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 September 7th, 2013, 12:20 PM #1 Senior Member   Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40 3d vector question If V were to be (3/2, 3/2, 6) and Q were to be (8, 15, -20), what would be the distance from V to Q or vice versa (in units)?
 September 7th, 2013, 12:43 PM #2 Senior Member   Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 Re: 3d vector question Are you familiar with the distance formula? In three dimensions, this is $\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2}$ where a=(a1,a2,a3) and b=(b1,b2,b3) are the two points.
 September 7th, 2013, 02:27 PM #3 Senior Member   Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40 Re: 3d vector question Yes I am familiar with it, but I'd like to have an answer, as my textbook may be wrong. It put 21.9 down as an answer, which sounds ridiculous.
 September 7th, 2013, 02:46 PM #4 Senior Member   Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 Re: 3d vector question It's approximately 30. Greater than though.
September 8th, 2013, 02:49 AM   #5
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Quote:
 Originally Posted by perfect_world It put 21.9 down as an answer
Are you sure the question gave Q as (8, 15, -20), not (8, 15, -10)? Using (8, 15, -10), the distance would be 21.9.

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