# Find the work W done by the force F in moving an object from a point A in space to a point B in space is defined as W = F.. Find the work done by a force of 3 newtons acting in the direction 2i + j +2k in moving an object 2 meters from (0, 0, 0) to (0, 2, 0).

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:

$\vec{ A }.\vec{ B } \ = \ a_1 b_1 \ + \ a_2 b_2 \ + \ a_3 b_3$

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:

$W \ = \ \vec{ F } . vec{ AB }$

$\Rightarrow W \ = \ \bigg ( 2 \hat{ i } \ + \ 4 \hat{ j } \ + \ 2 \hat{ k } \bigg ) . \bigg ( 7 \hat{ i } \ + \ 1 \hat{ j } \ + \ 2 \hat{ k } \bigg )$

$\Rightarrow W \ = \ ( 2 )( 7 ) \ + \ ( 4 )( 1 ) \ + \ ( 2 )( 2 )$

$\Rightarrow W \ = \ 22 \ J$