Given that `\vec {A}` = `\hat {i}` - 2`\hat {j}` + 3`\hat {k}`   and `\vec {B}` = 3`\hat {i}` - 4`\hat {k}`, find the projection of A on B.  (Ans: -5/9 )


Given:

`\vec {A}` = `\hat {i}` - 2`\hat {j}` + 3`\hat {k}` 
`\vec {B}` = 3`\hat {i}` - 4`\hat {k}`  

To Find:

The projection of vector `\vec {A}` on `\vec {B}` = A cos 𝞡 = ?

Solution: 

We know that the projection of `\vec {A}` is on  `\vec {B}` is A cos 𝜭, So, the scalar (dot) product is 

 

`\vec {B}` . `\vec {A}` =  B (A cos 𝜭)
 
or

 A cos 𝜭 = `\frac {vec {B} . vec {A}}{B}`


 A cos 𝜭 =  `\frac {( 3\hat {i} - 4 \hat {k} ) . (\hat {i} - 2\hat {j} + 3\hat {k} )}{sqrt {Bx^2 + By^2 + Bz^2}}`

A cos 𝜭 =  `\frac {(3 ｘ 1) hat {i} . hat {i} + (3 ｘ -2) hat {i} . hat {j} + (3 ｘ 3) hat {j} . hat {k} + (-4 ｘ 1) hat {k} . hat {i} + (-4 ｘ -2) hat {k} . hat {j} + (-4 ｘ 3) hat {k} . hat {k} }{sqrt {3^2 + (0)^2 + (-4)^2}}`

[∴ `hat {i}` . `hat {i}` = `hat {j}` . `hat {j}` = `hat {k}` . `hat {k}` =1
and
`hat {i}` . `hat {j}` = `hat {j}` . `hat {i}` = `hat {j}` . `hat {k}` = `hat {k}` . `hat {j}` = `hat {k}` . `hat {i}` = `hat {i}` . `hat {k}`= 0 ]

by putting the above values and calculating 


A cos 𝜭 =  `\frac {3 - 12}{sqrt {9 + 16}}`

A cos 𝜭 =  `\frac {-9}{sqrt {25}}`

A cos 𝜭 =  `\frac {-9}{5}`  ------ Ans



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Numerical Problem 2.10
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List of 11th Physics Numerical Problems of Chapter - 2
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