Mathematics Times – July 2019

a 0  0 so

3 1^2010

n 2

a



. As we have 3 1^5 

mod (11), that is 32010 (3 ) 15 402 mod(11).

Therefore 3 1 0^2010   mod(11). hence the

reminder is 0.

Vectors
1.Sol: We get around the problem a bit. If we
can take the circum-centre (0) as origin and

we simply write centroid

3

a b c

G

 



 

. Wee
then form a relation between centriod(G),
Circum- center(O), and Ortho-center(H).
That is the distance between the centroid and
the ortho-center is twice the distance between
the centroid and the circum-center.Therefore
H is a b c 
 
. Since N is a mid point of OH,

we get

2

Na b c 

 

.

Now, we have

2

a b c

NA

 



  

, similarly

2

b a c

NB

 



  



and

2

c a b

NC

 



  



by

adding these three vectors ,we get the desired

vector

2 2

a b c b a c

NA NB NC

   

   

       

1

2 2 2

c a b a b c

HO

    

  

     

2.Sol: We take (^) v to be the circum-centre but
with respect to any origin. Reader may
observe the given information carefully ,you
can see interesting fact that v is equidistant
from the vectors (^) i i,2
 
and j


. Let v x y( , )



circum-center, and the position vectors be

vertices of triangle ABC. That is A(1,0),

B(2,0) and C(0,1). Now we have

( 1)x       ^2 y^2 ( 2)x^2 y x^22 ( 1)y^2.

after simplifying, we get

3

2

x and

3

2

y ,

therefore the desired vector is

3 3

2 2

v i j

  

.

Hence

3

2

v 



which lies in the interval

2,3.

3.Sol: Given a i j k  6 3 6

   

and d i j k  

   

We want to write a b c 

 

. Where (^) b

is
parallel to d

and (^) c is perpendicular to d

.
For this , we need (^) b

to be the projection of
a

onto d

.
So
a d
b d
d d
 
 
 
 
 
 
Now a d     6 3 6 3
 
and
(^) d d    1 1 1 3
 
So,
Vectorsectors
1.Sol:
2.Sol:
3.Sol:
4.Sol:
3
1
3
a d
d d
 
  

 
 
Therefore b d i j k     
  
and
c a b        (6 3 6 ) (i j k i j k)
       
(^)   7 2 5i j k
  
4.Sol: We take center to be the origin and two of
the vertices to be
1 1
0,1, , 0, 1,
2 2
   
    
   
.
This gives us an isosceles triangle with sides
3 3
,
2 2
,2 and we can us the cosine rule from
elementary trigonometry
(. .,i e c a b^2  ^22 2 cos )ab  to get
3 3 3
4 2 cos
2 2 2
      
 
.