Question

If A=2i+3j and B=2j+3k , the component of B along A is

Answer 1

so, this is answer for the given question

I hope you can understand

Answer 2

Given:

A=2i+3j and B=2j+3k

To Find:

The component of B along A is

Solution:

Let the angle between vector A and vector B be ∅

A = 2i+3j

B=2j+3k

As we know that the component of B along A = ΙBΙ cos∅

Now,

vector A dot vector B

A.B = IAIIBIcos∅       (Dot product of A and B)

IBIcos∅ = (A.B)/IAI

Now we will find IAI which is the magnitude of A

IAI = √2²+3²+0²    

    =√13

Now,

The dot product of A and B

A.B = (2i+3j).(2j+3k)

      =2i×0 + 3j×2j + 0×3k

      = 6                              [∵i.0 = 0, j.j = 1]

So, the component

IBIcos∅ = (A.B)/IAI

            = 6/√13

Hence, The component of B along A is 6/√13.