Show that the vectors A → = 2 i ^ - 3 J ^ - k ~ and B → = - 6 i ^ + 9 j ^ + 3 k ^ are parallel.

Given a vector A → = 2 i ^ - 3 J ^ - k ~ . and B → = - 6 i ^ + 9 j ^ + 3 k ^ . As we know that A → × B → = ABsin θ . If these vectors are parallel then the angle between them will be zero. So the value of their cross product should be zero. Here we have, A → × B → = i ^ j ^ k ^ 2 - 3 - 1 - 6 9 3 = i ^ - 3 × 3 - - 1 × 9 + j ^ ( - 1 ) × 6 - 2 × 3 + k ^ ( 2 × 9 ) - - 3 - 6 = i ^ ( - 9 + 9 ) + j ^ ( 6 - 6 ) + k ^ ( 18 - 18 ) = 0 . So we get that AB sinθ = 0 . since A ≠ 0 , B ≠ 0 , so sinθ = 0 or θ = 0 . Since the angle between the vectors is zero, hence these vectors are parallel.

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Gamma Question Bank for Medical Physics>Chapter 7 - System of Particles and Rotational Motion>EXERCISE>Q 49

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A thin circular ring first slips down a smooth incline then rolls down a rough incline of identical geometry from same height. Ratio of time taken in the two motion is

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Important Questions on System of Particles and Rotational Motion

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Gamma Question Bank for Medical Physics>Chapter 7 - System of Particles and Rotational Motion>EXERCISE>Q 50

The thin rod shown below has mass $M$ and length $L$ . A force $F$ acts at one end as shown and the rod is free to rotate about the other end in the plane of force. Initial angular acceleration of the rod is

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