In this presentation we will see about the Three Dimensional Geometry of class12.For more such presentation please comment below and follow me.

1. MATHEMATICS PROJECT
Class : XII
Subject : Mathematics
Chapter : Three Dimensional Geometry
By: Tarun Singh
Shubham Maurya
Sakshat Yadav
Vaibhav
OPGM SR.SEC. SCHOOL
1

2. • Introduction: The topic was introduced in class XI in which we
have studied the how to provide position to a point in space which
has been divided in octants made by three axes X, Y & Z, distance
between two points and section formula. In previous chapter of
Vector Algebra we have studied some basic concepts of vectors
which we will use in the topic of Three Dimensional Geometry. The
purpose of the topic is to develop imagination and visualisation in
three dimension and to learn the conversion of vector form to
cartesian form and vice versa.
• Content:
• Direction Cosines & Direction Ratios of a line in space (dc’s & dr’s)
• Equation of a line in space
2

3. Direction Cosines and Direction Ratios
i) Direction Angles: As we have
studied in two dimensional
geometry each line is having
slope in terms of tangent of the
angle made by the line with
positive direction of X axis.
Similarly in three dimensional
geometry if a directed line is
passing through origin makes
angles ,  and  with positive
directions of X, Y and Z axes
respectively then the angles are
known as direction angles.
(angels are shown in adjacent
figure)
3

4. ii) The cosine of these directed angles, cos, cos, and cos are
termed as direction cosines of the line with general notation l, m
and n respectively. These can be obtained using above figure as
follows-
Note-1: If the direction of the line is reversed then direction
angles will be replaced by  - ,  -  and  -  i.e. the signs of
direction cosines are reversed.
1
1
cos
,
cos
,
cos
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2


































n
m
l
z
y
x
z
y
x
z
y
x
z
z
y
x
y
z
y
x
x
n
m
l
as
obtained
be
also
can
relation
a
and
z
y
x
z
n
z
y
x
y
m
z
y
x
x
l 


4

5. Note- 2: If the line is not passing through origin then a line
passes through origin can be drawn which is parallel to the given
line then the direction cosines of both the lines will be
proportional.
Note- 3: Three numbers which are proportional to the direction
cosines of a line are said to be direction ratios of the line. If l, m
and n are dc’s of a line then its dr’s will be as a =l, b = m and c
= n where  R-{0}.
.
'
,
,
1
1
)
(
1
1
,
,
)
(
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
line
the
of
s
dc
are
these
c
b
a
c
n
c
b
a
b
m
c
b
a
a
l
c
b
a
k
c
b
a
k
k
c
k
b
k
a
n
m
l
that
know
We
ck
n
bk
m
ak
l
let
k
c
n
b
m
a
l







































5

6. Note-4: The dr’s of a line segment with end points at P(x1, y1, z1) and
Q(x2, y2, z2) are x2 – x1, y2 – y1, z2 – z1 or x1 – x2, y1 – y2, z1 – z2 after
getting dr’s of line segment as a, b, c we can find dc’s l, m, n.
Example-1: If a line make 90o,135 o,45 o with the x, y ,and z axes
respectively find it direction cosines.
Solution: The dc’s can be obtained as
l = cos90=0, m = cos135= , n = cos45=
Example-2: If dr’s of a line are 6, 2, 3 then find its dc’s.
Solution: a = 6, b = 2, c = 3 then
2
1

2
1
7
3
,
7
2
,
7
6
3
2
6
3
,
3
2
6
2
,
3
2
6
6
2
2
2
2
2
2
2
2
2














n
m
l
n
m
l
6

7. Example-3: Find the direction cosines of the line joining of the points (7,
-5, 9) & (5, -3, 8).
Solution: The dr’s of line joining of A(7, -5, 9) & B(5, -3, 8) are
 5 – 7, – 3 + 5, 8 – 9 i.e. –2, 2, –1
Now dc’s of line joining of A & B can be obtained as
Example-4: Show that the points (1, –2, 3), (2, 3, –4), (0, –7, 10) are
collinear.
Solution: The similar problem had been solved by us in class XI using
section formula as well as in Vector Algebra. Now we will see how to
show using dr’s & dc’s. The points A (1, –2, 3), B(2, 3, –4), C(0, –7, 10)
will be collinear if dr’s of AB will be proportional to the dr’s of BC.
dr’s of AB 1, 5, – 7 and dr’s of BC  – 2, – 10, 14
Since
therefore A, B & C are collinear.
3
1
,
3
2
,
3
2
3
1
,
3
2
,
3
2









 n
m
l
or
n
m
l
14
7
10
5
2
1 




7

8. Equation of line in space
1. Equation of a line passes
through a point and parallel to
given vector:
Let l be a line passes through A
whose position vector is 𝑂𝐴 = 𝑎
and it is parallel to a vector 𝑏.
If P is a general point on the line
whose position vector is 𝑂𝑃 = 𝑟
then 𝐴𝑃 = 𝑏 𝑂𝑃 − 𝑂𝐴 = 𝑏
𝑟 − 𝑎 = 𝑏 or 𝒓 = 𝒂 + 𝒃 is
the required equation of the line
and known as vector form of
straight line.
8

9. If 𝑟 = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘, 𝑎 = 𝑥1𝑖 + 𝑦1𝑗 + 𝑧1𝑘 and
𝑏 = 𝑎𝑖 + 𝑏𝑗 + 𝑐𝑘 then the cartesian form the line will be
as
𝑥−𝑥1
𝑎
=
𝑦−𝑦1
𝑏
=
𝑧−𝑧1
𝑐
.
Example-5: Find the equation of line passes through
(5, - 2, 4) and parallel to the vector 2𝑖 − 𝑗 + 𝑘.
Solution:
Vector form: 𝑟 = 𝑎 + 𝑏
 𝑟 = 5𝑖 − 2𝑗 + 4𝑘 +  2𝑖 − 𝑗 + 𝑘
Cartesian form:
𝑥−𝑥1
𝑎
=
𝑦−𝑦1
𝑏
=
𝑧−𝑧1
𝑐

𝑥−5
2
=
𝑦+2
−1
=
𝑧−4
1
Note-5: In above example we can observe the conversion of
vector form to Cartesian or vice versa.
9

10. If 𝑂𝐴 = 𝑎 = 𝑥1𝑖 + 𝑦1𝑗 + 𝑧1𝑘 and
𝑂𝐵 = 𝑏 = 𝑥2𝑖 + 𝑦2𝑗 + 𝑧2𝑘 then catesian form of the line will be
as
𝒙−𝒙𝟏
𝒙𝟐−𝒙𝟏
=
𝒚−𝒚𝟏
𝒚𝟐−𝒚𝟏
=
𝒛−𝒛𝟏
𝒛𝟐−𝒛𝟏
Example-6: Find the equation of line passes through A(3, 4, -7) &
B(5, 1, 6).
Solution: A(3, 4, -7)  𝑎 = 3𝑖 + 4𝑗 − 7𝑘
B(5, 1, 6)  𝑏 = 5𝑖 + 𝑗 + 6𝑘
Vector form: 𝑟 − 𝑎 = 𝑏 − 𝑎
 𝑟 − 3𝑖 + 4𝑗 − 7𝑘 = 5𝑖 + 𝑗 + 6𝑘 − 3𝑖 + 4𝑗 − 7𝑘
𝑟 − 3𝑖 + 4𝑗 − 7𝑘 = 2𝑖 − 3𝑗 + 13𝑘
Cartesian form:
𝒙−𝒙𝟏
𝒙𝟐−𝒙𝟏
=
𝒚−𝒚𝟏
𝒚𝟐−𝒚𝟏
=
𝒛−𝒛𝟏
𝒛𝟐−𝒛𝟏
𝑥−3
5−3
=
𝑦−4
1−4
=
𝑧+7
6+7

𝑥−3
2
=
𝑦−4
−3
=
𝑧+7
13 10

11. Thank you