we get
a c
b
and also, we have b c c a That is
a c
c c a
after simplyfing , we get a c ( )c a That is
(1 )( ) 0 a c Which yields 1. Therefore a c 1 b
.
That is a c b 0
. Hence Centroid is 0.
8.Sol: It is a direct question about vector triple
product. First we find
2 1 1 4 6
0 3 2i j k
u v i j k
Let w ai bj
, but w is unit vector, that isa b^2 ^21. Which gives acos and
bsin. Now, we have
( )u v w a (1) 4( ) cos 4sinb Therefore maximum value of ( )u v w
is1 ( 4) 17^2 ^29.Sol: We get around the problem a bit. If we
can take the Point (P) as origin and we simply
write PA a PB b PC c , ,
. We then write
a given equation as
PA PB PC 2 3 0
as a b c 2 3 0
(1)
We have area of a triangle when position
vectors are given as
1
2a b b c c a
(2)
From (1) and (2), we get area of a triangle
as
3
2c a But we know area of triangle APC is
1
2c a
.Therefore
3
3
1ABC
APC
Statistics
1.Sol: Given that mean and standard deviation of
x x 1 , ,... 2 are μ and respectively. That isx 1 1 x 2 ... xn
n
and
2 2 2 2
2 ( )x xi xi xi xi 2
n n n n
(1)Also given that mean and standard deviations
of new list of numbers y 1 ,y 2 ,...yn are1 2 1 2
1 2 ... 2 2 3 ...
ˆ
n nx x x x
y y y x x
n n
x x 1 2 ... xn
n
and2 2 2
ˆ^2 y y^12 ... yn^2
n
2 2
1 2 1 2 2 2
3
2...
2 2 nx x x x x xn
(2)
From (1) and (2), we get
2 2 2 2(^2) ˆ 2 1 2 1 2 2 1 2
2
x x x x x x
n n
8.Sol:
9.Sol:
Statistics
1.Sol:
2.Sol:
2 2 2
1 2 2 1 2 ( 1 2 ) 0
2 2
x x x x x x
n n
That is ˆ and ˆ.
2.Sol: Given that mean and standard deviation of
x 1 , x 2 ,... are μ and respectively. That is