The angle between vector ( i + j ) and ( j + k ) is ( in Radian )

solution :  -

Let us consider

A = i + j

B = j + k

We know that

A . B = І A І . І B І Cos Ѳ

Therefore

Cos Ѳ = ( A . B ) / ( І A І . І B І )

 . . . . . . . . . . . . . . . . . . . . . ( 1 )

Now

A .B        = ( i + j ) . (j + k )

                = ( i . j ) + ( i . k ) + ( j . j  ) + ( j . k )

As Cos 90 = 0 ,

 ( i . j  )    = 0

Similarly ,

( i . j  ) = ( i . k ) = ( j . k ) = 0 

Therefore ,

A . B       = ( j . j  )

                = ( 1 . 1)

                = 1

. . . . . . . . . . . . . . . . . . . . . ( 2 )

The magnitude of the vector can be given as . . . . . .  .

I A I        = √  ( I i I ² + I j I ²   + I k I ² )  

                =  √  ( I 1 I ² + I 1 I ²   + I 0 I ² )

                = √  ( 1 + 1 + 0)

I A I        = √  ( 2 )

I B I         = √  ( 2 )

Therefore

I A I . I B I =  [ √  ( 2 ) ] .  [ √  ( 2 ) ]

                   = 2

. . . . . . . . . . . . . . . . . . . . . ( 3 )

Putting equation ( 2 ) and ( 3 ) in equation ( 1 ) we get ,

Cos Ѳ    = ( A . B ) / ( І A І . І B І )

Cos Ѳ    = ( 1 ) [ √  ( 2 ) ] .  [ √  ( 2 ) ]

Cos Ѳ    = 1 / 2

Therefore ,

Ѳ             = Cos ^{- 1} ( 1 / 2 )

Ѳ             = 60^O

Rad        =  π / 3

Answer:

The angle between vector ( i + j ) and ( j + k ) is  π / 3