Math, asked by yadavpratik7828, 1 year ago

Find the area of parallelogram formed by vectors a=3i+2j b=-3i+7j

Answers

Answered by pulakmath007
7

 \displaystyle\huge\red{\underline{\underline{Solution}}}

FORMULA TO BE IMPLEMENTED

1. The area of the parallelogram whose adjacent

 \sf{sides  \: are  \: given \:  by  \: the  \: vectors  \: \vec{a} \:  \: and \:  \vec{b} \:  \: is } \:

 =  \sf{ \:  |  \: \vec{a} \:  \times  \vec{b}  \: |  \: }

2. In vector cross product

  \sf{ \hat{i} \times \hat{i}  =  \hat{j}  \times  \hat{j} =  \hat{0}   \: , \:\hat{i} \times \hat{j}  =  \hat{k}  =  -  \hat{j} \times \hat{i} \:   }

TO DETERMINE

The area of the parallelogram formed by the

 \sf{vectors \:  \:  \vec{a}  = 3 \hat{i} + 2 \hat{j} \:  \: and \:  \:   \vec{b}  =  - 3 \hat{i} + 7\hat{j}\: }

CALCULATION

Here

 \sf{  \vec{a}  = 3 \hat{i} + 2 \hat{j} \:  \:  }

 \sf{   \vec{b}  =  - 3 \hat{i} + 7\hat{j}\: }

So

 \sf{   \: \vec{a}   \times  \:   \vec{b}  \: }

 =  \sf{  ( 3 \hat{i} + 2 \hat{j} \: )  \times (  - 3 \hat{i} + 7\hat{j}\:) }

 =  \sf{  - 9(  \: \hat{i} \times \hat{i} \: ) +  \: 21(  \: \hat{i} \times \hat{j} \: )   - 6(  \: \hat{j} \times \hat{i} \: )   \: +  14(  \: \hat{j} \times \hat{j} \: ) \:   }

 =  \sf{   \: 21 \: \hat{k}   + 6  \: \hat{k}  }

 =  \sf{   \: 27\: \hat{k}    }

Hence the area of the parallelogram is

 =  \sf{ \:  |  \: \vec{a} \:  \times  \vec{b}  \: |  \: }

 =  \sf{  |\: 27\: \hat{k}  \: |   }

 =  \sf{   27 \:  \:  \: sq \: unit  }

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