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The equation of motion of a point in space is x = 2t, y = – 4t, z = 4t where t measured in hours and the
co-ordinates of moving point in kilometers, then the distance of the point from the starting point O (0, 0, 0) in 10 hours is :

 

Option: 1

20 km 


Option: 2

40 km 


Option: 3

60 km 


Option: 4

55 km 


Answers (1)

best_answer

 

Distance formula -

The distance between two points A(x_{1},y_{1},z_{1})\, and \, B(x_{2},y_{2},z_{2}) is =\sqrt{\left ( x_{2}-x_{1} \right )^{2}+{\left ( y_{2}-y_{1} \right )^{2}}+{\left ( z_{2}-z_{1} \right )^{2}}}

 

- wherein

\vec{OA}= x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k}

\vec{OB}= x_{2}\hat{i}+y_{2}\hat{j}+z_{2}\hat{k}

\underset{AB}{\rightarrow}    = \underset{OB}{\rightarrow} - \underset{AB}{\rightarrow}

\underset{AB}{\rightarrow}  = \left ( x_{2}-x_{1} \right )\hat{i}+\left ( y_{2}-y_{1} \right )\hat{j}+\left ( z_{2}-z_{1} \right )\hat{k}

AB= \left | \underset{AB}{\rightarrow} \right |

AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}+\left ( z_{2}-z_{1} \right )^{2}}

 

 

 

 

Position of pt. after t hours is (2t, – 4t, 4t)

                Position of pt. after 10 hours is (20, – 40, 40)

                Distance from origin  

                = \sqrt{( 20 )^2 + ( -40 )^2 + ( 40 )^2} = 60 km

Posted by

Sayak

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