The position vector for a proton is initially r = 5.0i 6.0j + 2.0k and then

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The position vector for a proton is initially r = 5.0i – 6.0j + 2.0k and then later is r = – 2.0i + 6.0i + 2.0k, all in meters.

(a) What is the proton's displacement vector, and

(b) To what plane is that vector parallel?

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Fundamentals of Physics

ISBN: 978-0471758013

8th Extended edition

Authors: Jearl Walker, Halliday Resnick

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Question Posted: Jan 27, 2011 02:23 AM

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The position vector for a proton is initially r = 5.0i 6.0j + 2.0k and then

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We use Eq 42 and Eq 43 a With the initial position vector as a...View the full answer

Fundamentals of Physics

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