The position vector of two masses of 6 unit and 2 unit is `(6hati-7hatj)` & `(2hati + 5hatj)` respectively. Find out the position of their centre of mass.

The position vectors of two masses of 6 and 2 units are 6i-7j and 2i+5j-8k respectively. Deduce the position of their centre of mass.

Two particles of masses 3 kg and 2kg are located at (-6hati+4hatj-2hatk) and (2hati+5hatj+13hatk) respectively. Locate the position of centre of mass.

Velocity of two objets of mass 10 kg and 2 kg is (2hati-7hatj+3hatk) m//s and (-10hati+35hatj-3hatk) m//s respectively. Calculate the velocity of their centre of mass.

If the position vectors of two given points A and B be 4hati+hatj+2hatk and 2hati-5hatj-hatk respectively, then a unit vector in the direction of vec(AB) is -

The position vectors of the three vertices of a triangle are (-hati-3hatj+2hatk), (5hati+7hatj-5hatk) and (2hati+5hatj+6hatk) , then the position vector of the point of intersection of the medians of the triangle is -

The position vectors of the vertices A,B,C of the triangle ABC are (hati+hatj+hatk),(hati+5hatj-hatk) and (2hati+3hatj+5hatk) respectively. Find the greatest angle of the triangle.

In a system comprising of two particles of mass 2 kg and 5 kg, the positions of the two particles at t=0 are (4hati+3hatj) and (6hati-7hatj+7hatk) respectively and the velocities are (10hati-6hatk) and (3hati+6hatj) respectively. Deduce the velocity of the centre of mass and the position of the centre or mass at t=0 and t=4 s

The position vectors of points A,B and C are hati+hatj+hatk,hati + 5hatj -hatk and 2hati + 3hatj + 5hatk , respectively the greatest angle of triangle ABC is

Two objects of mass 10 kg and 2kg are moving with velocities (2hati-7hatj+3hatk) m//s and (-10hati+35hatj-3hatk) m/s respectively. Calculate the velocity of the centre of mass of the system.

The position vectors of the points A, B, C are (hati-3hatj-5hatk),(2hati-hatj+hatk) and (3hati-4hatj-4hatk) respectively. Then A, B, C form a/an -