Question

Two vectors A and B have precisely equal magnitudes. In order for the magnitude of A+ B to be 110 times larger than the magnitude of A-B , what must be the angle between them?

Answer 1

Thanks for pointing that out :)

Answer 2

yeah, it helps when trying to solve the problem to know what to solve :)

Answer 3

Are these vectors in any particular dimension? R^n ?

Answer 4

no particular direction but assume its positive.

Answer 5

\[A=<a_1,a_2,a_3,...,a_n>;\ |A|=\sqrt{(a_1)^2+(a_2)^2+(a_3)^2+...+(a_n)^2} \]
\[B=<b_1,b_2,b_3,...,b_n>;\ |B|=\sqrt{(b_1)^2+(b_2)^2+(b_3)^2+...+(b_n)^2}\]
so far

Answer 6

\[A+B=<a_1+b_1,a_2+b_2,a_3+b_3,...,a_n+b_n>\]
\[ |A+B|=\sqrt{(a_1+b_1)^2+(a_2+b_2)^2+(a_3+b_3)^2+...+(a_n+b_n)^2}\]

\[A-B=<a_1-b_1,a_2-b_2,a_3-b_3,...,a_n-b_n>\]
\[ |A+B|=\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2+...+(a_n-b_n)^2}\]
this might get messy

Answer 7

especially if i forget how to type ... that last ones |A-B|

Answer 8

but this is going towards angle between them; so i really should focus on that ...

Answer 9

\[A.B=a_1b_1+a_2b_2+a_3b_3+...+a_nb_n\]
and we determined the magnitudes already soo; the formula for the angle between them is:
\[cos(t)=\frac{A.B}{|A||B|}\]
\[t=cos^{-1}\frac{A.B}{|A||B|}\]
\[t=cos^{-1}\frac{a_1b_1+a_2b_2+a_3b_3+...+a_nb_n}{110((a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2+...+(a_n-b_n)^2)}\]

Answer 10

you sure this thing is any dimensional?

Answer 11

this is motion in 2 dimensions, if this answers your question.

Answer 12

well, it does; makes that contraption up there a bit smaller :)

Answer 13

i think so too.

Answer 14

\[t=cos^{-1}\frac{a_1b_1+a_2b_2}{110((a_1-b_1)^2+(a_2-b_2)^2)}\]

\[110t=cos^{-1}\frac{a_1b_1+a_2b_2}{(a_1)^2-2a_1b_1+(b_1)^2+(a_2)^2-2a_2b_2+(b_2)^2}\]

how that helps tho I cant tell; might have to wolf it to see how or if it simplifies

Answer 15

thanks, i'll solve and let you know.

Answer 16

this might be an easier thing to construct and see a solution with

Answer 17

|A+B| = 110|A-B|