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The distance of the point having position vector -\hat{i}+2\hat{j}+6\hat{k}

from the straight line passing through the point ( 2 , 3 , -4 ) 

and parallel to the vector, 6\hat{i}+3\hat{j}-4\hat{k} is :

  • Option 1)

    7

  • Option 2)

    4\sqrt3

  • Option 3)

    6

  • Option 4)

    2\sqrt{13}

 

Answers (1)

Point P(-1,2,6) & A(2,3,-4)

and \vec{n}=6\hat{i}+3\hat{j}-4\hat{k}

                                            \vec{x}= 6\hat{i}+3\hat{j}-4\hat{k}

PD^{2}=AP^{2}-AD^{2}

AD=|\frac{\vec{AP}.\vec{n}}{|\vec{n}|}|

AD=|\frac{(3\hat{i}+\hat{j}-10\hat{k}).(6\hat{i}+3\hat{j}-4\hat{k})}{|6\hat{i}+3\hat{j}-4\hat{k}|}|

        =\sqrt{61}

PD=\sqrt{110-61}=7

 CORRECT OPTION IS (1).

 


Option 1)

7

Option 2)

4\sqrt3

Option 3)

6

Option 4)

2\sqrt{13}

Posted by

Vakul

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