The angle between vector ( i + j ) and ( j + k ) is ( in Radian )
solution : -
Let us consider
A = i + j
B = j + k
To find out the
angle between the two vectors we must have to find out the dot ( scalar ) product of vector A and vector B.
The dot ( scalar ) product of these vectors can be represented as
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 i , j and k are the unit vectors along with the X axis , Y axis and Z axis respectively. So unit vector i , j and k are perpendicular to each other .
Therefore ,
( i . j ) = І i І І j І ( Cos Ѳ )
= І i І І j І ( Cos 90 )
As Cos 90 = 0 , above equation becomes
( i . j ) = І i І І j І ( 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 )
Similarly ,
I B I = √ ( I i I 2 + I j I 2 + I k I 2 )
= √ ( I 0 I 2 + I 1 I 2 + I 1 I 2 )
= √ ( 0 + 1 + 1 )
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 )
Cos Ѳ = 1 / 2
Therefore ,
Ѳ = Cos - 1 ( 1 / 2 )
Ѳ = 60O
This value is in degrees , we have to convert it in to the radian .
The formula for conversion is as follows
Rad = Deg X ( π / 180 )
= 60 X ( π / 180 )
Rad = π / 3
Answer:
The angle between vector ( i + j ) and ( j + k ) is π / 3
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