.-jJIMNII H.

PB½it4it§i ;t1MiMla
Objective ·
•
rify that the relat ion R in the set L of all Imes in a plan e, defin ed by R = {(a, b) : a\\ ban d a, b E L} 1s
Tove . alen ce relat ion •
an equiv

Method of Constructio n
0
II

e it on the table and fiIX a wh 1·t e pape r s h eet t ~ ;-

Take the· drawaling boar d. Plac -;--
d • h Material Required
On it. With sc
e an. penc il draw.
som e para llel 11nes
- an d nam e t em a 1 ,
1 icula r to give n draw n lines and nam e it b . A draw ing board
f½ and a~. Dra ~ a. me perp end_
0
1
. See figur e. ■ A white pape r sheet
Draw a line c1 1ncl med to the give n ·draw n lines II
■ Board pins, penc il and

Demonstration, Observation and Conclusion

scale
II

1. R is refl exiv e rela tion : Sinc e ever

y line is para llel to itsel f as II

a1 II a1, l½ 11 l½ and a3 II a3 etc. Ther efore , the relat ion R is reflexive.

2. Rela tion R is sym met ric:
(a) By cons truc tion , we know that a 1 II l½
and l½ II a 1
⇒ (a 1 , l½) ERan d (l½, ai) E R
⇒ Rela tion R is sym metr ic.
(b) Also by cons truc tion , we know that a 1
II a3 and a3 II a 1
(a 1 , a 3 ) E Ran d (a3 , a 1) E R
⇒ Rela tion R is sym metr ic.
thre e lines .
3. Rela tion R is tran sitiv e: a 1 , ~ and a 3 are
By cons truc tion - a 1 II~ and ~ II a3 , then 1
a II a 3 .
Hen ce (a , (½) E Ran d((½ , a 3 ) E R
1
⇒ (a 1 , (½) e R
Rela tion R is tran sitiv e.
:a II ban d a, b L} is refle xive , sym met ric and tran sitiv e
We have veri fied that the rela tion R = {(a, b)
E

Hen ce, rela tion R is an equi vale nce relat ion.

Applica tion k whe ther a give

activ ity is usef ul (z) in und erst and ing the conc ept of an equi vale nce rela tion . (ii) to chec
This _ _ _ _ _ _ __
_ __ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _
relat ion is an equi vale nce rela tion o_r _n_o_t_. _
 4

f'JoT
Activity 1.4 On e-O ne bu t On to Fu nc tio n

Objective
To dem onst rate a func tion whic h is one- one but not
onto .
Method of Constructio n Q

1. Take the draw ing boar d and place it on a table and ~ 0

fix the whit e pape r
. ed~ ;-
Q

shee t on it with the help of boar d pins . M aten·a1 R eqwr _

2. Take two sets A= {1, 2, 3} and B = {a, b, c, d}.
3. Take three poin ts on the left hand side of the whit
nam e them as 1, 2, 3. It repr esen ts set A. Take
e pape r shee t ~d
■ A draw ing board
■ A white pape r sheet --
0

hand side on the whit e pape r shee t and nam e them

It repr esen ts set B. (see figur e)
4 poin ts on ngh t
as a, b, c and d.
w
■ Board pins, pencil and
scale
-
4. Join the poin t 1 of the set A to poin t a of the set ~

B, poin t 2 of the set A

□
to poin t b of the set B and poin t 3 of the set A to poin
t c of the set B.
Demonstrat ion, Obs ervation and Conclusion
1. The imag e of the elem ent 1 of the set A
is elem ent a of the set B.
The imag e of the elem ent 2 of the set A is elem ent
The imag e of the elem ent 3 of the set A is elem ent
b of the set B.
c of the set B.
SetA

Function f
D
Set B

This dem onst rate that each elem ent of set A has
only one imag e in
set B. Henc e, we conc lude that the func tion fis
one- one.
2. Elem ent a of the set B has pre-i mag e elem ent
1 of the set A.
Elem ent b of the set B has pre-i mag e elem ent 2
of the set A. 3
Elem ent c of the set B has pre-i mag e elem ent 3
of the set A.
And elem ent d of the set B has no pre-i mag e in Set A Set B
the set A.
This dem onst rates that all the elem ents but one
elem ent of the set B has pre- imag e in set A. So,
conc lude that the func tion is not onto . w
Hen ce,th e func tion from set A= {1, 2, 3} to set B
but not onto . = {a, b, c, d} defin ed as {(l, a), (2, b), (3, c)} is one-o
n

~ppli cation
'his activ ity can be used to verif y whe ther a give

IJ
n func tion is one- one but not onto or not.

V Viva Voce
 Objective
'fo verify that for a fu nction J to be continuous t • .
provided l1X is sufficien tly small. a given pomt Xo , ll y = If (Xo + t!.x) - J(.xo) I is arbitrarily small

Method of Construction
1. Paste a white sheet on the hardboard.
2. Draw the curve of the given
the figure.
t'
con muous function as represented in
■
.
Material Required
~
"
;-·
-..
Hardboard, white sheets, _• _
pencil, scale, calculator, •
3. T~_any ~oint A (Xo, 0) on the positive side of x-axis and corresponding •
adhesive.
to . is pomt, mark the point p (Xo, Yo) on the curve.
y

t
,..
I

!>i.x2
Y' L - - - - -Af1- - - -

Demonstration
1. Take one more point M 1 (Xo + t!.x1 , 0) to the right ot A, where llx1 is an increment in x.
2. Draw the perpendicula r from M 1 to meet the curve at N 1. Let the coordinates of N 1 be (Xo + ~x 1, y 0 + ~y 1).
3. Draw a perpendicula r from the point P (Xo , y 0 ) to meet N 1M 1 at T 1.
4. Now mea sure AM 1 = L\.x1 (say) and record it and also measure N 1T 1 = ~y 1 and record it.
5. Reduce the increment in xto ~ ( i.e., ~< M 1) to get another point M2 (Xo + A~ , 0). Get the corresponding
point N2 on the curve.
6. Let the perpendicula r P'l\ intersects N2 M2 at T2 .
7. Again measure AM 2 = ~ and record it.
Measure N2T2 = Ay 2 and record it.
8. Repeat the above s teps for some more points so that Ax becomes smaller and smaller .

Observation
~ - - - - - -- - - - - - -- - - - - - --,------- -- -- - - - ---- -- .

1. ~ 8.No. ' Value of increment in Xo _ - ~ - - Corresponding increment in y

1Ax1 I = 5 cm lt1y 1 I = 0.7 cm
I 1.
.__~~ I _= _3 _c_m_
2_. ---J._ _ _ _ ___!IA
f _-_
!11~ I = 1 cm
_ _ __ +--_ _ _ _,__ y 2l
111:::. ~i·4 cm __
li1y3 I = 0 .2 cm
-~~---=~---
3.
!- _ 4_. _.J..--_ _ __ I11_x.2.4 ..!...I_=_O_.S_ cm
__!_ _ _ _ --·----·
111--=y'-'-4 ..!...l_=_0_. l_ cm
_ ____ _--+- - - - - - ' - -
S.
__ __ ___ __ _ IAxg_l_ = 0 .2 c~- _ _ ___ _ _ J i1y5 I =0.05 cm_.
 ~
- -~ - - - - - - - - -- - l,1y6 I = 0.01 cm
6. l,1Xu I 0.1 cm 'i

--
-----.J
0cm
7.
8.
l-1~ I - 0.05 cm
IAxs I -0.03
j-1.xg I - 0.01
cm
1,1y7 \
l,1Ys I
l,1y9 I
0cm
0cm
-
----
I

I
9.
2. So '1y becomes O when .1x becomes smaller.
cm
J
3. Thus lim .1y = 0 for a continuous function.
LU" ➔ O

Application
This activity is helpful in explaining the concept of derivative (left hand or right hand) a t any point on th e
.
curve corresponding to a fu nct10n.

000
f@ iil IIt§ Uh ii!¥! If· ► ·IM HH5' i\ IiMM 111
ObJecttve ~
To understan d the concept of decreasin g and increasin g fu nctions .

..,. •:
•
t
Method of Construction
1. r ake the drawing board and place it
on a table . Fix a white-pap er II
Material Required I)

sheet on it with the help of board pins . t

2. Draw two perpendic ular lines X'OX and YOY' on the white paper
• A drawing board
i
• A white paper sheet
sheet intersecti ng at 0 . These lines represent s coordinate axes . X'OX • Trigonometric tables ~

is x-axis and YOY' is y-axis . Board pins, pencil and

scale
• a
3. Draw two curves on the white paper as shown in the figure . Firs t ij
II
curve is graph of y = f(x) and second curve is graph of y = ~(x) .
4. Draw tangents to these curves as three different points with the help
of scale and pencil in such a way that these tangents cut x-axis . The tangent to left curve makes angles
a. 1, a. 2 and a 3 with the positive direction of x-axis.
s. Tangents on right hand sid e curves makes , p1, p2 and p3 , angles with positive direction of x-axis .
y

Y'

Demonstration
1. It is very much clear in the figure, that the angles a 1, a 2 and a 3 which the tangents on
left side curve
and
and making with positive direction of x-axis , are obtuse angles. Hence the values tan a 1 , tan a 2 ,
tan a 3 have negative values.
This means the derivatives of the function y = f(x), P1 , P2 and P3 are negative.
Hence, this curve of y = f(x) (on left) represent s a decreasin g function .
three
2. Take curve of the function y = ~(x) (on the right side). The three tangents to this curve makes
angles P1, p2 and p3 with the positive direction of x-axis . All these angles are acute.
Therefore the values of tan p1, tan P2 and tan P3 are positive. So the derivative s of the function y =
~(x)
g
at these points are positive. Thus the function y = ~(x) whose graph is on right side is an increasin
fu nction .

Observation
You can ~erify by actual measuremen t of valu ~~ of angles a 1 , a. 2 , a. 3 , P1, P2 a nd p3 tha t tan a. 1, tan a. 2 , tan a. 3
are negative and tan p1, tan p2 , tan p3 are pos1t1ve.
 1. a,1=(_ _)>goo, 0.2 =(_ _)>goo, 0.3 = ( ) > goo
:. tan a 1 = ( ) (negative), tan a 2 = ( ) (negative), tan a3 = (_ _). (negative).
⇒ The derivatives of y = f(x) at these three points are negative and hence the given function y == f{x)
is an decreasing function.
2 . ~ 1 = (_ _), < goo' ~2 = ( ) < goo' ~3 = ( ) < goo
:. tan ~1 =(_ _)(positive), tan ~ =(_ _)(positive), tan ~3 = (_ _) (positive).
2
~ T~e derivatives of y = q,(x) at these three points are positive and hence the given function y == ${x)
1s an increasing function.

Application

.q l
This activity may be used in explaining the concept of decreasing and increasing functions . J}
✓ Viv:1 Vn~~
 Activity 3.6 W or d Pr ob le m s
. . M .
M 1n 1m a on ax im a an d .
> .
'
~
Objective
To con struct an ope n box of max imu
from eac h com er. m ·
vo1um e from a given rect ang ular she
et by cutt ing equal squ

.
are pieces
Method of Construction
l. Tak e a rec tan gul ar cha rt-p
as rect ang le ABC a er O f .
D. p size 45 cm x 24 cm and nam e it .,,. ;-
2. Cut-ou t fou r equ al squ are pie Material Required • ·
Now fold up the fl f .
t f ces o size 1 cm x 1 cm from eac h corn • Chart papers of size _
.th th aps O
orm an ope n rect ang ular box from er.
this cha rt 45 cm x 24 cm, ce\\otape •
pha?ehrtwif th" e help of cell otap e.
e1g o 1s ope b Me asu re the leng th bre adth and ■ pencil, scissor, calculator - .-
n ox · h the
wit '
help of scale. Work out the volume
of the ope n box usi ng the form ula
V = leng th x bre adth x height.
3. The
b ano. the r rec tan gul ar cha rt pap
er s h eet of sam e size
· (45
cm x 24 cm) and rep eat the above
Y.cut ting out squ are s of size (2 pro cess
heig ht and find the vol ume of the cm x 2 cm) from eac h corn er and mea sure its leng th,
ope n box. bre adth and

D 45cm
C
X
X
X
X
E
u I\.)
~ .i,.
C"I 0

X
3
X
X
X
A
45cm B
4. Now tak e a thir d rec tan gul ar
cha rt pap er sheet of sam e size (45
squ are pie ces of size 3 cm x3 cm cm x 24 cm). Again rem ove equ al
from eac h cor ner and mea sur e its
find its vol ume . leng th, bre adt h and heig ht and
5. Go on r epe atin g this pro ces s
by cu ttin g out equ al squ ares piec
cm, 5 cm x 5 cm , 6 cm x 6 cm and es from eac h com er of size s 4 cm
7 cm x 7 cm . In eac h cas e mak e an x 4
up the flap s . Now me asu re the leng ope n rect ang ular box by folding
th , bre adth and h eigh t of the eac
h box and calc ulat e their vol~me.
Demonstration
1. Wh en x = 1, z = (4 5 - 2) cm =
43 cm, b = (24 - 2) cm = 22 cm and
:. Vol ume V = (43 x 22 x 1) cm3 heig ht = 1 cm
1 = 946 cm3
2. Wh en x = 2 cm, z = (45 - 4) cm
= 41 cm; 6 = (24 - 4) cm = 20 cm
:. Vol ume V = (41 x 20 x 2)cm 3 and h = 2 cm
2 = 164 0 cm 3
3. Wh en x = 3cm , z = (45 - 6) cm
= 39 cm, b = (24 - 6) cm = 18 cm
and heig ht = 3 cm
:. Vo lum e= V = 39 x 18 x 3 cm3
3 = 210 6 cm3
4. Wh en x = 4 cm, z = (45 - 8) cm
= 37 cm, b = (24 - 8) cm = 16 cm
and h = 4 cm
:. Vo lum e= V = 37 x 16 x 4 cm3 3
4 = 236 8 cm
5. Wh en x = 5 cm, z = (45 - 10)
cm = 35 cm, b = (24 .,.. 10) cm =
14 cm and h = 5 cm
:. Vo lum e= v = 35 x 14 x 5 cm3 3
5 = 245 0 cm
6. Wh en x = 6 cm, z = (45 - 12)
cm = 33 cm, b = (;4 - 12) cm =
12 cm and h = 6 cm
. . Vol ume = v = 33 x 12 x 6 cm3
6 = 237 6 cm
7. Wh en x = 7 cm, z = (45 - 14)
cm = 31 cm, b = (;4 - 14) cm =
10 cm and h = 7 cm
:. Vo lum e= v = 31 x 10 x 7 3
7 cm = 217 0 cm
Observation
('t n, tl w n \ ' \
- ,).H, \'n /
1. Whn , x .... 1 h ) l' \\\ J
2. Whc'n \' - 2 . :'
m. tlw n Y., -- l ()·
t>n \ ';~- '..' 1Ot , rn
\' 1
'n \'" ,, 'm
~ . th
3 . Wht
tn . tht'n \' ..
- -~J t,~ t' t\ \J
-i. Wh,'n x " ~i \..' l'l l \' 1 (Mmdm
un, Vnh1111<')
\ , tlw n \ ', - !
:.-1 :>o
5. \\'hr1, x • !'i \'.U 'nr'
hc 'n x • (' n n. tlw n \' c, ·· ~~J 7b 1.
~- W l) "'n \'' to be
n., tlwn V ~. • 2 17 he n size of sqti are
i. W ht\t'\ x • 7 n 1ximum " 24 S 0 cm 3 w
w ,· < 't> tw lu d, nt1.vnlu111c I" 1111
· ~thcn
whlueo nw r i~ ~ em
r ,·n ,,,a _' ,
Fr om th,: ,tbtwc-~
t~ Jlh) \'W from

Application m in in m m va lu e of
a fi m c-tion .
m.:i11n1 01 or
\'O t\('<' p t pf m
ti, ity is ns d u l in c'..-,:p ln in in~ th<'
is ae
 jtMMMMMl41'64Ftl¼BM·l·li8Pi5i,d¥!!l·IDdlii1l1'!JI
~
ObJettlve ~
~ mat amongst all the rectangles of the same perimeter. t11e square has a maximum area .
10
Metllod of Construction
l. Take the drawing board and fix the white paper sheet ·with the help
of ooard pins on it.
2. Take perimeter of the rectangle= 48 cm and draw the rectangles on Material Required
~ -;--
;- •
white paper sheet in such a way that perimeter of rectangles is 48 ■ A drawing board
cm and different lengths and breadths. ■ A white paper sheet
■ Scale, pencil and board
3. Cut all the rectangles of following dimensions : @

pins
4. p = 48 cm , 11 = 16 cm, b 1 = 8 cm ⇒ A1 = 11 x b1 = 16 cm x 8cm
2 ~
= 128 cm .
5. p = 48 cm,~= 15 cm , b2 = 9 cm ⇒ A2 = l2 x b2 = 15 cm x 9 cm = 135 cm 2 .
6. p = 48 cm,~= 14 cm , b3 = 10 cm ⇒ A3 = 13 x b3 = 14 cm x 10 cm = 140 cm 2 .
7. p = 48 cm, i. = 13 cm , b4 = 11 cm ⇒ ~ = 4 x b4 = 13 cm x 11 cm= 143 cm 2 .
8. p = 48 cm, ls= 12 cm , b5 = 12 cm ⇒ A5 = ls x b5 = 12 cm x 12 cm= 144 cm 2 .
9. p = 48 cm , 4, = 11 cm , b6 = 13 cm ⇒ ~ = 4, x b6 = 11 cm x 13 cm= 143 cm2 .
10. p = 48 cm, l, = 10 cm, b-, = 14 cm ⇒ A7 = 17 x b-, = 10 cm x 14 cm = 140 cm 2 .
P = 48 cm P=48crn P=48crn P = 48 cm

~ ~ R3 R4

~ i I A, !~ a 135 cm• A3 = 140 cm

2 0
()

3
A4 = 143 cm
2 ::::
()
3
16 cm
15cm
14 cm
13 cm

P=48 cm P=48crn P=48cm

R7

2 w ~
~ = 143 cm n ~ 7 = 140 cm n
3 3

12 cm
11 cm
10 cm

Demonstration
1. Area of rectangle R 1 = 128 cm 2 2 . Area of rectangle R2 = 13 5 cm 2
2
3. Area of rectangle R3 = 140 cm 4 . Area of rectangle~ = 143 cm 2
5. Area of rectangle Rs = 144 cm2 6. Area of rectangle R6 = 143 cm 2
2
7. Area ofrectangle R7 = 140 cm
8. Perimeter of each rectangle is same but their areas are different. Area of rectangle R 5 = 144 cm2 , this
IB maximum area . This can be verified using theoretical description in the note (given at last).

Observation
L Perimeter of each rectangle R 1, ~ , R3 , R4 , Rs, R6 , R7 = 48 cm .
2. Area of the rectangle ~ is greater than the area of the rectangle R 1 .
3. Area of the rectangle R3 is greater than the area of the rectangle R .
2
4 • Area of the rectangle R4 is greater than the area of the rectangle R .
3
5. Area of the rectangle Rs is greater than the area of the rectangle~ -
 6. Area of the rectangle Rti is smaller t hon tlw un·u nf thr ret·1m,t,th.· k "
7. Area of the rectangle R7 is stn u.ller tho.11 the areu vi th r r-r~tnnHk K,i
8. The rectangle R5 ha~ the di rnen~ion 12 cm " 12 r rn nud h r rw r 1t 1~ 11 M1w u t'
9. Of al.I the rectangles with sume perimr tt•r the sq1 lf\H· hu~ the m 1t.\11tnu11 llfl•n.

Application
This activity e.xplai n the ronc: t pl of mnxi11n1111 "llh ir ol n fu nr ttun ThtR com r pt 1·nn he WH•d in pr
economical packages . t' J1nri11
Notea: The following mnthem rmrnl prnuf vt•rln t•s th111. the nlmvr 1tr lt v 11v 1-t < CJ rrrct
Oi\'t~n Perimet('r .. ·lo t' ll \
2 (/ + b) • 48 (J ).
I ➔ I) • 2·l
Let
b • x c 111 , t hen I • (l ·l - .\) t rn
Hence area • A .. lb • x (2 4 - .\i I• 24 - I

⇒
A • 24x - .~
dA
- - • ? -l - ' J \'
dx ... "'
d JA
For ma..vJmu m or minimum vaJue of x, - 1 • Q
dx·

Now

--;,
, X .. 1 ). n1 HX 11lll ~ t' t> f\
1hu,. b • 12 cm and J .. 2·1 - x- 2·1 - 12 .. ] ). ern .
Htnce- tht" reetnnvle o ( 1,c · - H,
l') nm ctc r 'H > l ll'\ bn lJll\c' !\ 11 ifllllllt" t • . .
nrtn 1s nmx.imt1m . ' 1 pt r 1m, tc-1 ·IH t rn ,u1el 11t1 d ,·
12 r m whrn 11
 Activity 6.2 Shortest Distance between Two Skew
Lines

Objective
fo measure the shortest distance between two skew lines and verify it analytically.

Method of Construction O
•

,1 . Take the drawing board and fix a squared paper on it - - ~ ~ •

with the help of board pins. o _
Material Required
O ■ A drawing board, board pins
2. Q_g__tbe squared paper draw two perpendicular lines OA •
-and OB intersecting at 0, representing x-axis and y-axis -
· 1 ■ A white paper sheet •
respective y. ■ Three wooden blocks of size 2 cm x
and one wooden, -
3. Take the three blocks of the size 2 cm x 2 cm x 2 cm as 2 cm x 2 cm each
!, II and III. Name the 4th wooden block 2 cm x 2 cm x block of size 2 cm x 2 cm x 4 cm, wires~
4 cm as IV• of different lengths, set squares, adhesive,
4. Place blocks I, II, III such that their base centres are at pen/pencil, scale etc. •
the points (2 , 2) , (1 , 6) and (7, 6), respectively and block ----"'-...,_""'_ ,- •
0
IV with its base centre at (6, 2).
s. Place a straight wire joining the points P and Q, the centres of the bases of the block~ I ~d. 11 and
another wire joining the centres R and S of the tops of the blocks I and III and another wire Jommg the
centres Rand S of the tops of the blocks II and IV as shown in figure.
y
2 3 4 5 6 7 8 A
.x .
0

IV
·1 S(6, 2, 4)

2
2

(_
/
--
3

2
6
Q(7, 6)

7 ~ ? : r---- ----; !'

. l ,, l
. r~ ---- _[
i
I 11. ,, \
8
r I \[,-;:;..:JI . t
; I! .
B . t' i . .

;
I;
I
,1
/ "· ~
,'/
.. -.-- · ··-1
I I
y •
..
l
,,
• l,)/

6. These two wires represent two skew lines.

7. Take a wire and joih it perpendicularly with the skew lines and measure the actual distance.

Demonstration
1. A set-square is placed in such a way that its one perpendicular side is along the wire PQ.
2. Move the set-square along PQ till its other perpendicular side touches the other wire.
 ~'
-~
3. Measure the dis tan .
ce betwe~n the two lin in this position us in · g se t
-s qu ar e. Th.is is
· the sh
es
distance between two_ sk
4. Find the equation of ~~ ~e s. d Orte~t .
lineJommg P( 2 , 2 , O) an Q(7 , 6 , O) an d ot he r line joining R( 1, 6 , 2) an
drs of the line PQ : (5 , 4, d S(6 ,
0) an d fixed point (2 2
,A 2, 0~ -: A.
, 4).
Equation of line PQ : f =· (2i + 2j +Ok)+ A.(
51 + 4J + Ok)
,..,

drs of line RS : (5, -4 , 2)

an d fixed point (l , 6 2, )
Equation of line RS : A A ,..,

1
f = (i + 6 + 2k) + (S i-
~j + ~k ) ,..,
Here , ;;1 = 2i + 21A·+ ok , m1 =
Si + 4 J + Ok
f2 = i + 6] + 2k, ~ = S
i - 4] + 2k
A ,..,
J k
m1 X ~ = 5 4 o = i (8 _ 0) - ] (10 - 0) + k:( - 20 - 20 ) = Bi -
10 ] - 40 k
5 -4 2
Also Im1 x ~ I = ✓64 + 100 + 16 00 = -./17
64
ri -r2 (2i +2 ]+ ok )- (i +6 ] +2k)
=
= i- 4 ]- 2 k
(fi -r2 ) -(m x~ l = (i - 4] -2 k)
1 ·(B1-10]-4ok) = B + 40
+ Bo= 128
SD = w
1 - r2 ) • (m1 x ~ ll = 12 8 = 12 8
(m1 X ~ ) ✓1764 42
SD = -12 8 = -64 = 3 um.
42 ts
The di sta nc es between 21
two skew lines will appr
oximately sa m e.
Observations
1. Coordinates of po
in t Pa re (2, 2, 0).
2. Coordinates of point
Q ar e (7 , 6, 0).
3. Coordinates of po in t
R ar e (1 , 6, 2).
4. Coordinates of po in t
Sa re (6, 2, 4).
5. Equation of line PQ
is r = (2i + 2] +Ok)+ A.(
Si + 4] +O k).
6. Eq ua tio n of line RS
is r = (i + 6] + 2k) + µ(S
i - 4] + 2k).
7. Sh or te st di st an ce be
tw ee n PQ an d RS an al
8. Sh or te st di st an ce be yt ic ally = 3.
tw ee n PQ an d RS by ac
10 . The re su lts so ob tu al m ea su re m en t = 3.
ta in ed ar e equal.

Applicatio n
This ac tivity ca n be us ed
to explain th e concep t of sk
ew lines an d sh or te st di
st an ce be tw ee n two lin
es in space.

✓ Viva