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Particles starting at the corners start moving along the diagonal of the cube with a constant speed of 1 m/s. Taking the angle as the origin, find the position of the particle after 1 second.

Option: 1

\left(3,5,6\right)


Option: 2

\left(\frac{1}{\sqrt3},\frac{1}{\sqrt3},\frac{1}{\sqrt3}\right)


Option: 3

\left(\sqrt2,\sqrt3,\sqrt6\right)


Option: 4

\left(\frac{1}{2},\frac{5}{8},\frac{6}{4}\right)


Answers (1)

best_answer

Let the position of the particle after 1 second be (x, y, z).
Therefore,

Distance travelled from the origin is

  \\ =\sqrt{x^2+y^2+z^2} \\ = 01\ metre

Hence, x=y=z=\frac{1}{\sqrt3}

So the coordinates of the particle after 1 second is 

\left(\frac{1}{\sqrt3},\frac{1}{\sqrt3},\frac{1}{\sqrt3}\right)

Posted by

mansi

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