→ Two lines from a pair of equations in l, m, n
Coefficient of $x$ , $y$ , $z$ must be $+1$
→ The angle between two lines, perpendicular and parallel lines
Step with the UNIT vector
→ Writing a line: through a point, two points, or perpendicular to two lines
Half for a triangle, a sixth for a tetrahedron
→ Triangles and tetrahedra with coordinates

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