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

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then find its time period in seconds.

Find the magnitude of vector vecA=2hati-3hatj+4hatk

The coordinates of the initial point of a vector are (2,3) and those of the terminal point are (10,6).Find the magnitude and direction of the vector.

Position of a particle in a rectangular -co-ordinate (3,2,5) . Then its position vector will be

Position of a particle in a rectangular -co-ordinate (3,2,5) . Then its position vector will be

Position of a particle in a rectangular -co-ordinate (3,2,5) . Then its position vector will be

A particle is initially at point A(2,4,6)m moves finally to the point B(3,2,-3)m . Write the initial position vector,final position,and displacement vector of the particle.

A particle is initially at point A(2,4,6)m moves finally to the point B(3,2,-3)m . Write the initial position vector,final position,and displacement vector of the particle.

Vector vec(A) points N - E and its magnitude is 3kgms^(-1) it is multiplied by the scalar lamda such that lamda= - 4 second . Find the magnitude of the new vector quantity.