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**Unformatted text preview: **12 A&A 2. H 1 and H 2 begin their flights at time t = 0. Using just the z co-ordinate of its trajectory, we see that the time until H 2 breaks down is 1- (- 3) = 4 hours. So H 1 changes its heading at t = 6 hours, by which point its co-ordinates are: (6 + 40 6 ,- 3 + 10 6 ,- 3 + 2 6) = (246 , 57 , 9) . The distance from here to H 1 s position at (446 , 13 , 0) is p (246- 446) 2 + (57- 13) 2 + (9- 0) 2 = 42017 205 miles . At a speed of 150mph, the trip will therefore take (205 / 150) 1 . 37 hours. 12 A&A 12. (a) Choose any point Q ( x ,y ,z ) on the plane Ax + By + Cz- D = 0, and let n = A i + B j + C k . Then (as outlined in Example 12 . 5 . 11 in the text) the distance between P 1 and this plane is equal to the magnitude of the projection from the vector * QP 1 onto n . d = proj n * QP 1 = * QP 1 n | n | = | Ax 1 + By 1 + Cz 1- Ax- By- Cz | A 2 + B 2 + C 2 = | Ax 1 + By 1 + Cz 1- D | A 2 + B 2 + C 2 ....

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