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modphysnoob Group TitleBest ResponseYou've already chosen the best response.0
dw:1360298829035:dw find this vector
 one year ago

itsmylife Group TitleBest ResponseYou've already chosen the best response.0
your x axis coordinate is incorrect
 one year ago

itsmylife Group TitleBest ResponseYou've already chosen the best response.0
it should be (4,0)
 one year ago

modphysnoob Group TitleBest ResponseYou've already chosen the best response.0
yes , thanks
 one year ago

rebeccaskell94 Group TitleBest ResponseYou've already chosen the best response.0
That is a beautiful triangle you have there.
 one year ago

itsmylife Group TitleBest ResponseYou've already chosen the best response.0
then use mid point formula your gonna get, (2,1) as coordinates of bisector n then your vector is , V= 2i +j
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
modphys doesn't want the mid point. He want the vector that's perpendicular.
 one year ago

itsmylife Group TitleBest ResponseYou've already chosen the best response.0
well thats perpendicular bisector that means dividing equally that means mid point
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
What I would do, is write the line in the form \(y=x/2+2\). The slope of the intersecting line will be \[\frac{1}{\frac{1}{2}}=2.\] So you have a line that looks like \(y=2x\) as the line that intersects. Set them equal to each other to get \(2x=x/2+2\implies 5x/2=2\implies x=1\).
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Plug that in again, \(x=1\implies y=2(1)=2\), so the point of intersection is \((1,2)\).
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Which in turn means the vector is \[\vec{i}+2\vec{j}\]
 one year ago

modphysnoob Group TitleBest ResponseYou've already chosen the best response.0
I got same answer but here what my book did
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
If \(x=.8\), then the vector is \[.8\vec{i}+1.4\vec{j}\] Taking the dot product of that with the vector \[4\vec{i}2\vec{j},\] we get \[.8\cdot42\cdot1.4=0.4\neq0\]so that solution is wrong.
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.0
dw:1360299909016:dw just check the magnitude of vector A : (4*2)/sqrt(4^2+2^2) = 8/sqrt(20) = 8/2sqrt(5) = 4/5 * sqrt(5) 0.8 and 1.6 has the same magnitude
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
Just as in your book, we can say the direction vector B (from pt (0,2) to (4,0) is (4,2) let vector A be perpendicular to B. A dot B = 0 (x,y) dot (4,2) =0 we get 4x = 2y pick a nice number for x, like 1. then y =2 vector A = (1,2) is perpendicular to vector B make A unit length, by dividing by its length. A= sqrt(5) Au = (1,2)/sqrt(5) now to get the length to scale Au so it reaches the line between (0,2) and (4,0), we notice that the vector (4,0) projected onto Au has the correct length Au dot (4,0) = 4/sqrt(5) The vector we want is 4/sqrt(5) * (1,2)/sqrt(5) = 4/5 * (1,2) or (0.8, 1.6)
 one year ago
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