Mathematics Times – July 2019

So that

1

cos

3





 and finally, we get

cos^11

3

 

 

  

 

.

5.Sol: Given that v v v v1 2 3 4, , ,

   

are unit vectors.

Clearly we can understand that all these

vectors lie on the unit circle. That is v 1



makes

an angle  1 with the x - axis ,likewise v v 2 3,

and v 4 makes   2 , 3 and  4 with the x-axis

respectively. we can write

v 1 cos 1 isin 1 j

  

v 2 cos 2 isin 2 j

  

v 3 cos 3 isin 3 j

  

v 4 cos 4 isin 4 j

  

Consider option (a) , That is v v v v    1    2 3 4

 0. Which is not necessarily true for all

values of     1    2 3 4 , So we can eliminate

option (a).

Like wise, option (b) can be written as

v vi j cos cosi jisin sini j

 

given that v vi j

 

is in first quadrant i.e.

bothcos cos 0i j and sin sin 0i 

This is possible only when x-coordinate is

positive.

i.e.i is in first quadrant, jis in fourth

quadrant.

Whichis imposible, Since y-coordinate is

negative.

Simillarly, option (c) v vi j 0

 

, That is

cos cos sin sin 0i j i 

cos i j 0

i.e.

2 i j



    

      i j 2

Which is negative for all v v v v   1 2 3 4, , ,.

and finally, option (d) is given v vi j 0

 

i.e. cos i j 0

0

i j 2



    

i.e.

3

2

2 i j



    

Which is not possible for allvalues of

v v v v1 2 3 4, , ,

   

.

6.Sol: We get around the problem a bit. If we

can take the circum-centre (O) as origin and

we simply write centroid

5.Sol:

6.Sol:

7.Sol:

3

a b c

OG

 



  

.

We 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

OH



is a b c 

 

. Now , we have

HA a a b c      ( ) (b c)

     

,

HB b a b c      ( ) (a c)

     

,

and

HC c a b c      ( ) (a b)

     

,

By adding these vectors, we get

HA HB HC      2(a b c) 2HO

     

7.Sol:Given

a b b c c a    

    

i.e., a b b c    0

  

 a b c b    0

  

i.e., (a c b  ) 0

  

if the angle between the two vectors is zero

then they are parallel.That is

 a c b 

  