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Let  \vec{a},\vec{b},\: \: and \: \: \vec{c}  be three vectors such that \left | \vec{a} \right |= 1,\left | \vec{b} \right |= 2,\: and\: \left | \vec{c} \right |= 3.
If the projection of \vec{b}\: along\: \vec{a}   is equal to the projection of  \vec{c}\: along\: \vec{a};  and   \vec{b},\vec{c} are perpendicular to each other, then find \left | 3\, \vec{a}-2\, \vec{b}+2\, \vec{c} \right |.


 

 

 

 

 
 
 
 
 

Answers (1)
R Ravindra Pindel

\left | \vec{a} \right |= 1,\left | \vec{b} \right |= 2,\: and\: \left | \vec{c} \right |= 3
The projection of  \vec{b}\: along\: \vec{a}= \frac{\bar{b}\cdot \bar{a}}{\left | \vec{a} \right |}
 The projection of \vec{c}\: along\: \vec{a}= \frac{\bar{c}\cdot \bar{a}}{\left | \vec{a} \right |}
\Rightarrow \frac{\vec{b}.\vec{a}}{{\left | \vec{a} \right |}}= \frac{\vec{c}.\vec{a}}{{\left | \vec{a} \right |}}
\Rightarrow \vec{b}\cdot \vec{a}= \vec{c}\cdot \vec{a}---\left ( i \right )
\left ( 3a-2b+2c \right )\left ( 3a-2b+2c \right )= 9\left | \vec{a} \right |^{2}-6\bar{a}\bar{b}+6\bar{a}\bar{c}-6\bar{b}\bar{a}+4\left | \bar{b} \right |^{2}-4\bar{b}\bar{c}+6\bar{c}\tilde{a}-4\bar{c}\tilde{b}+4\left | \vec{c} \right |^{2}\left | 3\bar{a}-2\bar{b}+2\bar{c} \right |^{2}= 9\left | \vec{a} \right |^{2}+4\left | \vec{b} \right |^{2}+4\left | \vec{c} \right |^{2}-12\vec{a}\vec{b}+12\bar{a}\bar{c}-8\bar{b}\bar{c} 
\Rightarrow \left | 3\bar{a}-2\bar{b}+2\bar{c} \right |^{2}= 9\left | \vec{a} \right |^{2}+4\left | \bar{b} \right |^{2}+4\left | \bar{c} \right |^{2}  as  \vec{b}\cdot \vec{a}= \vec{c}\cdot \vec{a} and \vec{b} \ \& \ \vec{c} are perpendicular. 
\Rightarrow \left | 3\bar{a}-2\bar{b}+2\bar{c} \right |^2= 9\times 1+4\times 4+4\times 9= 61
\Rightarrow \left | 3\bar{a}-2\bar{b}+2\bar{c} \right |= \sqrt{61}

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