Question

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

Answer 1

We know that the magnitude of a vector is determined by the Pythagoras theorem.

Now, in our question the position vector of the given point will be 2i^+3j^+4k^

Now, |magnitude| = (2^2 +3^2 + 4^2) = 29units(Ans)

Answer 2

Answer: The magnitude of the position vector $\sqrt {(29)}$ units .

The position of particle is given by (2, 3, 4)

Hence the position vector is = (2i + 3j + 4k)

Magnitude of the position vector is “r” is given by, $\sqrt {(x^2 + y^2 + z^2)}$

x = 2, y = 3, z = 4

r = $\sqrt {(2^2 + 3^2 + 4^2) }$

= $\sqrt {(4+9+16)}$

= $\sqrt {(29)}$ units