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

\[\vec{A}.\vec{B}=\left |\vec{A} \right |\left |\vec{B} \right |\cos \theta\]

\[\cos \theta =\frac{\vec{A}.\vec{B}}{\left |\vec{A} \right |\left |\vec{B} \right |}\]

. . . . . . . . . . . . . . . . . . . . . ( 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 = 60^O

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

Milan Tomic

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