Given that AA = ˆiˆi - 2ˆjˆj + 3ˆkˆk   and BB = 3ˆiˆi - 4ˆkˆk, find the projection of A on B.  (Ans: -5/9 )


Given:

AA = ˆiˆi - 2ˆjˆj + 3ˆkˆk 
BB = 3ˆiˆi - 4ˆkˆk  

To Find:

The projection of vector AA on BB = A cos 𝞡 = ?

Solution: 

We know that the projection of AA is on  BB is A cos 𝜭, So, the scalar (dot) product is 

 

BB . AA B (A cos 𝜭)
 
or

 A cos 𝜭B.ABB.AB


 A cos 𝜭 =  (3ˆi-4ˆk).(ˆi-2ˆj+3ˆk)Bx2+By2+Bz2(3ˆi4ˆk).(ˆi2ˆj+3ˆk)Bx2+By2+Bz2

A cos 𝜭  (31)ˆi.ˆi+(3-2)ˆi.ˆj+(33)ˆj.ˆk+(-41)ˆk.ˆi+(-4-2)ˆk.ˆj+(-43)ˆk.ˆk32+(0)2+(-4)2(31)ˆi.ˆi+(32)ˆi.ˆj+(33)ˆj.ˆk+(41)ˆk.ˆi+(42)ˆk.ˆj+(43)ˆk.ˆk32+(0)2+(4)2

[∴ ˆiˆi . ˆiˆi = ˆjˆj . ˆjˆj = ˆkˆk . ˆkˆk =1
and
ˆiˆi . ˆjˆj = ˆjˆj . ˆiˆi = ˆjˆj . ˆkˆk = ˆkˆk . ˆjˆj = ˆkˆk . ˆiˆi = ˆiˆi . ˆkˆk= 0 ]

by putting the above values and calculating 


A cos 𝜭 =  3-129+163129+16

A cos 𝜭 =  -925925

A cos 𝜭 =  -9595  ------ Ans



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