A particle of mass 3 kg is located at a position vector r = (2 i - j + 3 k ) m . Its velocity vector is given by v = ( i + 2 j + k ) m s ⁻¹ , where is a constant. If the y -component and the z -component of the particle's angular momentum about the origin are equal in magnitude and sign, the value of is

Step-by-step solution

The angular momentum of the particle about the origin is given by L = m( r v ) . First, we compute the cross product r v : r v = vmatrix i & j & k 2 & -1 & 3 & 2 & 1 vmatrix Expanding the determinant: r v = i (-1 1 - 3 2) - j (2 1 - 3 ) + k (2 2 - (-1) ) r v = -7 i + (3 - 2) j + (4 + ) k Multiplying by the mass m = 3 kg : L = -21 i + 3(3 - 2) j + 3(4 + ) k The y -component of the angular momentum is L_y = 3(3 - 2) . The z -component of the angular momentum is L_z = 3(4 + ) . Given that L_y = L_z : 3(3 - 2) = 3(4 +