HW12AA

# HW12AA - 12 A&A 2 H 1 and H 2 begin their flights at...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

HW12AA - 12 A&A 2 H 1 and H 2 begin their flights at...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online