Question

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

Answer 1

The position vector is 2i+3j+4k
it's magnitude is \sqrt( {2}^{2} + {3}^{2} + {4}^{2} )
\sqrt{4 + 9 + 16}
\sqrt{29}
magnitude of position vector = \sqrt{29}

Answer 2

As per the question, the position of a particle in space is [2,3,4].

We are asked to calculate the magnitude of position vector of this point.

The position vector of this point r = 2i + 3j + 4k.

The magnitude of resultant of r is calculated as follows-

                                               $r= \sqrt{ x^2 +y^2+z^2}$

Here x, y and z are the position coordinates of a point in space.

                                                     $=\sqrt{2^2+3^2+4^2}\ unit$

                                                     $=\sqrt{4+9+16}\ unit$

                                                     $=\sqrt{29}\ unit$

Hence, the magnitude of its position vector is $\sqrt{29}\unit$