----------------------------------- Panphobia Tue Mar 05, 2013 10:38 pm Vectors question ----------------------------------- So I have a vectors test tomorrow, and I understand all the questions for review but could not get the final answer for one thinking question here is the question, [code]A and B are vectors Two vectors A + 3B and 4A - 3B make an angle of 90 degrees when tail-to-tail Find the angle between A and B if |A| = 3|B|. (Hint: u dot u = |u|^2)[/code] I started out by taking the dot product of A+3B and 4A-3B which gave me (period is dot in dot product) 4A . A + 4A . 3B - 3B . A - 3B . 3B = 0 , I am pretty sure you can simplify this more until you get to |a||b|cos(theta) but I am not sure how, can anyone steer me in the right direction? ----------------------------------- michaelp Tue Mar 05, 2013 10:55 pm RE:Vectors question ----------------------------------- Your first step is right. Try and use the hint to simplify what you have more. ----------------------------------- Panphobia Tue Mar 05, 2013 11:05 pm RE:Vectors question ----------------------------------- |4A|^2 + 4A . 3B - 3B . A - |9B|^2=0 lost now ----------------------------------- Dreadnought Tue Mar 05, 2013 11:10 pm Re: Vectors question ----------------------------------- Hopefully, you know that the dot product is linear and commutative, that is [code] For any vectors A, B and scalars s, t (real numbers) we have (sA).(tB) = (st)(A.B) and A.B = B.A [/code] As an exercise, use the definition of the dot product to prove these. ----------------------------------- Panphobia Tue Mar 05, 2013 11:17 pm RE:Vectors question ----------------------------------- that was actually one of my projects, um could |3B| be turned to 3|B| not sure ----------------------------------- michaelp Tue Mar 05, 2013 11:23 pm RE:Vectors question ----------------------------------- Note that |4A|^ 2 != 4|A|^2, and |9B|^2 != 9|B|^2. ----------------------------------- Dreadnought Tue Mar 05, 2013 11:26 pm Re: Vectors question ----------------------------------- that was actually one of my projects, um could |3B| be turned to 3|B| not sure Just go by definition We define |A|^2 := (A.A) then apply linearity to |sA| EDIT: Also, because of how |A| is defined, it is always nicer to work with |A|^2, in this case you can do this (if you are careful, though not required). ----------------------------------- Panphobia Tue Mar 05, 2013 11:33 pm RE:Vectors question ----------------------------------- No wonder most of the class didnt even get one mark on this question last year, psh ----------------------------------- Panphobia Tue Mar 05, 2013 11:44 pm RE:Vectors question ----------------------------------- ok so i did this so far, am I right? 4a.a - 3b.a + 12a.b - 9b.b=0 4|a|^2 - 15b.a - 9|b|^2=0 4|3b||3b| - 15b.a - 9|b|^2 = 0 36|b|^2 - 15b.a - 9|b|^2 = 0 ----------------------------------- Dreadnought Tue Mar 05, 2013 11:51 pm Re: Vectors question ----------------------------------- Careful - 3(B.A) + 12(A.B) = 15(A.B) ? ----------------------------------- Panphobia Tue Mar 05, 2013 11:54 pm RE:Vectors question ----------------------------------- oh yea ----------------------------------- Dreadnought Tue Mar 05, 2013 11:56 pm Re: Vectors question ----------------------------------- You have (-3)(A.B) + (12)(A.B) = (-3 + 12)(A.B) = ??? ----------------------------------- Panphobia Wed Mar 06, 2013 12:01 am RE:Vectors question ----------------------------------- ok i think I am completely done, point out errors please (4a-3b).(a+3b)=0 4a.a - 3b.a + 12a.b - 9b.b=0 4|a|^2 + 9b.a - 9|b|^2=0 4|3b||3b| + 9b.a - 9|b|^2 = 0 36|b|^2 + 9b.a - 9|b|^2 = 0 27|b|^2 = -9b.a -1/3 = b.a/b.b -1/3 = |b||a|cosQ/|b|^2 -1/3 = |a|cosQ/|b| -1/3 = 3|b|cosQ/|b| -1/3 = 3cosQ cosQ = -1/9 Q = 96.4 ----------------------------------- Dreadnought Wed Mar 06, 2013 12:02 am Re: Vectors question ----------------------------------- Almost, check this step carefully 27|b|^2 = -9b.a -1/3 = b.a/b.b ----------------------------------- Panphobia Wed Mar 06, 2013 12:10 am Re: Vectors question ----------------------------------- 27|b|^2 = -9b.a -1/3 = b.a/b.b ok look either way you put it |b|^2 = -9b.a/27 and then wouldnt b.a go b.a/|b|^2 = -9/27? or am I confused lol ----------------------------------- Dreadnought Wed Mar 06, 2013 12:14 am Re: Vectors question ----------------------------------- |b|^2 = -9b.a/27 and b.a/|b|^2 = -9/27 are not equal (you've flipped one side). ----------------------------------- Panphobia Wed Mar 06, 2013 12:16 am RE:Vectors question ----------------------------------- then you explain how rearranging lets say I = V/R and R = V/I see you divide on the top like if 2 = 5*x, and x = 2/5, then 5 = 2/x no? i used the same principle ----------------------------------- Dreadnought Wed Mar 06, 2013 12:21 am Re: Vectors question ----------------------------------- 27*(B.B) = (-9)*(B.A) ( 27 / -9 )*(B.B) = (B.A) (-3)*(B.B) = (B.A) -3 = (B.A)/(B.B) ----------------------------------- Panphobia Wed Mar 06, 2013 12:25 am RE:Vectors question ----------------------------------- then you would expand right side to be 3|b|cosQ/|b| and cos-1(-1) = 180, which is impossible??? ----------------------------------- Dreadnought Wed Mar 06, 2013 12:37 am Re: Vectors question ----------------------------------- I've been wondering that myself for over an hour now... (rather that finishing my own assignment :P) If you want my opinion, trust the math and don't worry too much about it. It's possible the only vectors that satisfy (4A - 3B).(A - 3B) = 0 and ||A|| = 3 ||B|| are A = B = 0 (the zero vector). ----------------------------------- Panphobia Wed Mar 06, 2013 12:44 am RE:Vectors question ----------------------------------- oh alright that is a strange answer, you see this was a test question last semester and my friend did not get it right saying it was 0, sooooo it has to be something else ----------------------------------- Dreadnought Wed Mar 06, 2013 1:24 am Re: Vectors question ----------------------------------- I think I've got a proof that A = B = 0 [code]Construct a right angle triangle with sides (4A - 3B) (A+ 3B) and H. Clearly H is the difference of the two sides, H = (3A - 6B) By the Pythagorean Theorem, ||H||^2 = ||4A - 3B||^2 + ||A + 3B||^2 ||H||^2 = 13||A||^2 -36(A.B) ||4A - 3B||^2 = 17||A||^2 - 24(A.B) ||A + 3B||^2 = 2||A||^2 +6(A.B) (note that I've used 9||B||^2 = ||A||^2 to simplify these) Then the theorem gives 13||A||^2 - 36(A.B) = 19||A||^2 - 18(A.B) 32||A||^2 - 18(A.B) = 0 32||A||^2 - 6||A||^2cos(theta) = 0 ( 6||A||^2 ) * ( 16/3 - cos(theta) ) = 0 So either ||A||^2 = 0 or 16/3 - cos(theta) = 0 Since |cos(theta)|