Question

A particle lies in space at point (2,3,4). Find the magnitude of its position vector.

Answer 1

hope the ans will helpful

Answer 2

Explanation:

The position vector is 2i+3j+4k

it's magnitude is squareroot( {2}^{2} + {3}^{2} + {4}^{2} )

\sqrt{4 + 9 + 16}

\sqrt{29}

magnitude of position vector = \sqrt{29}