The position vector of a particle in a circular motion about the origin sweeps out equal area in equal time

Answer is (a) π  \(\vec{A}\times\vec{B}\) = AB sinθ \(\hat{n}\) \(\vec{B}\times\vec{A}\) = AB sin(2π - θ)\(\hat{n}\) AB sinθ \(\hat{n}\) = AB sin(2π - θ)\(\hat{n}\) θ = 2π - θ 2θ = 2π θ = π 

(a) : vector a = 8vector b vector c = - 7vector b vector a and vector b are parallel and vector b and vector c are antiparallel. Thus vector a and vector c are...

The correct answer is (c) α < β if A > B 

(b) It is possible to have | vector C | < | vector A | and | vector C | < | vector B | 

(b) speed remains constant (d) tangential acceleration remains constant. Explanation:  Velocity and acceleration will not be constant because direction changes continuously. Hence option (a) and (c) are not correct. With a constant speed...

(c) A is correct but B is wrong

(a) vector r . vector T = 0 and vector F. vector T= 0 

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A +B=? How?

vector a || vector b ⇒ vector a × vector b = vector 0 ⇒ i(–9 – 3λ) + k(2λ + 6) = vector 0 ⇒ λ = – 3.