Summary Vector

MUST DO PRACTICE QUESTIONS
ON
VECTORS
CLASS XII
FOR CBSE 2025 EXAMINATION




1 |BY: SHASHANK VOHRA LECTURER MATHS: DOE, DELHI

, 
If a  2i  j  2
k , then Fi nd

( a ) Direction Ratio's of a

SOLUTION : Direction Ratio's of a are components of i , j &  k

s o , d .r of a  2,1,  2 
 
(b ) |a | i.e. Magnitude of a

SO LUTION : |a | 2 2  12  (  2) 2  4  1  4  9  3 units

( c ) D i rec tio n cosine's o f a
d .r .' s 2 1 2
SOLUTION : d .c ' s    , , 
|a | 3 3 3

( d ) Angle which a makes with x - axis , y  axis & z  axis
2 1 2
SOLUTION : As , d .c ' s are  , , 
3 3 3
2 1 2
 c os  , cos  , cos   , , 
3 3 3
2 1 2

Thus , cos      cos ( ) ( Angle w hich a makes with x - axis )
3 3
1 1 1

Thus , cos      cos ( ) ( Angle which a makes with y - axis )
3 3
2 2 2
Thus , cos      cos 1 ( )    cos 1 ( )
3 3 3

( Angle which a makes with z - axis )


2 |BY: SHASHANK VOHRA LECTURER MATHS: DOE, DELHI

, 
If a  2i  j  2
k , then Find

( a ) a unit vector in the direction of a

SOLUTION : As , |a | 2 2  12  (  2) 2  4  1  4  9  3 units

 a 2i  j  2
k
so , a unit vector in the direction of a   
|a | 3
 2 1 2
 A unit vector in t h e direction o f a  i  j   k
3 3 3

(b ) a unit vector OPPOSITE to the direction of a

SOLUTION : As , |a | 2 2  12  (  2) 2  4  1  4  9  3 units

 a  2i  j  2
k
s o , a unit vector opposite to direct ion o f a  (  )  
|a | 3
 2 1 2
 A un i t vec t or opposi te to direction of a   i  j  k.
3 3 3




3 |BY: SHASHANK VOHRA LECTURER MATHS: DOE, DELHI

, 
If a  2i  j  2
k , then Find

( a ) a vector of magnitude "5" in the direction of a

SOLUTION : As , |a | 2 2  12  (  2) 2  4  1  4  9  3 units

  a 
s o , a vector of magnitude "5" in the direction of a  5   
 |a | 
 
2i  j  2k
 5( )
3
 10 5 10
 A vector of magnitude "5" in the direction of a  i  j   k
3 3 3

(b ) a vector of magnitude "7" opposite to the direction of a

SOLUTION : As , |a | 2 2  12  (  2) 2  4  1  4  9  3 units

  a 
so , a vector of magnitude "7" opposite to the direction o f a   7   
 |a | 
 
2i  j  2k
  7( )
3

A vector of m agnitude "7" o p posite to the dire c t ion of a
 14 7 14
  i  j   k
3 3 3




4 |BY: SHASHANK VOHRA LECTURER MATHS: DOE, DELHI