Okay, let's solve this step-by-step: * Given: Length of line AB (TL) = 90mm θ (inclination with HP) = 45° TV makes an angle of 60° with VP * To find: Inclinations with planes (θ, Ø), projections of line AB * Since θ is given as 45°, draw FV making an angle of 45° with XY line. * TV makes an angle of 60° with VP. So draw TV making an angle of 60° with XY line. * TV gives the length of TL when it is parallel to XY line. So TL = 90mm. * This gives the projections of line AB.

1. ORTHOGRAPHIC PROJECTIONS
OF POINTS, LINES, PLANES, AND SOLIDS.
TO DRAW PROJECTIONS OF ANY OBJECT,
ONE MUST HAVE FOLLOWING INFORMATION
A) OBJECT
{ WITH IT’S DESCRIPTION, WELL DEFINED.}
B) OBSERVER
{ ALWAYS OBSERVING PERPENDICULAR TO RESP. REF.PLANE}.
C) LOCATION OF OBJECT,
{ MEANS IT’S POSITION WITH REFFERENCE TO H.P. & V.P.}
TERMS ‘ABOVE’ & ‘BELOW’ WITH RESPECTIVE TO H.P.
AND TERMS ‘INFRONT’ & ‘BEHIND’ WITH RESPECTIVE TO V.P
FORM 4 QUADRANTS.
OBJECTS CAN BE PLACED IN ANY ONE OF THESE 4 QUADRANTS.
IT IS INTERESTING TO LEARN THE EFFECT ON THE POSITIONS OF VIEWS ( FV, TV )
OF THE OBJECT WITH RESP. TO X-Y LINE, WHEN PLACED IN DIFFERENT QUADRANTS.
STUDY ILLUSTRATIONS GIVEN ON HEXT PAGES AND NOTE THE RESULTS.TO MAKE IT EASY
HERE A POINT A IS TAKEN AS AN OBJECT. BECAUSE IT’S ALL VIEWS ARE JUST POINTS.

2. NOTATIONS
FOLLOWING NOTATIONS SHOULD BE FOLLOWED WHILE NAMEING
DIFFERENT VIEWS IN ORTHOGRAPHIC PROJECTIONS.
OBJECT POINT A LINE AB
IT’S TOP VIEW a ab
IT’S FRONT VIEW a’ a’ b’
IT’S SIDE VIEW a” a” b”
SAME SYSTEM OF NOTATIONS SHOULD BE FOLLOWED
INCASE NUMBERS, LIKE 1, 2, 3 – ARE USED.

3. VP
2 nd
Quad. 1ST Quad.
Y
Observer
X Y HP
X
3rd Quad. 4th Quad.
THIS QUADRANT PATTERN,
IF OBSERVED ALONG X-Y LINE ( IN RED ARROW DIRECTION)
WILL EXACTLY APPEAR AS SHOWN ON RIGHT SIDE AND HENCE,
IT IS FURTHER USED TO UNDERSTAND ILLUSTRATION PROPERLLY.

4. Point A is POINT A IN VP POINT A IN
Placed In 2ND QUADRANT 1ST QUADRANT
VP a’
different A A
quadrants a’
and it’s Fv & Tv a
are brought in
same plane for HP OBSERVER
Observer to see
clearly. HP OBSERVER
Fv is visible as
it is a view on
VP. But as Tv is a
is a view on Hp,
it is rotated
downward 900,
In clockwise
direction.The
In front part of a
Hp comes below
xy line and the
part behind Vp HP
comes above. HP OBSERVER
OBSERVER
Observe and
note the a
process. a’
A a’
POINT A IN A POINT A IN
3 QUADRANT
RD
VP 4TH QUADRANT
VP

5. Basic concepts for drawing projection of point
FV & TV of a point always lie in the same vertical line
FV of a point ‘P’ is represented by p’. It shows position of the point
with respect to HP.
If the point lies above HP, p’ lies above the XY line.
If the point lies in the HP, p’ lies on the XY line.
If the point lies below the HP, p’ lies below the XY line.
TV of a point ‘P’ is represented by p. It shows position of the point with
respect to VP.
If the point lies in front of VP, p lies below the XY line.
If the point lies in the VP, p lies on the XY line.
If the point lies behind the VP, p lies above the XY line.

6. PROJECTIONS OF A POINT IN FIRST QUADRANT.
POINT A ABOVE HP POINT A ABOVE HP POINT A IN HP
& INFRONT OF VP & IN VP & INFRONT OF VP
For Tv
For Tv
PICTORIAL PICTORIAL For Tv
PRESENTATION A PRESENTATION
a’ a’
Fo
A
r
Y Y
Fv
Y
Fo
a a’
rFv
S
S . ON
a
X X X A
SE ATI
a
Fo
CA T
r
E EN
Fv
OV E S
AB PR
L IC
AL PH
OF RA
Fv above xy, Fv above xy, Fv on xy,
OG
Tv below xy. Tv on xy. Tv below xy.
TH
OR
VP VP VP
a’ a’
X Y X Y X
a’ Y
a
a a
HP HP HP

7. PROJECTIONS OF STRAIGHT LINES.
INFORMATION REGARDING A LINE means
IT’S LENGTH,
POSITION OF IT’S ENDS WITH HP & VP
IT’S INCLINATIONS WITH HP & VP WILL BE GIVEN.
AIM:- TO DRAW IT’S PROJECTIONS - MEANS FV & TV.
SIMPLE CASES OF THE LINE
1. A VERTICAL LINE ( LINE PERPENDICULAR TO HP & // TO VP)
2. LINE PARALLEL TO BOTH HP & VP.
3. LINE INCLINED TO HP & PARALLEL TO VP.
4. LINE INCLINED TO VP & PARALLEL TO HP.
5. LINE INCLINED TO BOTH HP & VP.
STUDY ILLUSTRATIONS GIVEN ON NEXT PAGE
SHOWING CLEARLY THE NATURE OF FV & TV
OF LINES LISTED ABOVE AND NOTE RESULTS .

8. For Tv Orthographic Pattern
(Pictorial Presentation) V.P.
a’
Note: a’
Fv is a vertical line
A Showing True Length Fv
1.
FV &
.
V .P
Tv is a point. b’
A Line b’
perpendicular Y
X Y
Fo
to Hp B r Fv
& TV a b
Tv a b
// to Vp X
H.P.
Orthographic Pattern
(Pictorial Presentation) For Tv Note: V.P.
Fv & Tv both are
2. // to xy a’ Fv b’
b’ &
.
A Line F.V B both show T. L.
// to Hp a’
.
V.P
Y
& A X Y
// to Vp Fo
rF
v
b a b
Tv
V.
T.
a
X
H.P.

9. Fv inclined to xy V.P.
Tv parallel to xy. b’
3. b’
.
F.V
A Line inclined to Hp B
F.V.
a’ θ
and Y
V.P.
θ
parallel to Vp a’ X Y
θ
(Pictorial presentation) A b
a
T.V. b
.
T.V
a
X H.P.
Orthographic Projections
Tv inclined to xy V.P.
4. Fv parallel to xy.
a’ Fv b’
A Line inclined to Vp F .V
. b’
.
V.P
and a’
parallel to Hp A
Ø
B X Y
(Pictorial presentation) a Ø
Ø Tv
a b
T.V.
b
H.P.

10. For Tv
For Tv
5. A Line inclined to both
b’ Hp and Vp b’
.
. (Pictorial presentation) V.P
V.P B
B
.
F.V
α
.
F.V
α Y
Y
On removal of object a’ For
a’ For i.e. Line AB Fv
Fv
Fv as a image on Vp.
A
A Tv as a image on Hp,
β
β X
X a T.V. b
a T.V. b
V.P.
b’
FV
a’ α
X Y
Orthographic Projections Note These Facts:-
Fv is seen on Vp clearly. Both Fv & Tv are inclined to xy.
To see Tv clearly, HP is a β (No view is parallel to xy)
rotated 900 downwards, Both Fv & Tv are reduced
Hence it comes below xy. TV lengths.
(No view shows True Length)
H.P. b

11. Orthographic Projections Note the procedure Note the procedure
Means Fv & Tv of Line AB When Fv & Tv known, When True Length is known,
are shown below, How to find True Length. How to locate FV & TV.
with their apparent Inclinations (Views are rotated to determine (Component a’b2’ of TL is drawn
True Length & it’s inclinations which is further rotated
α &β
with Hp & Vp). to determine FV)
V.P. V.P. V.P.
b’ b’ b 1’ b’
Fv
θ
FV FV TL
b1
TL α
’
a’ α a’ a’ b2 ’
θ
X Y X Y X Y
Ø b1
a β b1 a
a β TV β
TV TV
Tv
TL
H.P. b H.P. b H.P. b
b
2
Here TV (ab) is not // to XY line In this sketch, TV is rotated Here a’b1’ is component
Hence it’s corresponding FV and made // to XY line. of TL ab1 gives length of FV.
a’ b’ is not showing Hence it’s corresponding Hence it is brought Up to
True Length & FV a’ b1’ Is showing Locus of a’ and further rotated
True Inclination with Hp. True Length to get point b’. a’ b’ will be Fv.
& Similarly drawing component
True Inclination with Hp. of other TL(a’b1‘) TV can be drawn.

12. Projection of straight line
Line inclined to both HP & VP
Type-I
Given projections (FV & TV) of the line. To find True length & true
inclination of the line with HP (θ) and with VP(Φ).
PROBLEM
End A of a line AB is 20mm above HP & 20mm in front of VP while
its end B is 55mm above HP and 75mm in front of VP. The distance
between end projectors of the line is 50mm. Draw projections of the
line and find its true length and true inclination with the principal
planes. Also mark its traces.

13. b’ b1 ’
θ: True inclination of
the line with HP = 24º
TL
55
a’ b2’ α : Inclination of FV of
θ α the line with HP/XY
HT VT’ 20
X Y
v
h’
50
Ø: True inclination of
20
Φ β b1 the line with VP = 41º
a
β : Inclination of TV of
75 the line with VP/XY
TL
b b2

14. Line inclined to both HP & VP
Type –II
Given (i) T.L., θ and Ø,
(ii) T.L., F.V., T.V.
to draw projections, find α, β,H.T. and V.T.
PROBLEM
A line AB, 70mm long, has its end A 20 mm above HP and 20mm in
front of VP. It is inclined at 30° to HP and 45°to VP. Draw its projections
and mark its traces.

15. b’ b1’
70
a’ 30°
HT b2’
15 VT’ Y
X
h’ v
20
b1
a 45°
70
b2
b

16. Q10.11 The top view of a 75mm long line AB measures 65mm,while its front
view measures 50mm. Its one end A is in HP and12mm in front of VP. Draw the
projections of AB and determine its inclination with HP and VP
To draw FV &TV of the line
Given, Hint: Draw ab1=65mm // to XY.
AB
TL=75mm,TV=65mm,FV=50mm Because when TV is // to XY, FV
To find θ & Ø
gives TL.
A is in HP & 12mm VP
→
b’ b1 ’
Ans. θ=31º
50
75
Ans. Ø=49º
a’
X Y
31º
12
65 b1
a
49º
75
65
b b2

17. Q10.12 A line AB, 65mm long has its end A 20mm above H.P. and 25mm in
front of VP. The end B is 40mm above H.P. and 65mm in front of V.P. Draw the
projections of AB and show its inclination with H.P. and V.P.
Given, To draw FV &TV of the line Hint1:Mark a’ 20mm above
AB H.P & a 25mm below XY
TL=65mm
To find θ & Ø
A is 20mm ↑ HP & 25mm →V.P. b’ b1 ’ Hint2:Draw locus of b’ 40mm
B is 40mm ↑ & 65mm → V.P. above XY & locus of b 65 mm
below XY
a’ 65 b2 ’
40
18º
20
X Y
25
38º Ans. θ=18º
b1
a
Ans. Ø=38º
65
65
b b2

18. Q10.13:The projectors of the ends of a line AB are 5cm apart. The end A is
2cm above the H.P and 3cm in front of V.P. The end B is1cm below H.P. and
4cm behind the V.P. Determine the true length and traces of AB, and its
inclination with the two planes
Given,
To find,
A0B0=50mm True Length, θ,Ø, H.T. and V.T.
A is 20mm ↑ HP & 30mm →V.P.
B is 10mm ↓ & 40mm ←
V.P.
b b2
a’
40
HT
b2 ’
20
VT’ 91
X v
h’ Y
50
10
b’ 20º
30
Ans. θ=20º
a 50º b1
Ans. Ø=50º

19. Q10.14:A line AB, 90mm long, is inclined at 45 to the H.P. and its top view
makes an angle of 60 with the V.P. The end A is in the H.P. and 12mm in front
of V.P. Draw its front view and find its true inclination with the V.P.
b’
Given, b1’
T.L.=90mm, θ=45º, β=60º
A is in the H.P. & 12mm→V.P.
To find/draw,
F.V.,T.V. & Ø
90
Ans. Ø = 38º
a’
X Y
45º
b1
12
60º 38º
a
90
b b2

20. Q10.16:The end A of a line AB is 25 mm behind the V.P. and is below
the H.P. The end B is 12 mm in front of the VP and is above the HP The
distance between the projectors is 65mm. The line is inclined at 40 to
the HP and its HT is 20 mm behind the VP. Draw the projections of the
line and determine its true length and the VT
Given, To find/draw,
A0B0=65mm F.V., T.V., T.L., VT’
A is 25mm ←V.P.& is ↓H.P.
B is 12mm →V.P. & is above HP
θ = 40º b1 ’
b’ b2 ’
Ans. TL= a’b2’=123 mm
VT’
a
b2
b1
HT
25
20
X h’
v
Y
12
40º
a’
b
65

21. 10.17:A line AB, 90mm long, is inclined at 30 to the HP. Its end A is 12mm above the HP and
20mm in front of the VP. Its FV measures 65mm. Draw the TV of AB and determine its
inclination with the VP
b’ b1 ’
90
a’ 65
30°
12
X Y
20
44° b1
a
Ans: Ø = 44º
90
b b2

22. Q10.23:Two lines AB & AC make an angle of 120 between them in their FV & TV. AB is
parallel to both the HP & VP. Determine the real angle between AB & AC.
C
c2 ’ c1 ’
c’
112° Ans. 112º
b’ 120° a’
X Y
b a
c2 c1
120°
c

23. Q8:A line AB 65 mm long has its end A in the H.P. & 15 mm in front of the V.P. The end B is
in the third quadrant. The line is inclined at 30 to the H.P. and at 60 to the V.P. Draw its
projections.
VP
b b2
b
65
15
a’ b2’ X Y a”
X Y a’ 30º HP
30º
60º
15
60º 65 b1
a a
b’
b”
b’ b1’

24. Q10.19 A line AB, inclined at 40º to the V.P. has its end 50mm and 20mm above the
H.P. the length of its front view is 65mm and its V.T. is 10mm above the H.P.
determine .the true length of AB its inclination with the H.P. and its H.T.
Given, To find,
Ø = 40º, A is 20mm↑HP, TL, θ & HT
B is 50 mm ↑ HP, FV=65mm, VT is
10mm ↑ HP
b1 ’ b’
85
a’
50
b2 ’ 21º HT
VT’
20
10
X Y
40º v h’
b1 Ans,
a
TL = 85 mm,
θ = 21º &
HT is 17 mm
behind VP
b2

25. Q10.19 A line AB, inclined at 40º to the V.P. has its end 50mm and 20mm above the H.P. the length of its front view is
65mm and its V.T. is 10mm above the H.P. determine .the true length of AB its inclination with the H.P. and its H.T.
B1’
Step1: For solving the problem by
Given, To find, trapezoidal method, draw a line at 40º(Ø)
Ø = 40º, A is 20mm↑HP, from VT’. Then draw perpendiculars from
B is 50 mm ↑ HP, TL, θ & HT
a’ and b’ on this line.
FV=65mm, VT is 10mm ↑ HP
Step2: Then draw projectors from a’ and b’
and mark the distance of b’B1’ on the
projector of b’ below XY. Similarly mark
b’ the distance a’A1’ on the projector of a’
below XY
65 A1 ’
a’
50
40º
VT’ 21º HT
20
X
10
v h’ Y
a
Ans: A1’B1’=TL=85mm
Ans:HT is 17 mm behind VP
Ans:θ = 21º
b

26. Q6. The top view of a 75mm long line CD measures 50 mm. C is 50 mm in front of the VP &
15mm below the HP. D is 15 mm in front of the VP & is above the HP. Draw the FV of CD &
find its inclinations with the HP and the VP. Show also its traces.
Given, To draw,
TL = 75 mm, FV =50 mm, FV & to find θ & Ø
C is 15mm ↓ HP & 50 mm → VP, VT’ To mark HT & VT
D is 15 mm → VP
Hint 1: Cut an arc of 50 mm
d’ d1 ’ from c on locus of D to get
the TV of the line
Hint 2: Make TV (cd), // to XY
75 so that FV will give TL
h’
X Y
v
15
d d2
c’ θ=48º Ans: θ=48º
Locus of D
50
Ans: Ø=28º
50 75
HT
c
d1
Ø=28º

27. Q10.10 A line PQ 100 mm long is inclined at 30º to the H.P. and at 45º to the V.P. Its
mid point is in the V.P. and 20 mm above the H.P. Draw its projections, if its end P is in
the third quadrant and Q is in the first quadrant.
Given,
To draw,
TL = 100, θ = 30º, Mid point M is
20mm↑HP & in the VP FV & TV
End P in third quadrant &
End Q in first quadrant
q’ q1 ’
p2 p
p2’ 50 m’ 50 q2 ’
30º
20
p1 50 45º
X q1 Y
m
p1’ p’ 50
q q2

28. Problem 3: The front view of a 125 mm long line PQ measures 75 mm while its top view
measures 100 mm. Its end Q and the mid point M are in the first quadrant. M being 20 mm
from both the planes. Draw the projections of line PQ.

30. For Tv
For Tv
5. A Line inclined to both
b’ Hp and Vp b’
.
. (Pictorial presentation) V.P
V.P B
B
.
F.V
α
.
F.V
α Y
Y
On removal of object a’ For
a’ For i.e. Line AB Fv
Fv
Fv as a image on Vp.
A
A Tv as a image on Hp,
β
β X
X a T.V. b
a T.V. b
V.P.
B b’
FV
a’ α
A X Y
Orthographic Projections Note These Facts:-
Fv is seen on Vp clearly. Both Fv & Tv are inclined to xy.
To see Tv clearly, HP is a β (No view is parallel to xy)
rotated 900 downwards, Both Fv & Tv are reduced
Hence it comes below xy. TV lengths.
(No view shows True Length)
H.P. b