The aim of this question is to develop a concrete understanding of the key concepts related to vector algebra such as magnitude, direction, and the dot product of two vectors in cartesian form.
Given a vector $ \vec{ A } \ = \ a_1 \hat{ i } \ + \ a_2 \hat{ j } \ + \ a_3 \hat{ k } $, its direction and magnitude are defined by the following formulas:
\[ |A| \ = \ \sqrt{ a_1^2 \ + \ a_2^2 \ + \ a_3^2 } \]
\[ \hat{ A } \ = \ \dfrac{ \vec{ A } }{ |A| } \]
The dot product of two vectors $ \vec{ A } \ = \ a_1 \hat{ i } \ + \ a_2 \hat{ j } \ + \ a_3 \hat{ k } $ and $ \vec{ B } \ = \ b_1 \hat{ i } \ + \ b_2 \hat{ j } \ + \ b_3 \hat{ k } $ is defined as:
\[ \vec{ A }.\vec{ B } \ = \ a_1 b_1 \ + \ a_2 b_2 \ + \ a_3 b_3 \]
Expert Answer
Let:
\[ \vec{ A } \ = \ 2 \hat{ i } \ + \ \hat{ j } \ + \ 2 \hat{ k } \]
To find the direction of $ \vec{ A } $, we can use the following formula:
\[ \text{ Direction of } \vec{ A } = \ \hat{ A } \ = \ \dfrac{ \vec{ A } }{ |A| } \]
\[ \Rightarrow \hat{ A } \ = \ \dfrac{ 2 \hat{ i } \ + \ \hat{ j } \ + \ 2 \hat{ k } }{ \sqrt{ ( 2 )^2 \ + \ ( 1 )^2 \ + \ ( 2 )^2 } } \]
\[ \Rightarrow \hat{ A } \ = \ \dfrac{ 2 \hat{ i } \ + \ \hat{ j } \ + \ 2 \hat{ k } }{ \sqrt{ 4 \ + \ 1 \ + \ 4 } } \]
\[ \Rightarrow \hat{ A } \ = \ \dfrac{ 2 \hat{ i } \ + \ \hat{ j } \ + \ 2 \hat{ k } }{ \sqrt{ 9 } } \]
\[ \Rightarrow \hat{ A } \ = \ \dfrac{ 2 \hat{ i } \ + \ \hat{ j } \ + \ 2 \hat{ k } }{ 3 } \]
\[ \Rightarrow \hat{ A } \ = \ \dfrac{ 2 }{ 3 } \hat{ i } \ + \ \dfrac{ 1 }{ 3 } \hat{ j } \ + \ \dfrac{ 2 }{ 3 } \hat{ k } \]
Given that:
\[ \text{ Magnitude of Force } = \ |F| = 3 \ N \]
\[ \text{ Direction of Force } = \ \hat{ F } \ = \ \hat{ A } \ = \ \dfrac{ 2 }{ 3 } \hat{ i } \ + \ \dfrac{ 1 }{ 3 } \hat{ j } \ + \ \dfrac{ 2 }{ 3 } \hat{ k } \]
To find $ \vec{ F } $ we can use following formula:
\[ \vec{ F } \ = \ |F| . \hat{ F } \]
\[ \Rightarrow \vec{ F } \ = \ ( 3 ) . \bigg ( \dfrac{ 2 }{ 3 } \hat{ i } \ + \ \dfrac{ 1 }{ 3 } \hat{ j } \ + \ \dfrac{ 2 }{ 3 } \hat{ k } \bigg ) \]
\[ \Rightarrow \vec{ F } \ = \ 2 \hat{ i } \ + \ \hat{ j } \ + \ 2 \hat{ k } \]
To find $ \vec{ AB } $ we can use following formula:
\[ \Rightarrow \vec{ AB } \ = \ \bigg ( 0 \hat{ i } \ + \ 2 \hat{ j } \ + \ 0 \hat{ k } \bigg ) \ – \ \bigg ( 0 \hat{ i } \ + \ 0 \hat{ j } \ + \ 0 \hat{ k } \bigg ) \]
\[ \Rightarrow \vec{ AB } \ = \ 2 \hat{ j } \]
To find the work done $ W $, we can use the following formula:
\[ W \ = \ \vec{ F } . vec{ AB } \]
\[ \Rightarrow W \ = \ \bigg ( 2 \hat{ i } \ + \ \hat{ j } \ + \ 2 \hat{ k } \bigg ) . \bigg ( 2 \hat{ j } \bigg ) \]
\[ \Rightarrow W \ = \ ( 2 )( 0 ) \ + \ ( 1 )( 2 ) \ + \ ( 2 )( 0 ) \]
\[ \Rightarrow W \ = \ 2 \ J \]
Numerical Result
\[ W \ = \ 2 \ J \]
Example
Given $ \vec{ F } \ = \ 2 \hat{ i } \ + \ 4 \hat{ j } \ + \ 2 \hat{ k } $ and $ \vec{ AB } \ = \ 7 \hat{ i } \ + \ \hat{ j } \ + \ 2 \hat{ k } $, Find the work done $ \vec{ W }.
To find $ W $, we can use following formula:
\[ W \ = \ \vec{ F } . vec{ AB } \]
\[ \Rightarrow W \ = \ \bigg ( 2 \hat{ i } \ + \ 4 \hat{ j } \ + \ 2 \hat{ k } \bigg ) . \bigg ( 7 \hat{ i } \ + \ 1 \hat{ j } \ + \ 2 \hat{ k } \bigg )\]
\[ \Rightarrow W \ = \ ( 2 )( 7 ) \ + \ ( 4 )( 1 ) \ + \ ( 2 )( 2 ) \]
\[ \Rightarrow W \ = \ 22 \ J \]